fix(curriculum): clean-up Project Euler 121-140 (#42731)
* fix: clean-up Project Euler 121-140 * fix: corrections from review Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com> * fix: missing backticks Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com> * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> * fix: missing delimiter Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com> Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
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@ -18,10 +18,10 @@ Find the maximum prize fund that should be allocated to a single game in which f
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# --hints--
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`euler121()` should return 2269.
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`discGamePrize()` should return `2269`.
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```js
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assert.strictEqual(euler121(), 2269);
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assert.strictEqual(discGamePrize(), 2269);
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```
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# --seed--
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@ -29,12 +29,12 @@ assert.strictEqual(euler121(), 2269);
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## --seed-contents--
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```js
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function euler121() {
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function discGamePrize() {
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return true;
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}
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euler121();
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discGamePrize();
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```
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# --solutions--
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@ -8,28 +8,41 @@ dashedName: problem-122-efficient-exponentiation
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# --description--
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The most naive way of computing n15 requires fourteen multiplications:
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The most naive way of computing $n^{15}$ requires fourteen multiplications:
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n × n × ... × n = n15
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$$n × n × \ldots × n = n^{15}$$
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But using a "binary" method you can compute it in six multiplications:
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n × n = n2n2 × n2 = n4n4 × n4 = n8n8 × n4 = n12n12 × n2 = n14n14 × n = n15
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$$\begin{align}
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& n × n = n^2\\\\
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& n^2 × n^2 = n^4\\\\
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& n^4 × n^4 = n^8\\\\
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& n^8 × n^4 = n^{12}\\\\
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& n^{12} × n^2 = n^{14}\\\\
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& n^{14} × n = n^{15}
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\end{align}$$
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However it is yet possible to compute it in only five multiplications:
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n × n = n2n2 × n = n3n3 × n3 = n6n6 × n6 = n12n12 × n3 = n15
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$$\begin{align}
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& n × n = n^2\\\\
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& n^2 × n = n^3\\\\
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& n^3 × n^3 = n^6\\\\
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& n^6 × n^6 = n^{12}\\\\
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& n^{12} × n^3 = n^{15}
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\end{align}$$
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We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
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We shall define $m(k)$ to be the minimum number of multiplications to compute $n^k$; for example $m(15) = 5$.
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For 1 ≤ k ≤ 200, find ∑ m(k).
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For $1 ≤ k ≤ 200$, find $\sum{m(k)}$.
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# --hints--
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`euler122()` should return 1582.
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`efficientExponentation()` should return `1582`.
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```js
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assert.strictEqual(euler122(), 1582);
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assert.strictEqual(efficientExponentation(), 1582);
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```
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# --seed--
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@ -37,12 +50,12 @@ assert.strictEqual(euler122(), 1582);
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## --seed-contents--
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```js
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function euler122() {
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function efficientExponentation() {
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return true;
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}
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euler122();
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efficientExponentation();
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```
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# --solutions--
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@ -8,20 +8,20 @@ dashedName: problem-123-prime-square-remainders
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# --description--
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Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)n + (pn+1)n is divided by pn2.
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Let $p_n$ be the $n$th prime: 2, 3, 5, 7, 11, ..., and let $r$ be the remainder when ${(p_n−1)}^n + {(p_n+1)}^n$ is divided by ${p_n}^2$.
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For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
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For example, when $n = 3, p_3 = 5$, and $4^3 + 6^3 = 280 ≡ 5\\ mod\\ 25$.
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The least value of n for which the remainder first exceeds 109 is 7037.
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The least value of $n$ for which the remainder first exceeds $10^9$ is 7037.
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Find the least value of n for which the remainder first exceeds 1010.
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Find the least value of $n$ for which the remainder first exceeds $10^{10}$.
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# --hints--
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`euler123()` should return 21035.
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`primeSquareRemainders()` should return `21035`.
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```js
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assert.strictEqual(euler123(), 21035);
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assert.strictEqual(primeSquareRemainders(), 21035);
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```
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# --seed--
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@ -29,12 +29,12 @@ assert.strictEqual(euler123(), 21035);
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## --seed-contents--
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```js
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function euler123() {
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function primeSquareRemainders() {
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return true;
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}
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euler123();
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primeSquareRemainders();
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```
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# --solutions--
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@ -8,42 +8,118 @@ dashedName: problem-124-ordered-radicals
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# --description--
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The radical of n, rad(n), is the product of the distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.
