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Re M2329 [#permalink]
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16 Sep 2014, 01:19

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Re: M2329 [#permalink]
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27 May 2015, 22:40
Hello,
Please explain stmt. 2 in detail  Is it possible to get more than one solution for stmt.2?
Regards, Mahuya
Statements (1) and (2) combined are insufficient. There is no limitation on the number of books. Without this limitation, the question cannot be answered: there is an infinite number of integers that satisfy the setup and the statements.
Answer: E[/quote]

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Re: M2329 [#permalink]
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05 Jul 2015, 17:44
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mahuya78Basically the question is saying that X (being the number of books) divided by 5 gives you remainder of 1. Question is, can we determine X? A) X/7 gives us remainder of 1. So X can be: 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99 etc (just keep adding 7 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 71. So insufficient. B) X/8 gives us remainder of 4. So X can be: 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108 etc (again, just keep adding 8 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 76. So again, insufficient. C) Apparently there seems to be only one option: 36, which would make you believe you have enough information. But the truth is that if you keep on trying, you will find other options that will fit both restrictions (for instance: 316), making both together insufficient. Since testing all these numbers is somewhat impossible in the given time, I recommend you to learn this type of question so whenever you are asked, you know how to tackle it. Cheers!

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@michaelyb wrote: @mahuya78Basically the question is saying that X (being the number of books) divided by 5 gives you remainder of 1. Question is, can we determine X? A) X/7 gives us remainder of 1. So X can be: 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99 etc (just keep adding 7 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 71. So insufficient. B) X/8 gives us remainder of 4. So X can be: 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108 etc (again, just keep adding 8 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 76. So again, insufficient. C) Apparently there seems to be only one option: 36, which would make you believe you have enough information. But the truth is that if you keep on trying, you will find other options that will fit both restrictions (for instance: 316), making both together insufficient. Since testing all these numbers is somewhat impossible in the given time, I recommend you to learn this type of question so whenever you are asked, you know how to tackle it. Cheers! Hi, you have missed lot many numbers in between ... after 36, the next will not be 316.. the way to find the next number after first number is .. first number +LCM of the two numbers.. in this case 36 + LCM of 7 & 8= 36 + 56 = 92.. and next 92 + 56 = 148 and so on..
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17 Oct 2015, 15:40
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chetan2uYou forgot to include 5 in your LCM calculation since X divided by 5 gives you a remainder of 1 according to the question stem, and 148 does not. So the next number after 36 is indeed 36 + the LCM of 7, 8 and 5, or 36+280 .

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Given: 5x+1 > 1,6,11,16... S1) 7x+1 Therefore the combine sequence will be x+35 Let's start with 7x+1, here LCM(7,5)=35 gives the repeating term "35". Now we need to figure out variable term. 7x+1 x=0 => 1 => Divide this output of 1 [7*0+1 = 1] by 5 => 1/5 => remainder(1). Note that this remainder 1 matches with the repeating term of original sequence = 5x+1 That's how you know that we have received the variable term of the merged sequence => x + 35 Now common terms => x+35 => 35,36,37,38,39,40,41..... 5x+1 => 5*6+1=36, 5*8+1=41 Not sufficient
S2) 8x+4 X=4 => 32+4 = 36 => 36/5 = r(1) => variable term = 36 Merged sequence = 36x+40 NOTE:initially you will take some time to figure out this merged sequence but with practice you will get better. I prefer this method because in using simply number from the sequence many times i miss some and am not able to get correct results as to if statement is sufficient or not. 36x+40 => 40,76,112,148,184,220,256 5x+1 > 5*15+1=76, & 5*51+1=256 Not sufficient
S(1+2): 5x+1 & 7x+1 & 8x+4 => common terms between x+35 & 36x+40 36x+40 = (35x+6)+(x+35) Whenever, we have something as this => one sequence can be made part of other => there are ALWAYS common terms.
Also, x+35 is a sequence of adding 1 to all the numbers after 35 => 36,37,38... So we can definitely say that 40, 76 and other terms of 36x+40 are also a part of x+35. So not sufficient. E
Last edited by manishtank1988 on 01 Apr 2017, 09:05, edited 1 time in total.

