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16 Sep 2014, 01:46
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If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 smaller than the remainder when \(b\) is divided by 11?
(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number.
(2) The remainder when \(b\) is divided by 12 is \(b\)
(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number.
(2) The remainder when \(b\) is divided by 12 is \(b\)
16 Sep 2014, 01:46
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Official Solution:
If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 smaller than the remainder when \(b\) is divided by 11?
First, observe that the possible remainders when a positive integer is divided by 11 are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10.
(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:
The above implies that \(a = 69q + prime^5\) and \(prime^5 < 69\) (since the remainder must be less than the divisor). The only prime number with a fifth power less than 69 is 2: \((2^5 = 32) < 69\). Thus, we have \(a = 69q + 32\), which can be 32, 101, 170, ... As \(a\) is a two-digit integer, \(a = 32\).
Dividing \(a = 32\) by 11 results in a remainder of 10. Since 10 is the maximum possible remainder when dividing by 11, this remainder cannot be smaller than the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10). Therefore, the answer to the question is NO.
Sufficient.
(2) The remainder when \(b\) is divided by 12 is \(b\):
As the remainder must be less than the divisor, \(b\) must be less than 12. Since we know that \(b\) is a two-digit integer greater than 10, \(b\) must be 11.
Dividing \(b = 11\) by 11 results in a remainder of 0. Since 0 is the minimum possible remainder, the remainder when \(a\) is divided by 11 cannot be smaller than 0. Therefore, the answer to the question is NO.
Sufficient.
Answer: D
If \(a\) and \(b\) are two-digit positive integers greater than 10, is the remainder when \(a\) is divided by 11 smaller than the remainder when \(b\) is divided by 11?
First, observe that the possible remainders when a positive integer is divided by 11 are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10.
(1) The remainder when \(a\) is divided by 69 is the fifth power of a prime number:
The above implies that \(a = 69q + prime^5\) and \(prime^5 < 69\) (since the remainder must be less than the divisor). The only prime number with a fifth power less than 69 is 2: \((2^5 = 32) < 69\). Thus, we have \(a = 69q + 32\), which can be 32, 101, 170, ... As \(a\) is a two-digit integer, \(a = 32\).
Dividing \(a = 32\) by 11 results in a remainder of 10. Since 10 is the maximum possible remainder when dividing by 11, this remainder cannot be smaller than the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10). Therefore, the answer to the question is NO.
Sufficient.
(2) The remainder when \(b\) is divided by 12 is \(b\):
As the remainder must be less than the divisor, \(b\) must be less than 12. Since we know that \(b\) is a two-digit integer greater than 10, \(b\) must be 11.
Dividing \(b = 11\) by 11 results in a remainder of 0. Since 0 is the minimum possible remainder, the remainder when \(a\) is divided by 11 cannot be smaller than 0. Therefore, the answer to the question is NO.
Sufficient.
Answer: D
General Discussion
shashanksagar
GMAT 1: 720 Q50 V37
Posts: 62
09 Jul 2016, 03:09
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DAMN!! how m I suppose to solve this within 2 minutes.. 
yeah +1 for 800 level classification. 
