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27 Apr 2016, 11:20
Kudos
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A
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Difficulty:
25%
(medium)
Question Stats:
78% (01:43) correct
22%
(02:00)
wrong
based on 319
sessions
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If p is a positive integer, and r is the units digit of \(3^{4p}×5^{5p}×6^{6p}×7^{7p}\), then r + 2 is divisible by which of the following?
A. 2
B. 3
C. 5
D. 7
E. 9
A. 2
B. 3
C. 5
D. 7
E. 9
27 Apr 2016, 11:44
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Bunuel
the number will be an even number as it contains 6 as a term in its product, so units digit will be even..
EVEN +2 will be another EVEN and will be div by 2..
ans A
As a matter of FACT, r+2 will be div ONLY by 2 as r is 0 as the product contains both 5 and 6..
General Discussion
27 Nov 2017, 19:12
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Bunuel
Without calculating the units digits of each value in the expression, we see that since 5^5p has a prime factor of 5 and 6^6p has a prime factor of 2, we know that r is 0, and thus r + 2 = 2, so r + 2 is divisible by 2.
Alternate Solution:
Let’s focus just on 5^5p and 6^6p. Note that the units digit of 5^5p is 5, no matter what the value of p, because 5^1 = 5, 5^2 = 25, and 5^3 = 125, and this pattern of 5 as the units digit of any power of 5 continues. Similarly, consider powers of 6: 6^1 = 6, 6^2 = 36, and 6^3 = 216, and we see that 6 raised to any power has a 6 as the units digit in the answer. The product 5 x 6 is 30, and we see that the units digit of the product 5^5p x 6^6p is 0.
Thus, no matter what the other numbers are in the product, we see that the long product 3^4p x 5^5p x 6^6p x 7^7p will end in 0, because 5^5p x 6^6p = 0. Thus, r = 0.
We can also see, then, that r + 2 = 2, which is divisible by 2.
Answer: A
