; two simple tasks
 
(define-syntax fold-of
  (syntax-rules (range in is)
    ((_ "z" f b e) (set! b (f b e)))
    ((_ "z" f b e (v range fst pst stp) c ...)
      (let* ((x fst) (p pst) (s stp)
             (le? (if (positive? s) <= >=)))
        (do ((v x (+ v s))) ((le? p v) b)
          (fold-of "z" f b e c ...))))
    ((_ "z" f b e (v range fst pst) c ...)
      (let* ((x fst) (p pst) (s (if (< x p) 1 -1)))
        (fold-of "z" f b e (v range x p s) c ...)))
    ((_ "z" f b e (v range pst) c ...)
      (fold-of "z" f b e (v range 0 pst) c ...))
    ((_ "z" f b e (x in xs) c ...)
      (do ((t xs (cdr t))) ((null? t) b)
        (let ((x (car t)))
          (fold-of "z" f b e c ...))))
    ((_ "z" f b e (x is y) c ...)
      (let ((x y)) (fold-of "z" f b e c ...)))
    ((_ "z" f b e p? c ...)
      (if p? (fold-of "z" f b e c ...)))
    ((_ f i e c ...)
      (let ((b i)) (fold-of "z" f b e c ...)))))
 
(define-syntax list-of (syntax-rules ()
  ((_ arg ...) (reverse (fold-of
    (lambda (d a) (cons a d)) '() arg ...)))))
 
(define (sum xs) (apply + xs))
 
(define (square x) (* x x))
 
(define (digits n . args)
  (let ((b (if (null? args) 10 (car args))))
    (let loop ((n n) (d '()))
      (if (zero? n) d
          (loop (quotient n b)
                (cons (modulo n b) d))))))
 
(define (f limit)
  (let loop ((n 1) (m 5))
    (when (<= n limit)
      (display n) (newline)
      (loop (* n m) (/ 10 m)))))
 
(display (f 10000)) (newline)
 
(display  (list-of n
    (n range 110 1000 11)
    (= (sum (map square (digits n))) (/ n 11))))

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