1995 AMC 12 Problem 27

Attempt Problem 27 of the 1995 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

Consider the triangular array of numbers with 0,0, 1,1, 2,2, 3,3, …\ldots along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 11 through 66 are shown.
0112223443478745111515115 \begin{array}{cccccc} &&0&&&\\ &&1&1&&\\ &2&2&2&&\\ &3&4&4&3&\\ 4&7&8&7&4&\\ 5&11&15&15&11&5 \end{array}
Let f(n)f(n) denote the sum of the numbers in row n.n. What is the remainder when f(100)f(100) is divided by 100?100?
1212
3030
5050
6262
7474
Answer: E
Concepts:recurrencesmodular arithmetic
Small hint:
Relate a row’s sum to the previous row’s sum by counting how often each old entry contributes
Big hint:
Solve the resulting recurrence, then compute the power of 22 modulo 100100
Solution:
Every previous entry contributes to two entries of the next row, and the two new boundary entries contribute an additional 2.2. Thus f(n)=2f(n−1)+2,f(n)=2f(n-1)+2, with f(1)=0.f(1)=0. Hence f(n)=2n−2. f(n)=2^n-2. Since 220≡76(mod100)2^{20}\equiv76\pmod{100} and 762≡76(mod100),76^2\equiv76\pmod{100}, we have 2100≡76(mod100).2^{100}\equiv76\pmod{100}. Therefore f(100)≡74(mod100),f(100)\equiv74\pmod{100}, and the correct answer is E.
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