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The radical of $n, rad(n)$, is the product of the distinct prime factors of $n$. For example, $504 = 2^3 × 3^2 × 7$, so $rad(504) = 2 × 3 × 7 = 42$.
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If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:
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If we calculate $rad(n)$ for $1 ≤ n ≤ 10$, then sort them on $rad(n)$, and sorting on $n$ if the radical values are equal, we get:
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Unsorted
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<div style="text-align: center;">
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<table cellpadding="2" cellspacing="0" border="0" align="center">
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<tbody>
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<tr>
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<td colspan="2">$Unsorted$</td>
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<td></td>
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<td colspan="3">$Sorted$</td>
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</tr>
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<tr>
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<td>$n$</td>
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<td>$rad(n)$</td>
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<td></td>
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<td>$n$</td>
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<td>$rad(n)$</td>
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<td>$k$</td>
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</tr>
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<tr>
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<td>1</td>
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<td>1</td>
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<td></td>
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<td>1</td>
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<td>1</td>
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<td>1</td>
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</tr>
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<tr>
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<td>2</td>
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<td>2</td>
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<td></td>
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<td>2</td>
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<td>2</td>
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<td>2</td>
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</tr>
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<tr>
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<td>3</td>
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<td>3</td>
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<td></td>
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<td>4</td>
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<td>2</td>
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<td>3</td>
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</tr>
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<tr>
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<td>4</td>
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<td>2</td>
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<td></td>
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<td>8</td>
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<td>2</td>
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<td>4</td>
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</tr>
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<tr>
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<td>5</td>
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<td>5</td>
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<td></td>
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<td>3</td>
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<td>3</td>
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<td>5</td>
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</tr>
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<tr>
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<td>6</td>
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<td>6</td>
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<td></td>
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<td>9</td>
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<td>3</td>
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<td>6</td>
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</tr>
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<tr>
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<td>7</td>
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<td>7</td>
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<td></td>
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<td>5</td>
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<td>5</td>
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<td>7</td>
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</tr>
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<tr>
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<td>8</td>
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<td>2</td>
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<td></td>
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<td>6</td>
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<td>6</td>
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<td>8</td>
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</tr>
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<tr>
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<td>9</td>
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<td>3</td>
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<td></td>
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<td>7</td>
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<td>7</td>
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<td>9</td>
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</tr>
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<tr>
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<td>10</td>
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<td>10</td>
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<td></td>
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<td>10</td>
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<td>10</td>
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<td>10</td>
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</tr>
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</tbody>
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</table>
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</div><br>
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Sorted n rad(n)
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n rad(n) k 11
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111 22
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222 33
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423 42
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824 55
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335 66
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936 77
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557 82
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668 93
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779 1010
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101010 Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 and E(6) = 9. If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).
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Let $E(k)$ be the $k$th element in the sorted $n$ column; for example, $E(4) = 8$ and $E(6) = 9$. If $rad(n)$ is sorted for $1 ≤ n ≤ 100000$, find $E(10000)$.
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# --hints--
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`euler124()` should return 21417.
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`orderedRadicals()` should return `21417`.
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```js
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assert.strictEqual(euler124(), 21417);
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assert.strictEqual(orderedRadicals(), 21417);
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```
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# --seed--
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@ -51,12 +127,12 @@ assert.strictEqual(euler124(), 21417);
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## --seed-contents--
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```js
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function euler124() {
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function orderedRadicals() {
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return true;
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}
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euler124();
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orderedRadicals();
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```
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# --solutions--
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@ -8,18 +8,18 @@ dashedName: problem-125-palindromic-sums
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# --description--
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The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 62 + 72 + 82 + 92 + 102 + 112 + 122.
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The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: $6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2$.
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There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 02 + 12 has not been included as this problem is concerned with the squares of positive integers.
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There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that $1 = 0^2 + 1^2$ has not been included as this problem is concerned with the squares of positive integers.
