1996 AHSME Problems/Problem 12
Contents
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 .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 13 | |
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