Engineering Board Exam Reviewer · Number & Digit Problems · hard
For any positive integer k, let f1(k) denote the square of the sum of the digits of k. For n ≥ 2, let fn(k) = f1(fn-1(k)). Find f1988(11).
- 321
- 177
- 108
- 169
Answer
D. 169
Solution
Repeatedly applying f1 to 11 gives f1=4, f2=16, f3=49, f4=169, f5=256, f6=169, and it cycles between 169 and 256 from there on. Since 1988 lands on the 169 side of the cycle, f1988(11) = 169.
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