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Find the sum of all the numbers less than 108 that are both palindromic and can be written as the sum of consecutive squares.
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Find the sum of all the numbers less than $10^8$ that are both palindromic and can be written as the sum of consecutive squares.
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# --hints--
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`euler125()` should return 2906969179.
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`palindromicSums()` should return `2906969179`.
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```js
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assert.strictEqual(euler125(), 2906969179);
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assert.strictEqual(palindromicSums(), 2906969179);
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```
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# --seed--
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@ -27,12 +27,12 @@ assert.strictEqual(euler125(), 2906969179);
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## --seed-contents--
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```js
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function euler125() {
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function palindromicSums() {
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return true;
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}
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euler125();
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palindromicSums();
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```
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# --solutions--
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@ -10,14 +10,24 @@ dashedName: problem-126-cuboid-layers
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The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.
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If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face. However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes. We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8. It turns out that 154 is the least value of n for which C(n) = 10. Find the least value of n for which C(n) = 1000.
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<img class="img-responsive center-block" alt="3x2x1 cuboid covered by twenty-two 1x1x1 cubes" src="https://cdn.freecodecamp.org/curriculum/project-euler/cuboid-layers.png" style="background-color: white; padding: 10px;">
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If we add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.
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However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly, the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.
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We shall define $C(n)$ to represent the number of cuboids that contain $n$ cubes in one of its layers. So $C(22) = 2$, $C(46) = 4$, $C(78) = 5$, and $C(118) = 8$.
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It turns out that 154 is the least value of $n$ for which $C(n) = 10$.
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Find the least value of $n$ for which $C(n) = 1000$.
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# --hints--
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`euler126()` should return 18522.
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`cuboidLayers()` should return `18522`.
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```js
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assert.strictEqual(euler126(), 18522);
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assert.strictEqual(cuboidLayers(), 18522);
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```
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# --seed--
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@ -25,12 +35,12 @@ assert.strictEqual(euler126(), 18522);
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## --seed-contents--
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```js
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function euler126() {
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function cuboidLayers() {
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return true;
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}
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euler126();
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cuboidLayers();
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```
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# --solutions--
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@ -8,38 +8,32 @@ dashedName: problem-127-abc-hits
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# --description--
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The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.
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The radical of $n$, $rad(n)$, is the product of distinct prime factors of $n$. For example, $504 = 2^3 × 3^2 × 7$, so $rad(504) = 2 × 3 × 7 = 42$.
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We shall define the triplet of positive integers (a, b, c) to be an abc-hit if:
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GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
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a < b
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a + b = c
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rad(abc) < c
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1. $GCD(a, b) = GCD(a, c) = GCD(b, c) = 1$
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2. $a < b$
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3. $a + b = c$
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4. $rad(abc) < c$
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For example, (5, 27, 32) is an abc-hit, because:
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GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
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1. $GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1$
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2. $5 < 27$
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3. $5 + 27 = 32$
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4. $rad(4320) = 30 < 32$
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5 < 27
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It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for $c < 1000$, with $\sum{c} = 12523$.
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5 + 27 = 32
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rad(4320) = 30 < 32
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It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for c < 1000, with ∑c = 12523.
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Find ∑c for c < 120000.
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Find $\sum{c}$ for $c < 120000$.
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# --hints--
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`euler127()` should return 18407904.
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`abcHits()` should return `18407904`.
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```js
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assert.strictEqual(euler127(), 18407904);
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assert.strictEqual(abcHits(), 18407904);
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```
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# --seed--
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@ -47,12 +41,12 @@ assert.strictEqual(euler127(), 18407904);
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## --seed-contents--
|
||||
|
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```js
|
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function euler127() {
|
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function abcHits() {
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|
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return true;
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}
|
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|
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euler127();
|
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abcHits();
|
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```
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# --solutions--
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@ -12,14 +12,26 @@ A hexagonal tile with number 1 is surrounded by a ring of six hexagonal tiles, s
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New rings are added in the same fashion, with the next rings being numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows the first three rings.
