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Joined: 02 Sep 2009
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Re M2135 [#permalink]
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16 Sep 2014, 01:15
Official Solution: (1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient. (2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient. (1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient. Answer: E
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Re: M2135 [#permalink]
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03 Jul 2015, 22:40
Bunuel wrote: Official Solution:
(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient. (2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient. (1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.
Answer: E Thanks for the explanation Bunuel. I too used the same method. However, I was wondering if there is any faster method to determine the two numbers (13 and 41). In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17 and for (2), this will be 6,13, 20, 27, 34. Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer. Any suggestions?



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Re: M2135 [#permalink]
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03 Jul 2015, 23:03
samuraijack256 wrote: Bunuel wrote: Official Solution:
(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient. (2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient. (1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.
Answer: E Thanks for the explanation Bunuel. I too used the same method. However, I was wondering if there is any faster method to determine the two numbers (13 and 41). In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17 and for (2), this will be 6,13, 20, 27, 34. Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer. Any suggestions? Hi, the fatser method in these Q is to find the first common multiple and thereafter add the LCM of two numbers.. in this case when we have got the fiest number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41.. the next 41+28=69 and so on... so 13 gives remainder 1 41 gives remainder 2 69 gives remainder 0... and so on hope it helped..
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Re: M2135 [#permalink]
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03 Jul 2015, 23:10
samuraijack256 wrote: Bunuel wrote: Official Solution:
(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient. (2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient. (1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.
Answer: E Thanks for the explanation Bunuel. I too used the same method. However, I was wondering if there is any faster method to determine the two numbers (13 and 41). In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17 and for (2), this will be 6,13, 20, 27, 34. Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer. Any suggestions? another way would be to equate two eq from two statements.. \(x=4q+1\), \(x=7p+6\)... \(4q+1=7p+6\).... \(4q=7p+5\).... now substitute p as 1,3,5... as p has to be odd to make the eq even .. where we getan integer value of q... those are the numbers we rae looking for.. for ex p=1,q=3... so number=4q+1=13.. p=3, q is not an int.. p=5, q=10 number =4q+1=41.. and so on
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Re: M2135 [#permalink]
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03 Jul 2015, 23:22
chetan2u wrote: samuraijack256 wrote: Bunuel wrote: Official Solution:
(1) The remainder when \(x\) is divided by 4 is 1. Given: \(x=4q+1\), so \(x\) cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient. (2) The remainder when \(x\) is divided by 7 is 6. Given: \(x=7p+6\), so \(x\) cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient. (1)+(2) If \(x=13\) then the remainder is 1 but if \(x=41\) then the remainder is 2. Not sufficient.
Answer: E Thanks for the explanation Bunuel. I too used the same method. However, I was wondering if there is any faster method to determine the two numbers (13 and 41). In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17 and for (2), this will be 6,13, 20, 27, 34. Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer. Any suggestions? Hi, the fatser method in these Q is to find the first common multiple and thereafter add the LCM of two numbers.. in this case when we have got the fiest number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41.. the next 41+28=69 and so on... so 13 gives remainder 1 41 gives remainder 2 69 gives remainder 0... and so on hope it helped.. This is exactly what i had been looking for. Thanks Chetan! Earlier, I too had tried looking at this from an LCM perspective, but after a few steps, I got confused and let it go. I looked through your other proposed method as well (the one involving equating the two expressions), but I would prefer your first suggestion for the ease of application ( identifying if the variable has to be even/odd will again take a few 10s of seconds ).



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Re: M2135 [#permalink]
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22 Nov 2015, 07:15
am i correct that as long as no limit is set on how far we can go calculating the potential values of n the answer will always be E? e.g. if we were set some limit: say 2 digit numbers or even etc  then we need to be accurate and find the values?
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Re: M2135 [#permalink]
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14 Oct 2016, 11:59
Thinking in retrospect how the time for this question could have been reduced. Clearly either statement is not sufficient on its own, which can be quickly tested.. Now for together: x = LCM of (7,4) * (some variable integer K) + some constant remainder, lets say R => x = 28*K + R We can find R by finding the common number as shown in the official solution. Now remainder of 28K/3 will depend on the value of K, since 28 is not a multiple of 3. So no matter what the value of R is, when we divide x by 3, the reminder obtained will depend on K hence we do not have a unique solution. Bunuel  do you think it is valid line of reasoning. If yes, then we would not need to find R and can save some time.



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Re M2135 [#permalink]
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28 Aug 2017, 09:34
I think this is a highquality question and I agree with explanation.










