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Example: How to convert list: '(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15) Into list of lists: '((0 1 2 3) (4 5 6 7) (8 9 10 11) (12 13 14 15)) Based on answers provided here so far, this is what I've come up with: First define function to take up to 'n' elements from beginning of the list: (define (take-up-to n xs) (define (iter xs n taken) (cond [(or (zero? n) (empty? xs)) (reverse taken)] [else (iter (cdr xs) (- n 1) (cons (car xs) taken))])) (iter xs n '())) Second is similar function for the rest of list: (define (drop-up-to n xs) (define (iter xs n taken) (cond [(or (zero? n) (empty? xs)) xs] [else (iter (cdr xs) (- n 1) (cons (car xs) taken))])) (iter xs n '())) This could have been done as one function that returns two values and Racket has a function 'split-at' that produces same result, but I did this as an exercise. ps. Is this correct use of tail recursion ? Than split-into-chunks can be written like this: (define (split-into-chunks n xs) (if (null? xs) '() (let ((first-chunk (take-up-to n xs)) (rest-of-list (drop-up-to n xs))) (cons first-chunk (split-into-chunks n rest-of-list))))) pps. Can this one be improved even more or is it 'good enough' ?

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