|
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By finding the difference between tile n and each of its six neighbours we shall define PD(n) to be the number of those differences which are prime. For example, working clockwise around tile 8 the differences are 12, 29, 11, 6, 1, and 13. So PD(8) = 3. In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and 10, hence PD(17) = 2. It can be shown that the maximum value of PD(n) is 3. If all of the tiles for which PD(n) = 3 are listed in ascending order to form a sequence, the 10th tile would be 271. Find the 2000th tile in this sequence.
|
||||
<img class="img-responsive center-block" alt="three first rings of arranged hexagonal tiles with numbers 1 to 37, and with highlighted tiles 8 and 17" src="https://cdn.freecodecamp.org/curriculum/project-euler/hexagonal-tile-differences.png" style="background-color: white; padding: 10px;">
|
||||
|
||||
By finding the difference between tile $n$ and each of its six neighbours we shall define $PD(n)$ to be the number of those differences which are prime.
|
||||
|
||||
For example, working clockwise around tile 8 the differences are 12, 29, 11, 6, 1, and 13. So $PD(8) = 3$.
|
||||
|
||||
In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and 10, hence $PD(17) = 2$.
|
||||
|
||||
It can be shown that the maximum value of $PD(n)$ is $3$.
|
||||
|
||||
If all of the tiles for which $PD(n) = 3$ are listed in ascending order to form a sequence, the 10th tile would be 271.
|
||||
|
||||
Find the 2000th tile in this sequence.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler128()` should return 14516824220.
|
||||
`hexagonalTile()` should return `14516824220`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler128(), 14516824220);
|
||||
assert.strictEqual(hexagonalTile(), 14516824220);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -27,12 +39,12 @@ assert.strictEqual(euler128(), 14516824220);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler128() {
|
||||
function hexagonalTile() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler128();
|
||||
hexagonalTile();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,20 +8,20 @@ dashedName: problem-129-repunit-divisibility
|
||||
|
||||
# --description--
|
||||
|
||||
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
|
||||
A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.
|
||||
|
||||
Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
|
||||
Given that $n$ is a positive integer and $GCD(n, 10) = 1$, it can be shown that there always exists a value, $k$, for which $R(k)$ is divisible by $n$, and let $A(n)$ be the least such value of $k$; for example, $A(7) = 6$ and $A(41) = 5$.
|
||||
|
||||
The least value of n for which A(n) first exceeds ten is 17.
|
||||
The least value of $n$ for which $A(n)$ first exceeds ten is 17.
|
||||
|
||||
Find the least value of n for which A(n) first exceeds one-million.
|
||||
Find the least value of $n$ for which $A(n)$ first exceeds one-million.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler129()` should return 1000023.
|
||||
`repunitDivisibility()` should return `1000023`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler129(), 1000023);
|
||||
assert.strictEqual(repunitDivisibility(), 1000023);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -29,12 +29,12 @@ assert.strictEqual(euler129(), 1000023);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler129() {
|
||||
function repunitDivisibility() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler129();
|
||||
repunitDivisibility();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,22 +8,22 @@ dashedName: problem-130-composites-with-prime-repunit-property
|
||||
|
||||
# --description--
|
||||
|
||||
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
|
||||
A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.
|
||||
|
||||
Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
|
||||
Given that $n$ is a positive integer and $GCD(n, 10) = 1$, it can be shown that there always exists a value, $k$, for which $R(k)$ is divisible by $n$, and let $A(n)$ be the least such value of $k$; for example, $A(7) = 6$ and $A(41) = 5$.
|
||||
|
||||
You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
|
||||
You are given that for all primes, $p > 5$, that $p − 1$ is divisible by $A(p)$. For example, when $p = 41, A(41) = 5$, and 40 is divisible by 5.
|
||||
|
||||
However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
|
||||
|
||||
Find the sum of the first twenty-five composite values of n for whichGCD(n, 10) = 1 and n − 1 is divisible by A(n).
|
||||
Find the sum of the first twenty-five composite values of $n$ for which $GCD(n, 10) = 1$ and $n − 1$ is divisible by $A(n)$.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler130()` should return 149253.