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Re: M2329 [#permalink]
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01 Apr 2017, 09:01
Hello moderators, Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahnaPlease let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated. Thanks

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Re: M2329 [#permalink]
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01 Apr 2017, 09:22
manishtank1988 wrote: Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated. Thanks Hey, the best way to solve such questions is as follows: Question is saying N = 5k + 1; Number could be 1,6,... Statement 1: N = 7k+1 If I want to find the common numbers, I will use First Common Number + LCM method. So, N = 1, 1st Term+(LCM of 5,7), 2nd Term + (LCM of 5,7), and so on. or N = 1, 36,72. Since, we can multiple values of N. It means Insufficient. Statement 2: N = 8k+4 First common N = 36. Next term = 36 + LCM(5,8), and so on. or N = 36 + 40 = 76 So, we have more than 1 common values, hence, Insufficient. Combining: N = 7k+1 and N = 8k'+4 First common = 36. Second = 36 + LCM(7,8)=92. Hence, again more than 1. Insufficient. Hence, E
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Re: M2329 [#permalink]
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02 Apr 2017, 10:15
abhimahna wrote: manishtank1988 wrote: Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated. Thanks Hey, the best way to solve such questions is as follows: Question is saying N = 5k + 1; Number could be 1,6,... Statement 1: N = 7k+1 If I want to find the common numbers, I will use First Common Number + LCM method. So, N = 1, 1st Term+(LCM of 5,7), 2nd Term + (LCM of 5,7), and so on. or N = 1, 36,72. Since, we can multiple values of N. It means Insufficient. Statement 2: N = 8k+4 First common N = 36. Next term = 36 + LCM(5,8), and so on. or N = 36 + 40 = 76 So, we have more than 1 common values, hence, Insufficient. Combining: N = 7k+1 and N = 8k'+4 First common = 36. Second = 36 + LCM(7,8)=92. Hence, again more than 1. Insufficient. Hence, E Thanks abhimahna

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Re: M2329 [#permalink]
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03 Apr 2017, 07:14
manishtank1988 wrote: Hello moderators, Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahnaPlease let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated. Thanks Check out these divisibility posts: https://www.veritasprep.com/blog/2011/0 ... unraveled/https://www.veritasprep.com/blog/2011/0 ... yapplied/https://www.veritasprep.com/blog/2011/0 ... emainders/https://www.veritasprep.com/blog/2011/0 ... spartii/Now the question can be done in seconds. Data in the Question stem: N = (5a + 1) or (5a'  4) Stmnt 1: N = (7b + 1) or (7b'  6) Using Stmnt 1 alone (and data in the question stem), The common remainder is 1 so N = 35m + 1. (because 35 is the LCM of 5 and 7) There will be infinite values of N such as 1, 36, 71 etc Stmnt 2: N = (8c + 4) or (8c'  4) Using Stmnt 2 alone (and data in the question stem), The common remainder is 4 so N = 40p  4 (because 40 is the LCM of 5 and 8) There will be infinite values of N such as 36, 76 etc Using both stmnts, we see that the first common value is 36. So N = 280q + 36 (because 280 is the LCM of 35 and 40). There will be infinite numbers of this form. Answer (E)
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Re: M2329 [#permalink]
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16 Apr 2017, 18:56
VeritasPrepKarishma wrote: manishtank1988 wrote: Hello moderators, Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahnaPlease let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated. Thanks Check out these divisibility posts: https://www.veritasprep.com/blog/2011/0 ... unraveled/https://www.veritasprep.com/blog/2011/0 ... yapplied/https://www.veritasprep.com/blog/2011/0 ... emainders/https://www.veritasprep.com/blog/2011/0 ... spartii/Now the question can be done in seconds. Data in the Question stem: N = (5a + 1) or (5a'  4) Stmnt 1: N = (7b + 1) or (7b'  6) Using Stmnt 1 alone (and data in the question stem), The common remainder is 1 so N = 35m + 1. (because 35 is the LCM of 5 and 7) There will be infinite values of N such as 1, 36, 71 etc Stmnt 2: N = (8c + 4) or (8c'  4) Using Stmnt 2 alone (and data in the question stem), The common remainder is 4 so N = 40p  4 (because 40 is the LCM of 5 and 8) There will be infinite values of N such as 36, 76 etc Using both stmnts, we see that the first common value is 36. So N = 280q + 36 (because 280 is the LCM of 35 and 40). There will be infinite numbers of this form. Answer (E) Great solution: VeritasPrepKarishmaThanks a lot

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