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Arithmetic

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Source: — Problem Solving |

by GMATGuruNY » Sun Jan 07, 2018 4:28 am
Roland2rule wrote:What is the last digit of $$3^{987}$$ ?

A. 1

B. 7

C. 9

D. 2
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7� --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
The units digits repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 7 is raised to a power that is a multiple of 4, the units digit will be 1.

Thus:
7��� will have a units digit of 1.
From here, the cycle of units digits will repeat: 7, 9, 3, 1...
Thus:
7��� --> units digit of 7.
7��� --> units digit of 9.
7��� --> units digit of 3.

The last digit of 7��� is 3.
The correct answer is not among the answer choices.
Please confirm that you have posted the answer choices correctly (especially since the problem as posted includes only four answer choices, while an actual GMAT problem always includes five answer choices.)
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