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- ;;; Factorial digit sum
- ;;; Problem 20
- ;;; n! means n × (n - 1) × ... × 3 × 2 × 1
- ;;; For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
- ;;; and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
- ;;; Find the sum of the digits in the number 100!
- ;; Idea: Seems simple. Just calculate 100! and take the digit sum. I
- ;; do not know, whether there is a trick in there to calculate the
- ;; digit sum more efficiently, but it is definitely very simple to
- ;; calculate the digit sum of large integers in GNU Guile.
- (import
- (except (rnrs base) let-values map)
- (only (guile)
- lambda* λ)
- (prefix (math) math:)
- (debug-utils))
- (displayln "sum of digits of 100!:"
- (math:digits-sum
- (math:factorial-linear 100)))
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