|
||||
`compositeRepunit()` should return `149253`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler130(), 149253);
|
||||
assert.strictEqual(compositeRepunit(), 149253);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -31,12 +31,12 @@ assert.strictEqual(euler130(), 149253);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler130() {
|
||||
function compositeRepunit() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler130();
|
||||
compositeRepunit();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,20 +8,20 @@ dashedName: problem-131-prime-cube-partnership
|
||||
|
||||
# --description--
|
||||
|
||||
There are some prime values, p, for which there exists a positive integer, n, such that the expression n3 + n2p is a perfect cube.
|
||||
There are some prime values, $p$, for which there exists a positive integer, $n$, such that the expression $n^3 + n^{2}p$ is a perfect cube.
|
||||
|
||||
For example, when p = 19, 83 + 82×19 = 123.
|
||||
For example, when $p = 19,\\ 8^3 + 8^2 × 19 = {12}^3$.
|
||||
|
||||
What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.
|
||||
What is perhaps most surprising is that the value of $n$ is unique for each prime with this property, and there are only four such primes below one hundred.
|
||||
|
||||
How many primes below one million have this remarkable property?
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler131()` should return 173.
|
||||
`primeCubePartnership()` should return `173`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler131(), 173);
|
||||
assert.strictEqual(primeCubePartnership(), 173);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -29,12 +29,12 @@ assert.strictEqual(euler131(), 173);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler131() {
|
||||
function primeCubePartnership() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler131();
|
||||
primeCubePartnership();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,18 +8,18 @@ dashedName: problem-132-large-repunit-factors
|
||||
|
||||
# --description--
|
||||
|
||||
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k.
|
||||
A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$.
|
||||
|
||||
For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these prime factors is 9414.
|
||||
For example, $R(10) = 1111111111 = 11 × 41 × 271 × 9091$, and the sum of these prime factors is 9414.
|
||||
|
||||
Find the sum of the first forty prime factors of R(109).
|
||||
Find the sum of the first forty prime factors of $R({10}^9)$.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler132()` should return 843296.
|
||||
`largeRepunitFactors()` should return `843296`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler132(), 843296);
|
||||
assert.strictEqual(largeRepunitFactors(), 843296);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -27,12 +27,12 @@ assert.strictEqual(euler132(), 843296);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler132() {
|
||||
function largeRepunitFactors() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler132();
|
||||
largeRepunitFactors();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,20 +8,20 @@ dashedName: problem-133-repunit-nonfactors
|
||||
|
||||
# --description--
|
||||
|
||||
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
|
||||
A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.
|
||||
|
||||
Let us consider repunits of the form R(10n).
|
||||
Let us consider repunits of the form $R({10}^n)$.
|
||||
|
||||
Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).
|
||||
Although $R(10)$, $R(100)$, or $R(1000)$ are not divisible by 17, $R(10000)$ is divisible by 17. Yet there is no value of n for which $R({10}^n)$ will divide by 19. Remarkably, 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of $R({10}^n)$.
|
||||
|
||||
Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).
|
||||
Find the sum of all the primes below one-hundred thousand that will never be a factor of $R({10}^n)$.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler133()` should return 453647705.
|
||||
`repunitNonfactors()` should return `453647705`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler133(), 453647705);
|
||||
assert.strictEqual(repunitNonfactors(), 453647705);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -29,12 +29,12 @@ assert.strictEqual(euler133(), 453647705);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler133() {
|
||||
function repunitNonfactors() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler133();
|
||||
repunitNonfactors();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,18 +8,18 @@ dashedName: problem-134-prime-pair-connection
|
||||
|
||||
# --description--
|
||||
|
||||
Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p1 whilst also being divisible by p2.
|
||||
Consider the consecutive primes $p_1 = 19$ and $p_2 = 23$. It can be verified that 1219 is the smallest number such that the last digits are formed by $p_1$ whilst also being divisible by $p_2$.
|
||||
|
||||
In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive primes, p2 > p1, there exist values of n for which the last digits are formed by p1 and n is divisible by p2. Let S be the smallest of these values of n.
|
||||
In fact, with the exception of $p_1 = 3$ and $p_2 = 5$, for every pair of consecutive primes, $p_2 > p_1$, there exist values of $n$ for which the last digits are formed by $p_1$ and $n$ is divisible by $p_2$. Let $S$ be the smallest of these values of $n$.
