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M30-18

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16 Sep 2014, 01:46
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Difficulty:

95% (hard)

Question Stats:

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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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.

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09 Jul 2016, 03:09
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DAMN!! how m I suppose to solve this within 2 minutes..
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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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15 Aug 2016, 22:21
I think this is a high-quality question and I agree with explanation.
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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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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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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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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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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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04 Jun 2018, 04:05
Great quality question! very tricky
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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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