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25 Nov 2014, 21:48
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43%
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If a is an integer, what is the units digit of a^18?
(1) a^2 has a units digit of 9
(2) a^7 has a units digit of 3
(1) a^2 has a units digit of 9
(2) a^7 has a units digit of 3
25 Nov 2014, 23:02
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The cyclicity of the last digit is at most 4.
e.g.
Last digit 2 - 2, 4, 8, 6 ... (cycle of 4)
Last digit 3 - 3, 9, 7, 1 ... (cycle of 4)
Last digit 4 - 4, 6, 4, 6 ... (cycle of 2)
Last digit 5 - 5, 5, 5, 5 ... (cycle of 1)
Last digit 6 - 6, 6, 6, 6 ... (cycle of 1)
Last digit 7 - 7, 9, 3, 1 ... (cycle of 4)
Last digit 8 - 8, 4, 2, 6 ... (cycle of 4)
Last digit 9 - 9, 1, 9, 1 ... (cycle of 2)
If last digit is 0 or 1, it remains 0 or 1 - cycle of 1
1. \(a^2\) has a units digit of 9
In the cycle of 4, \(a^2\) represents the second term. Last digit of a square is 9 in two cases.
Last digit 3 - 3, 9, 7, 1
Last digit 7 - 7, 9, 3, 1
So we don't know the last digit of a. But we are not asked the last digit of a. We are asked the last digit of \(a^{18}\). It is also the second term in the cycle of 4. So it doesn't matter whether last digit of a is 3 or 7, the last digit of \(a^{18}\) will be 9 in either case. Sufficient alone.
2. \(a^7\) has a units digit of 3
In the cycle of 4, \(a^7\) represents the third term. Last digit of the third term will be 3 in only one case - when the last digit of a is 7. So we can say that the last digit of \(a^{18}\) must be 9. Sufficient alone.
Answer (D)
General Discussion
(1) Sufficient. There are only two numbers -> 3 and 7 that can end in similar units digits when raised to a power of some number
3^2 = 9
3^3 = ..7 (multiplying just units digit 9 by 3)
3^4 = ..1 (multiplying just units digit 7 by 3)
3^5 = ..3 (multiplying just units digit 1 by 3)
3^6 = ..9
3^18 = ..9
7^2 = 49
7^3 = ..3 (multiplying just units digit 9 by 7)
7^4 = ..1 (multiplying just units digit 3 by 7)
7^5 = ..7
7^6 = ..9
7^9 = ..7
7^18 = ..9
(2) Sufficient. Only 7 has a units digit of 3 when raised to the power of 7.
7^1=7
7^2 = 49
7^3 = ..3 (multiplying just units digit 9 by 7)
7^4 = ...1 (multiplying just units digit by 7)
7^7 = ...3
3^1 = 3
3^2 = 9
3^3 = ..7
3^4 = ..1
3^7 = ..7
Answer D
3^2 = 9
3^3 = ..7 (multiplying just units digit 9 by 3)
3^4 = ..1 (multiplying just units digit 7 by 3)
3^5 = ..3 (multiplying just units digit 1 by 3)
3^6 = ..9
3^18 = ..9
7^2 = 49
7^3 = ..3 (multiplying just units digit 9 by 7)
7^4 = ..1 (multiplying just units digit 3 by 7)
7^5 = ..7
7^6 = ..9
7^9 = ..7
7^18 = ..9
(2) Sufficient. Only 7 has a units digit of 3 when raised to the power of 7.
7^1=7
7^2 = 49
7^3 = ..3 (multiplying just units digit 9 by 7)
7^4 = ...1 (multiplying just units digit by 7)
7^7 = ...3
3^1 = 3
3^2 = 9
3^3 = ..7
3^4 = ..1
3^7 = ..7
Answer D