|
||||
|
||||
Find ∑ S for every pair of consecutive primes with 5 ≤ p1 ≤ 1000000.
|
||||
Find $\sum{S}$ for every pair of consecutive primes with $5 ≤ p_1 ≤ 1000000$.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler134()` should return 18613426663617120.
|
||||
`primePairConnection()` should return `18613426663617120`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler134(), 18613426663617120);
|
||||
assert.strictEqual(primePairConnection(), 18613426663617120);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -27,12 +27,12 @@ assert.strictEqual(euler134(), 18613426663617120);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler134() {
|
||||
function primePairConnection() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler134();
|
||||
primePairConnection();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,20 +8,20 @@ dashedName: problem-135-same-differences
|
||||
|
||||
# --description--
|
||||
|
||||
Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n, for which the equation, x2 − y2 − z2 = n, has exactly two solutions is n = 27:
|
||||
Given the positive integers, $x$, $y$, and $z$, are consecutive terms of an arithmetic progression, the least value of the positive integer, $n$, for which the equation, $x^2 − y^2 − z^2 = n$, has exactly two solutions is $n = 27$:
|
||||
|
||||
342 − 272 − 202 = 122 − 92 − 62 = 27
|
||||
$$34^2 − 27^2 − 20^2 = 12^2 − 9^2 − 6^2 = 27$$
|
||||
|
||||
It turns out that n = 1155 is the least value which has exactly ten solutions.
|
||||
It turns out that $n = 1155$ is the least value which has exactly ten solutions.
|
||||
|
||||
How many values of n less than one million have exactly ten distinct solutions?
|
||||
How many values of $n$ less than one million have exactly ten distinct solutions?
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler135()` should return 4989.
|
||||
`sameDifferences()` should return `4989`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler135(), 4989);
|
||||
assert.strictEqual(sameDifferences(), 4989);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -29,12 +29,12 @@ assert.strictEqual(euler135(), 4989);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler135() {
|
||||
function sameDifferences() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler135();
|
||||
sameDifferences();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,20 +8,20 @@ dashedName: problem-136-singleton-difference
|
||||
|
||||
# --description--
|
||||
|
||||
The positive integers, x, y, and z, are consecutive terms of an arithmetic progression. Given that n is a positive integer, the equation, x2 − y2 − z2 = n, has exactly one solution when n = 20:
|
||||
The positive integers, $x$, $y$, and $z$, are consecutive terms of an arithmetic progression. Given that $n$ is a positive integer, the equation, $x^2 − y^2 − z^2 = n$, has exactly one solution when $n = 20$:
|
||||
|
||||
132 − 102 − 72 = 20
|
||||
$$13^2 − 10^2 − 7^2 = 20$$
|
||||
|
||||
In fact there are twenty-five values of n below one hundred for which the equation has a unique solution.
|
||||
In fact, there are twenty-five values of $n$ below one hundred for which the equation has a unique solution.
|
||||
|
||||
How many values of n less than fifty million have exactly one solution?
|
||||
How many values of $n$ less than fifty million have exactly one solution?
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler136()` should return 2544559.
|
||||
`singletonDifference()` should return `2544559`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler136(), 2544559);
|
||||
assert.strictEqual(singletonDifference(), 2544559);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -29,12 +29,12 @@ assert.strictEqual(euler136(), 2544559);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler136() {
|
||||
function singletonDifference() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler136();
|
||||
singletonDifference();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,30 +8,38 @@ dashedName: problem-137-fibonacci-golden-nuggets
|
||||
|
||||
# --description--
|
||||
|
||||
Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + ..., where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ; that is, Fk = Fk−1 + Fk−2, F1 = 1 and F2 = 1.
|
||||
Consider the infinite polynomial series $A_{F}(x) = xF_1 + x^2F_2 + x^3F_3 + \ldots$, where $F_k$ is the $k$th term in the Fibonacci sequence: $1, 1, 2, 3, 5, 8, \ldots$; that is, $F_k = F_{k − 1} + F_{k − 2}, F_1 = 1$ and $F_2 = 1$.
