71 lines
2.1 KiB
Markdown
71 lines
2.1 KiB
Markdown
|
---
|
|||
|
|
id: 5900f4051000cf542c50ff18
|
|||
|
|
title: 'Problem 153: Investigating Gaussian Integers'
|
|||
|
|
challengeType: 5
|
|||
|
|
forumTopicId: 301784
|
|||
|
|
dashedName: problem-153-investigating-gaussian-integers
|
|||
|
|
---
|
|||
|
|
|
|||
|
|
# --description--
|
|||
|
|
|
|||
|
|
As we all know the equation x2=-1 has no solutions for real x.
|
|||
|
|
|
|||
|
|
If we however introduce the imaginary number i this equation has two solutions: x=i and x=-i.
|
|||
|
|
|
|||
|
|
If we go a step further the equation (x-3)2=-4 has two complex solutions: x=3+2i and x=3-2i. x=3+2i and x=3-2i are called each others' complex conjugate.
|
|||
|
|
|
|||
|
|
Numbers of the form a+bi are called complex numbers.
|
|||
|
|
|
|||
|
|
In general a+bi and a−bi are each other's complex conjugate. A Gaussian Integer is a complex number a+bi such that both a and b are integers.
|
|||
|
|
|
|||
|
|
The regular integers are also Gaussian integers (with b=0).
|
|||
|
|
|
|||
|
|
To distinguish them from Gaussian integers with b ≠ 0 we call such integers "rational integers."
|
|||
|
|
|
|||
|
|
A Gaussian integer is called a divisor of a rational integer n if the result is also a Gaussian integer.
|
|||
|
|
|
|||
|
|
If for example we divide 5 by 1+2i we can simplify in the following manner:
|
|||
|
|
|
|||
|
|
Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i.
|
|||
|
|
|
|||
|
|
The result is .
|
|||
|
|
|
|||
|
|
So 1+2i is a divisor of 5.
|
|||
|
|
|
|||
|
|
Note that 1+i is not a divisor of 5 because .
|
|||
|
|
|
|||
|
|
Note also that if the Gaussian Integer (a+bi) is a divisor of a rational integer n, then its complex conjugate (a−bi) is also a divisor of n. In fact, 5 has six divisors such that the real part is positive: {1, 1 + 2i, 1 − 2i, 2 + i, 2 − i, 5}.
|
|||
|
|
|
|||
|
|
The following is a table of all of the divisors for the first five positive rational integers:
|
|||
|
|
|
|||
|
|
n Gaussian integer divisors with positive real partSum s(n) of these
|
|||
|
|
|
|||
|
|
divisors111 21, 1+i, 1-i, 25 31, 34 41, 1+i, 1-i, 2, 2+2i, 2-2i,413 51, 1+2i, 1-2i, 2+i, 2-i, 512 For divisors with positive real parts, then, we have: . For 1 ≤ n ≤ 105, ∑ s(n)=17924657155. What is ∑ s(n) for 1 ≤ n ≤ 108?
|
|||
|
|
|
|||
|
|
# --hints--
|
|||
|
|
|
|||
|
|
`euler153()` should return 17971254122360636.
|
|||
|
|
|
|||
|
|
```js
|
|||
|
|
assert.strictEqual(euler153(), 17971254122360636);
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
# --seed--
|
|||
|
|
|
|||
|
|
## --seed-contents--
|
|||
|
|
|
|||
|
|
```js
|
|||
|
|
function euler153() {
|
|||
|
|
|
|||
|
|
return true;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
euler153();
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
# --solutions--
|
|||
|
|
|
|||
|
|
```js
|
|||
|
|
// solution required
|
|||
|
|
```
|