* fix: clean-up Project Euler 301-320 * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
1.7 KiB
id, title, challengeType, forumTopicId, dashedName
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f4a81000cf542c50ffbb | Problem 316: Numbers in decimal expansions | 5 | 301972 | problem-316-numbers-in-decimal-expansions |
--description--
Let p = p_1 p_2 p_3 \ldots be an infinite sequence of random digits, selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
It can be seen that p corresponds to the real number 0.p_1 p_2 p_3 \ldots.
It can also be seen that choosing a random real number from the interval [0,1) is equivalent to choosing an infinite sequence of random digits selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
For any positive integer n with d decimal digits, let k be the smallest index such that p_k, p_{k + 1}, \ldots p_{k + d - 1} are the decimal digits of n, in the same order.
Also, let g(n) be the expected value of k; it can be proven that g(n) is always finite and, interestingly, always an integer number.
For example, if n = 535, then
for p = 31415926\mathbf{535}897\ldots, we get k = 9
for p = 35528714365004956000049084876408468\mathbf{535}4\ldots, we get k = 36
etc and we find that g(535) = 1008.
Given that \displaystyle\sum_{n = 2}^{999} g\left(\left\lfloor\frac{{10}^6}{n}\right\rfloor\right) = 27280188, find \displaystyle\sum_{n = 2}^{999\\,999} g\left(\left\lfloor\frac{{10}^{16}}{n}\right\rfloor\right).
Note: \lfloor x\rfloor represents the floor function.
--hints--
numbersInDecimalExpansion() should return 542934735751917760.
assert.strictEqual(numbersInDecimalExpansion(), 542934735751917760);
--seed--
--seed-contents--
function numbersInDecimalExpansion() {
return true;
}
numbersInDecimalExpansion();
--solutions--
// solution required