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If \(a\) and \(b\) are twodigit 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
Official Solution:If \(a\) and \(b\) are twodigit 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 twodigit 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 twodigit 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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09 Jul 2016, 03:09
DAMN!! how m I suppose to solve this within 2 minutes..



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03 Aug 2016, 23:28
I think this is a highquality question and I agree with explanation. Great question this has to be classified a "800level" problem lol!



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15 Aug 2016, 22:21
I think this is a highquality question and I agree with explanation.
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22 Sep 2016, 06:25
This is a very highquality 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: ifaandbaretwodigitpositiveintegersgreaterthan172886.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 highquality question and I agree with explanation.



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25 Sep 2018, 02:23
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, 23:57
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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16 Oct 2018, 00:01
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. Please reread the question and solution more carefully.
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16 Oct 2018, 00:06
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, 05:18
god level question under test circumstance!



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18 Jan 2019, 11:36
Bunuel wrote: Official Solution:
If \(a\) and \(b\) are twodigit 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 twodigit 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 twodigit 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 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, 01:12
bcal95 wrote: Bunuel wrote: Official Solution:
If \(a\) and \(b\) are twodigit 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 twodigit 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 twodigit 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 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 twodigit integer, then \(a=32\). Hope it's clear.
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19 Jan 2019, 11:41
Bunuel wrote: bcal95 wrote: Bunuel wrote: Official Solution:
If \(a\) and \(b\) are twodigit 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 twodigit 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 twodigit 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 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 twodigit 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.







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