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# If p is a positive integer, and r is the units digit of 3^4p×5^5p×6^6p

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Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135543 [1], given: 12697

If p is a positive integer, and r is the units digit of 3^4p×5^5p×6^6p [#permalink]

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27 Apr 2016, 10:20
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25% (medium)

Question Stats:

74% (01:11) correct 26% (01:46) wrong based on 132 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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 135543 [1], given: 12697

Math Expert
Joined: 02 Aug 2009
Posts: 5346

Kudos [?]: 6117 [2], given: 121

Re: If p is a positive integer, and r is the units digit of 3^4p×5^5p×6^6p [#permalink]

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27 Apr 2016, 10:44
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Bunuel wrote:
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

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..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6117 [2], given: 121

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1802

Kudos [?]: 981 [1], given: 5

Re: If p is a positive integer, and r is the units digit of 3^4p×5^5p×6^6p [#permalink]

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27 Nov 2017, 18:12
1
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Expert's post
Bunuel wrote:
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

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.

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Jeffery Miller

GMAT Quant Self-Study Course
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Kudos [?]: 981 [1], given: 5

Re: If p is a positive integer, and r is the units digit of 3^4p×5^5p×6^6p   [#permalink] 27 Nov 2017, 18:12
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