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    <p>1) For integer division in scheme, use the <i>quotient</i> function, as in (quotient 9 2) gives 4.<BR> 2) For integer modulus in scheme, use the <i>remainder</i> function, as in (remainder 9 2) gives 1.</p> <p>Scheme is based on recursion, as you stated. To solve a recursive problem, you have to state the solution in terms of a smaller instances of the problem: "The number of 6 digits in a number N is ... " and for ... you state the solution in terms of the solution of part of N.</p> <p>The easiest part of N to recur on is N/10, which is all but the last digit of N. Then, to solve this recursively, you should assume that you know the answer to (num (quotient N 10)). Call that value X. How does knowing N and X tell you how many digits is in N?</p> <p>If the last digit of N is 6, then the solution is X+1. Otherwise, the solution is X. How can you tell if the last digit is 6? Use the remainder function.</p> <pre><code>(define (num N) ; check cases (cond ; case 1) negative condition ((&lt; N 0) (num (- N))) ; case 2) zero condition, the terminal case ((= N 0) 0) ; case 3) recursive case with a 6 as the last digit ((= (remainder N 10) 6) (+ X 1)) ; case 4) recursive case without a 6 as the last digit (#t X) )) </code></pre> <p>Now you just have to substitute the recursive assumption in for X, so for example case 3 becomes</p> <pre><code>; case 3 ((= (remainder N 10) 6) (+ (num (quotient N 10)) 1)) </code></pre> <p>Change X for case 4 as well and you'll get the solution.</p>
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