1961 AMC 12 Problem 28

Attempt Problem 28 of the 1961 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 1961 AMC 12 solutions, or check the answer key.

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

28.

If 21377532137^{753} is multiplied out, the units’ digit in the final product is:
11
33
55
77
99
Answer: D
Concepts:units digitmodular exponentiationpattern recognition
Small hint:
Only the units digit of the base matters
Big hint:
List the four-term cycle of powers of 77 modulo 1010
Solution:
Powers of 77 have units digits 7,7, 9,9, 3,3, 11 in a cycle of length 4.4. Since 753≡1(mod4),753\equiv1\pmod4, the units digit of 21377532137^{753} is 7.7.
Therefore, the correct answer is D.
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