45 lines
809 B
Markdown
45 lines
809 B
Markdown
---
|
|
id: 5900f4291000cf542c50ff3b
|
|
title: 'Problem 188: The hyperexponentiation of a number'
|
|
challengeType: 5
|
|
forumTopicId: 301824
|
|
dashedName: problem-188-the-hyperexponentiation-of-a-number
|
|
---
|
|
|
|
# --description--
|
|
|
|
The hyperexponentiation or tetration of a number a by a positive integer b, denoted by a↑↑b or ba, is recursively defined by:
|
|
|
|
a↑↑1 = a,
|
|
|
|
a↑↑(k+1) = a(a↑↑k).
|
|
|
|
Thus we have e.g. 3↑↑2 = 33 = 27, hence 3↑↑3 = 327 = 7625597484987 and 3↑↑4 is roughly 103.6383346400240996\*10^12. Find the last 8 digits of 1777↑↑1855.
|
|
|
|
# --hints--
|
|
|
|
`euler188()` should return 95962097.
|
|
|
|
```js
|
|
assert.strictEqual(euler188(), 95962097);
|
|
```
|
|
|
|
# --seed--
|
|
|
|
## --seed-contents--
|
|
|
|
```js
|
|
function euler188() {
|
|
|
|
return true;
|
|
}
|
|
|
|
euler188();
|
|
```
|
|
|
|
# --solutions--
|
|
|
|
```js
|
|
// solution required
|
|
```
|