Maths Olympiad Prep

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Number theory Difficulty 3.9 AMC 10/12 Find the answer Japan

Determine ten's place of 11121311^{12^{13}}. (121312^{13}th power of 1111, not 1313th power of 111211^{12}.)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

For positive integer nn,
11n=(10+1)n=k=0nCnk10k=n10n+n100=n10n+1(mod100). 11^n = (10 + 1)^n = \sum_{k=0}^{n} C_n^k 10^k = n \cdot 10^n + n \cdot 10^0 = n \cdot 10^n + 1 \pmod{100}.
Therefore, the ten's place of 11n11^n is equal to the last digit of nn. The last digit of 12n12^n for n=1,2,n = 1, 2, \dots is 2,4,8,6,2,4,2, 4, 8, 6, 2, 4, \dots, with period 44, so the last digit of 121312^{13} is 22 and the answer is 22.

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