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# M30-18

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:46
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Difficulty:

95% (hard)

Question Stats:

25% (01:22) correct 75% (01:52) wrong based on 197 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, 00:46
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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, 02:09
1
DAMN!! how m I suppose to solve this within 2 minutes..
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03 Aug 2016, 22:28
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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, 21:21
I think this is a high-quality question and I agree with explanation.
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22 Sep 2016, 05:25
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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, 09:06
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, 01:58
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, 13:57
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, 20:10
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, 03:05
Great quality question! very tricky
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03 Sep 2018, 21:14
I think this is a high-quality question and I agree with explanation.
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25 Sep 2018, 01:23
That's a brilliant question!! yeah +1 for 800 level classification. Bunuel if you're listening. ha ha.
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15 Oct 2018, 22:57
I believe the answer would be C. It is asked that whether the remainder when a is divided by 11 less than the remainder when b is divided by 11.

We got the answer to the question that what is the value of a. But there is no where mentioned that the value of a cannot be equal to b.
So lets take the case when b=11 and a=11, then remainder when b/12 and a/12 is same & is not less.

Let me know if I am missing anything here
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15 Oct 2018, 23:01
Akashkdh wrote:
I believe the answer would be C. It is asked that whether the remainder when a is divided by 11 less than the remainder when b is divided by 11.

We got the answer to the question that what is the value of a. But there is no where mentioned that the value of a cannot be equal to b.
So lets take the case when b=11 and a=11, then remainder when b/12 and a/12 is same & is not less.

Let me know if I am missing anything here

No.

The question asks: is the remainder when a is divided by 11 less than the remainder when b is divided by 11?

Each statement gives a NO answer to the question: the remainder when a is divided by 11 is NOT less than the remainder when b is divided by 11.

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15 Oct 2018, 23:06
ohh got it thanks.

So it is either equal & less than but NOT Less than. Got it right. Thanks
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20 Oct 2018, 04:18
god level question under test circumstance!
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18 Jan 2019, 10:36
Bunuel wrote:
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.

I'm not sure that I understand the explanation of statement 1. I the issue that I am having is that a=32. My current understanding of statement one is as follows:

Statement one translates into a/69 = q + (prime)^5 which then becomes a = 69q + (prime)^5. From there I understand the part where prime must = 2, but I’m not sure how a = 32. My understanding is that a /= 32, and a should = 32 + 69q.

My reasoning is the following example:

a/22 = 22q + 16. In this case a=92 would satisfy the condition. For that matter any multiple of 22 and adding four would satisfy the condition in statement 1. If we use the same language as statement 1 just substituting for the new values, we have: The remainder when a is divided by 12 is the fourth power of a prime number. From the logic in the above problem it would follow that a=16, and a/=16. Thus a could be multiple numbers. Which produce a different remainder when divided by 11.

Could someone help explain where I am going wrong with my logic/how a = 32? I could be re arranging the question to a point where the logic breaks down, but I'm just not sure how a= 32.
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19 Jan 2019, 00:12
bcal95 wrote:
Bunuel wrote:
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.

I'm not sure that I understand the explanation of statement 1. I the issue that I am having is that a=32. My current understanding of statement one is as follows:

Statement one translates into a/69 = q + (prime)^5 which then becomes a = 69q + (prime)^5. From there I understand the part where prime must = 2, but I’m not sure how a = 32. My understanding is that a /= 32, and a should = 32 + 69q.

My reasoning is the following example:

a/22 = 22q + 16. In this case a=92 would satisfy the condition. For that matter any multiple of 22 and adding four would satisfy the condition in statement 1. If we use the same language as statement 1 just substituting for the new values, we have: The remainder when a is divided by 12 is the fourth power of a prime number. From the logic in the above problem it would follow that a=16, and a/=16. Thus a could be multiple numbers. Which produce a different remainder when divided by 11.

Could someone help explain where I am going wrong with my logic/how a = 32? I could be re arranging the question to a point where the logic breaks down, but I'm just not sure how a= 32.

So, you understand that a must be 2. Then, "The remainder when $$a$$ is divided by 69 is the fifth power of a prime number" gives $$a=69q+32$$: 32, 101, 170, ... Since $$a$$ is a two-digit integer, then $$a=32$$.

Hope it's clear.
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19 Jan 2019, 10:41
Bunuel wrote:
bcal95 wrote:
Bunuel wrote:
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.

I'm not sure that I understand the explanation of statement 1. I the issue that I am having is that a=32. My current understanding of statement one is as follows:

Statement one translates into a/69 = q + (prime)^5 which then becomes a = 69q + (prime)^5. From there I understand the part where prime must = 2, but I’m not sure how a = 32. My understanding is that a /= 32, and a should = 32 + 69q.

My reasoning is the following example:

a/22 = 22q + 16. In this case a=92 would satisfy the condition. For that matter any multiple of 22 and adding four would satisfy the condition in statement 1. If we use the same language as statement 1 just substituting for the new values, we have: The remainder when a is divided by 12 is the fourth power of a prime number. From the logic in the above problem it would follow that a=16, and a/=16. Thus a could be multiple numbers. Which produce a different remainder when divided by 11.

Could someone help explain where I am going wrong with my logic/how a = 32? I could be re arranging the question to a point where the logic breaks down, but I'm just not sure how a= 32.

So, you understand that a must be 2. Then, "The remainder when $$a$$ is divided by 69 is the fifth power of a prime number" gives $$a=69q+32$$: 32, 101, 170, ... Since $$a$$ is a two-digit integer, then $$a=32$$.

Hope it's clear.

Bunuel,

Thanks for clarifying that bit; that's a huge help! Overall, I think this is a great question.
Re: M30-18   [#permalink] 19 Jan 2019, 10:41
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# M30-18

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