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M30-18

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M30-18  [#permalink]

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New post 16 Sep 2014, 01:46
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23% (01:52) correct 77% (01:55) wrong based on 64 sessions

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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 less 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 \(b\)

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Re M30-18  [#permalink]

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New post 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 less than the remainder when \(b\) is divided by 11?

First of all, note that the remainder when a positive integer is divided by 11 could be: 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:

\(a=69q+prime^5\) and \(prime^5 < 69\) (the remainder must be less than the divisor). The only prime number whose fifth power is less than 69 is 2: \((2^5=32) < 69\). So, we have that \(a=69q+32\): 32, 101, 170, ... Since \(a\) is a two-digit integer, then \(a=32\).

Now, \(a=32\) divided by 11 gives the remainder of 10. Since 10 is the maximum remainder possible when divided by 11, then this remainder cannot be less then the remainder when \(b\) is divided by 11 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10). So, the answer to the question is NO.

Sufficient.

(2) The remainder when \(b\) is divided by 12 is \(b\):

The remainder must be less than the divisor, hence \(b\) must be less than 12 and since we are told that \(b\) is a two-digit integers greater than 10, then \(b\) must be 11.

Now, \(b=11\) divided by 11 gives the remainder of 0. Since 0 is the minimum remainder possible, than the remainder when \(a\) is divided by 11 cannot possible be less than 0. So, the answer to the question is NO.

Sufficient.


Answer: D
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Re: M30-18  [#permalink]

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New post 09 Jul 2016, 03:09
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DAMN!! how m I suppose to solve this within 2 minutes.. :( :(
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Re M30-18  [#permalink]

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New post 03 Aug 2016, 23:28
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I think this is a high-quality question and I agree with explanation. Great question this has to be classified a "800-level" problem lol!
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Re M30-18  [#permalink]

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New post 15 Aug 2016, 22:21
I think this is a high-quality question and I agree with explanation.
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Re: M30-18  [#permalink]

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New post 22 Sep 2016, 06:25
This is a very high-quality question but I guess it is too tricky to be solved under pressure in about 2 mins.
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Re: M30-18  [#permalink]

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New post 30 Sep 2016, 10:06
Is it not possible for a & b to be same two digits positive integers or am I reading the question incorrectly
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Re: M30-18  [#permalink]

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New post 01 Oct 2016, 02:58
vibhav_gupta08 wrote:
Is it not possible for a & b to be same two digits positive integers or am I reading the question incorrectly


Different variables can represent the same number, if it's not explicitely stated otherwise. But this does not change the answer here. For more check this: if-a-and-b-are-two-digit-positive-integers-greater-than-172886.html#p1669869
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M30-18  [#permalink]

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New post 10 Mar 2018, 14:57
Great question and easy to fall victim to the common mistake type - failing to recognise that "no" is a sufficient answer. We need to be careful and note we are not asked what the remainder is, but whether r of b can be larger

I got to a=2 but and r=10 and then assumed that because I didn't know b, (i) was insufficient. In fact, we don't need to know the remainder of B at all because we can't have a remainder larger than 10 anyway
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Re: M30-18  [#permalink]

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New post 23 Mar 2018, 21:10
tinyinthedesert

You in fact do need to know the remainder of B, hence why the answer to the question is D and not (C) as it would need to be if you are referencing "the remainder cant be greater than 10 anyway". That is contingent upon info from statement 1.
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Re: M30-18  [#permalink]

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New post 04 Jun 2018, 04:05
Great quality question! very tricky
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Re M30-18  [#permalink]

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New post 03 Sep 2018, 22:14
I think this is a high-quality question and I agree with explanation.
Re M30-18 &nbs [#permalink] 03 Sep 2018, 22:14
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