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1988 AIME Problems/Problem 2

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Problem

For any positive integer k, let f_1(k) denote the square of the sum of the digits of k. For n \ge 2, let f_n(k) = f_1(f_{n - 1}(k)). Find f_{1988}(11).

Solution

We see that f(11)=4

f(4)=16

f(16)=49

f(49)=169

f(169)=256

f(256)=169

Note that this revolves between the two numbers. f_{1988}(169)=169\implies f_{1988}(11)=\boxed{169}

See also

1988 AIME (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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