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Problem 12¶

Highly Divisible Triangular Number¶

The sequence of triangle numbers is generated by adding the natural numbers. So the $7$^th triangle number would be $1+2+3+4+5+6+7=28$. The first ten terms would be:
$$1,3,6,10,15,21,28,36,45,55,\dots$$
Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that $28$ is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?

高度整除三角数¶

三角数の数列は自然数の和で表わされ, 7番目の三角数は 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 である. 三角数の最初の10項は:
$$1,3,6,10,15,21,28,36,45,55,\dots$$
となる.

最初の7項について, その約数を列挙すると, 以下のとおり.

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

これから, 7番目の三角数である28は, 5個より多く約数をもつ最初の三角数であることが分かる.
では, 500個より多く約数をもつ最初の三角数はいくつか.

#!r6rs
#!chezscheme

(import (chezscheme))

(define (triangular-numbers end-condition)
  (let loop ([i 1] [current 0] [result '()])
    (if (and
          (not (null? result))
          (end-condition result))
      result
      (let ([next (+ current i)])
        (loop (add1 i) next (cons next result))))))

(define (divisors num)
  (let ([check-max (isqrt num)])
    (let loop ([i 1] [result '()])
      (cond
        [(< check-max i) result]
        [(zero? (mod num i))
         (loop (add1 i) (cons* i (div num i) result))]
        [else (loop (add1 i) result)]))))

(define over500-divisors-triangular-number
  (triangular-numbers
    (lambda (lis)
      (< 500 (length (divisors (car lis)))))))

(define answer-12
  (car over500-divisors-triangular-number))

(printf "12: ~D~%" answer-12)