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)

\textdollar{}f(f(f(k))) = 27\textdollar{}

We will alternate k between even and odd.

If k is odd, then \textdollar{}f(k) in terms of k = (k+3)/2

Next, we have \textdollar{}f((k+3)/2) and \textdollar{}(k+3)/2 is even so \textdollar{}f((k+3)/2)= (k+3)/4 \textdollar{}

We are given that \textdollar{}f((k+3)/4)= 27\textdollar{}

\textdollar{}(k+3)/4= 27\textdollar{} so \textdollar{}k= 105\textdollar{}. The sum of it's digits is 1+0+5= \textdollar{}6\textdollar{}

See also

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