1996 AHSME Problems/Problem 12

Problem 12

A function from the integers to the integers is defined as follows:

Suppose is odd and . What is the sum of the digits of ?

Solution

First iteration

To get , you could either have and add , or and divide by .

If you had the former, you would have , and the function's rule would have you divide. Thus, is the only number for which .

Second iteration

Going out one step, if you have , you would have to have . For , you would either have and add , or and divide by .

Both are possible: and return values of . Thus, , and .

Third iteration

Going out the final step, if you have , you would have to have or .

If you doubled either of these, would not be odd. So you must subtract .

If you subtract from , you would compute , which would halve it, and not add the back.

If you subtract from , you would compute , which would add the back.

Thus, , and is odd. The desired sum of the digits is , and the answer is .

Solution 2 (rigorous but easy)

We will work from the inside to the outside and alternate \( k \) between even and odd.

If \( k \) is odd, then \( f(k) \) in terms of \( k \) is \(\frac{k+3}{2}\).

Next, we have \( f\left(\frac{k+3}{2}\right) \), and \(\frac{k+3}{2}\) is even, so \( f\left(\frac{k+3}{2}\right) = \frac{k+3}{4} \).

We are given that \( f\left(\frac{k+3}{4}\right) = 27 \).

The sum of its digits is \( 1 + 0 + 5 = 6 \).

{gnv12}

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.