|
||||
|
||||
For this problem we shall be interested in values of x for which AF(x) is a positive integer.
|
||||
For this problem we shall be interested in values of $x$ for which $A_{F}(x)$ is a positive integer.
|
||||
|
||||
Surprisingly AF(1/2)
|
||||
Surprisingly
|
||||
|
||||
=
|
||||
$$\begin{align}
|
||||
A_F(\frac{1}{2}) & = (\frac{1}{2}) × 1 + {(\frac{1}{2})}^2 × 1 + {(\frac{1}{2})}^3 × 2 + {(\frac{1}{2})}^4 × 3 + {(\frac{1}{2})}^5 × 5 + \cdots \\\\
|
||||
& = \frac{1}{2} + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{5}{32} + \cdots \\\\
|
||||
& = 2
|
||||
\end{align}$$
|
||||
|
||||
(1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + ...
|
||||
The corresponding values of $x$ for the first five natural numbers are shown below.
|
||||
|
||||
= 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
|
||||
| $x$ | $A_F(x)$ |
|
||||
|---------------------------|----------|
|
||||
| $\sqrt{2} − 1$ | $1$ |
|
||||
| $\frac{1}{2}$ | $2$ |
|
||||
| $\frac{\sqrt{13} − 2}{3}$ | $3$ |
|
||||
| $\frac{\sqrt{89} − 5}{8}$ | $4$ |
|
||||
| $\frac{\sqrt{34} − 3}{5}$ | $5$ |
|
||||
|
||||
= 2 The corresponding values of x for the first five natural numbers are shown below.
|
||||
We shall call $A_F(x)$ a golden nugget if $x$ is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.
|
||||
|
||||
xAF(x) √2−11 1/22 (√13−2)/33 (√89−5)/84 (√34−3)/55
|
||||
|
||||
We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690. Find the 15th golden nugget.
|
||||
Find the 15th golden nugget.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler137()` should return 1120149658760.
|
||||
`goldenNugget()` should return `1120149658760`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler137(), 1120149658760);
|
||||
assert.strictEqual(goldenNugget(), 1120149658760);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -39,12 +47,12 @@ assert.strictEqual(euler137(), 1120149658760);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler137() {
|
||||
function goldenNugget() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler137();
|
||||
goldenNugget();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,16 +8,22 @@ dashedName: problem-138-special-isosceles-triangles
|
||||
|
||||
# --description--
|
||||
|
||||
Consider the isosceles triangle with base length, b = 16, and legs, L = 17.
|
||||
Consider the isosceles triangle with base length, $b = 16$, and legs, $L = 17$.
|
||||
|
||||
By using the Pythagorean theorem it can be seen that the height of the triangle, h = √(172 − 82) = 15, which is one less than the base length. With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b ± 1. Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 and b, L are positive integers.
|
||||
<img class="img-responsive center-block" alt="isosceles triangle with edges named as L - two edges with the same length and base of the triangle as b; and height of the triangle - h from the base of the triangle to the angle between L edges" src="https://cdn.freecodecamp.org/curriculum/project-euler/special-isosceles-triangles.png" style="background-color: white; padding: 10px;">
|
||||
|
||||
By using the Pythagorean theorem, it can be seen that the height of the triangle, $h = \sqrt{{17}^2 − 8^2} = 15$, which is one less than the base length.
|
||||
|
||||
With $b = 272$ and $L = 305$, we get $h = 273$, which is one more than the base length, and this is the second smallest isosceles triangle with the property that $h = b ± 1$.
|
||||
|
||||
Find $\sum{L}$ for the twelve smallest isosceles triangles for which $h = b ± 1$ and $b$, $L$ are positive integers.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler138()` should return 1118049290473932.
|
||||
`isoscelesTriangles()` should return `1118049290473932`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler138(), 1118049290473932);
|
||||
assert.strictEqual(isoscelesTriangles(), 1118049290473932);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -25,12 +31,12 @@ assert.strictEqual(euler138(), 1118049290473932);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler138() {
|
||||
function isoscelesTriangles() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler138();
|
||||
isoscelesTriangles();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -12,14 +12,18 @@ Let (a, b, c) represent the three sides of a right angle triangle with integral
|
||||
|
||||
For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 square can be tiled with twenty-five 1 by 1 squares.
|
||||
|
||||
However, if (5, 12, 13) triangles were used then the hole would measure 7 by 7 and these could not be used to tile the 13 by 13 square. Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?
|
||||
<img class="img-responsive center-block" alt="two 5 x 5 squares: one with four 3x4x5 triangles placed to create 1x1 hole in the middle; second with twenty-five 1x1 squares" src="https://cdn.freecodecamp.org/curriculum/project-euler/pythagorean-tiles.png" style="background-color: white; padding: 10px;">
|
||||
|
||||
However, if (5, 12, 13) triangles were used, the hole would measure 7 by 7. These 7 by 7 squares could not be used to tile the 13 by 13 square.
|
||||
|
||||
Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to occur?
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler139()` should return 10057761.
|
||||
`pythagoreanTiles()` should return `10057761`.
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler139(), 10057761);
|
||||
assert.strictEqual(pythagoreanTiles(), 10057761);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -27,12 +31,12 @@ assert.strictEqual(euler139(), 10057761);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler139() {
|
||||
function pythagoreanTiles() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler139();
|
||||
pythagoreanTiles();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
@ -8,22 +8,28 @@ dashedName: problem-140-modified-fibonacci-golden-nuggets
|
||||
|
||||
# --description--
|
||||
|
||||
Consider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk−1 + Gk−2, G1 = 1 and G2 = 4; that is, 1, 4, 5, 9, 14, 23, ... .
|
||||
Consider the infinite polynomial series $A_G(x) = xG_1 + x^2G_2 + x^3G_3 + \cdots$, where $G_k$ is the $k$th term of the second order recurrence relation $G_k = G_{k − 1} + G_{k − 2}, G_1 = 1$ and $G_2 = 4$; that is, $1, 4, 5, 9, 14, 23, \ldots$.
|
||||
|
||||
For this problem we shall be concerned with values of x for which AG(x) is a positive integer.
|
||||
For this problem we shall be concerned with values of $x$ for which $A_G(x)$ is a positive integer.
|
||||
|
||||
The corresponding values of x for the first five natural numbers are shown below.
|
||||
The corresponding values of $x$ for the first five natural numbers are shown below.
|
||||
|
||||
xAG(x) (√5−1)/41 2/52 (√22−2)/63 (√137−5)/144 1/25
|
||||
| $x$ | $A_G(x)$ |
|
||||
|-----------------------------|----------|
|
||||
| $\frac{\sqrt{5} − 1}{4}$ | $1$ |
|
||||
| $\frac{2}{5}$ | $2$ |
|
||||
| $\frac{\sqrt{22} − 2}{6}$ | $3$ |
|
||||
| $\frac{\sqrt{137} − 5}{14}$ | $4$ |
|
||||
| $\frac{1}{2}$ | $5$ |
|
||||
|
||||
We shall call AG(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365. Find the sum of the first thirty golden nuggets.
|
||||
We shall call $A_G(x)$ a golden nugget if $x$ is rational because they become increasingly rarer; for example, the 20th golden nugget is 211345365. Find the sum of the first thirty golden nuggets.
|
||||
|
||||
# --hints--
|
||||
|
||||
`euler140()` should return 5673835352990.
|
||||
`modifiedGoldenNuggets()` should return `5673835352990`
|
||||
|
||||
```js
|
||||
assert.strictEqual(euler140(), 5673835352990);
|
||||
assert.strictEqual(modifiedGoldenNuggets(), 5673835352990);
|
||||
```
|
||||
|
||||
# --seed--
|
||||
@ -31,12 +37,12 @@ assert.strictEqual(euler140(), 5673835352990);
|
||||
## --seed-contents--
|
||||
|
||||
```js
|
||||
function euler140() {
|
||||
function modifiedGoldenNuggets() {
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
euler140();
|
||||
modifiedGoldenNuggets();
|
||||
```
|
||||
|
||||
# --solutions--
|
||||
|
||||
Reference in New Issue
Block a user