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If 1 book was left when a pile of books was stacked in rows
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07 Aug 2009, 16:04
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If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all? (1) When the books were stacked in rows of 7, 1 book was left (2) When the books were stacked in rows of 8, 4 books were left M2329
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Re: books
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07 Aug 2009, 16:24
E for me.
From original st, # of books would be x, where x = 5(n) +1, where n is a positive integer.
x could be 6, 11, 16, 21......36, 71, 76
St.1: Number of books would be: x = 7(n) +1, where n is a positive integer. Now, x could be: 8, 15....36...71 INSUFF
St.2: Number of books would be: x = 8(n) +4, where n is a positive integer. Now, x could be: 12 15....36...68, 76 INSUFF
Combined: x could be 36 or 76, and it will satisfy all the conditions. So E.
PS. If there was a upper bound, something like the total # of books is less than 50 or there are less than 10 rows in the bookshelf...or something on those lines.



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Re: books
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07 Aug 2009, 16:39
E for me as well. we have two equations but 3 unknown variables, insufficient data to solve it:
5x + 1 = 7y + 1 8z + 4 = 5x + 1



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Re: books
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27 Mar 2011, 03:28
Jivana wrote: E for me.
From original st, # of books would be x, where x = 5(n) +1, where n is a positive integer.
x could be 6, 11, 16, 21......36, 71, 76
St.1: Number of books would be: x = 7(n) +1, where n is a positive integer. Now, x could be: 8, 15....36...71 INSUFF
St.2: Number of books would be: x = 8(n) +4, where n is a positive integer. Now, x could be: 12 15....36...68, 76 INSUFF
Combined: x could be 36 or 76, and it will satisfy all the conditions. So E.
PS. If there was a upper bound, something like the total # of books is less than 50 or there are less than 10 rows in the bookshelf...or something on those lines. i think there is a mistake since 76:7 remainder is not 1 as per statement 1 So why not C?



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Re: books
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27 Mar 2011, 03:55
tinki wrote: Jivana wrote: E for me.
From original st, # of books would be x, where x = 5(n) +1, where n is a positive integer.
x could be 6, 11, 16, 21......36, 71, 76
St.1: Number of books would be: x = 7(n) +1, where n is a positive integer. Now, x could be: 8, 15....36...71 INSUFF
St.2: Number of books would be: x = 8(n) +4, where n is a positive integer. Now, x could be: 12 15....36...68, 76 INSUFF
Combined: x could be 36 or 76, and it will satisfy all the conditions. So E.
PS. If there was a upper bound, something like the total # of books is less than 50 or there are less than 10 rows in the bookshelf...or something on those lines. i think there is a mistake since 76:7 remainder is not 1 as per statement 1 So why not C? There is a mistake in the statement but conclusion is still correct Clearly, individual statements are not enough. We are looking for a number that leaves remainder 1 when divided by 5 and 7 and remainder 4 when divided by 8. First such number is 36 and which is of the form \(35*k+1\) with \(k =1\). Now, for all positive integer k, this number will leave remainder of 1 when divided by 5 or 7. If we can find 2 k's for which the remainder is 4 on division by 8, then answer would be E. Since \(k=1\) works, lets see if \(k=1+8\) works as well, \(35*9+1 = 316\) which has remainder 1 when divided by 5 and 7 and remainder 4 when divided by 8. So, we have found two numbers that satisfy given conditions, so insufficient. In fact, all numbers of form \(35k+1\) with \(k = 1,8,17,25\) etc. should satisfy this condition. Answer is indeed E.



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Re: books
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27 Mar 2011, 04:14
I'm also getting E. B = 5x + 1 = 6, 11, 16, 21, 26, 31, 36 From (1), B = 7x + 1 = 8, 15, 22, 29, 36 So B = 35x + 36 = 36, 71, 106, not sufficient From (2) B = 8x + 4 = 12, 20, 28, 36 B = 40x + 36 = 36, 116 , not sufficient From (1) and (2) 36 is a common choice but LCM of 35 and 40 = 7 * 8 * 5 = 280 B = 280x + 36, so again we have multiple choices as 36, 316 ... Answer  E
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Re: books
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27 Mar 2011, 04:42
noboru wrote: If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
a When the books were stacked in rows of 7, 1 book was left b When the books were stacked in rows of 8, 4 books were left
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient
3 days to go!!!!!!!! Let the number of books be \(x\) According to the stem; \(x = 5K+1\), where K is some integer Possible values: 6,11,16,21,26,31, 36,41 1. \(x = 7L+1\), where L is some integer Possible values: 8,15,22,29, 36,48 Find the LCM of the divisors: \(LCM(5,7) = 35\) From the stem and statement 1, we know that the first common digit that satisfies the condition is 36. Hence, \(x = 35M+36\) Thus, possible values of x: \(M=0; x=35*0+36=36\) \(M=1; x=35*1+36=71\) \(M=2; x=35*2+36=106\) \(M=3; x=35*3+36=141\) Not Sufficient. 2. \(x = 8N+4\), where N is some integer Possible values: 12,20,28, 36,44 Find the LCM of the divisors: \(LCM(5,8) = 40\) From the stem and statement 1, we know that the first common digit that satisfies the condition is 36. Hence, \(x = 40Q+36\) Thus, possible values of x: \(Q=0; x=40*0+36=36\) \(Q=1; x=40*1+36=76\) \(Q=2; x=40*2+36=116\) \(Q=3; x=40*3+36=156\) Not Sufficient. Combining 1 and 2: \(x = 35M+36\) \(x = 40Q+36\) \(35M+36 = 40Q+36\) \(35M=40Q\) \(7M=8Q\) Which has infinite solutions, three of which follows \(M=8, Q=7; x= 35M+36 = 35*8+36 = 316\) \(M=8*2=16, Q=7*2=14; x= 35M+36 = 35*16+36 = 596\) \(M=8*3=24, Q=7*3=21; x= 40Q+36 = 40*21+36 = 876\) Not Sufficient. Ans: "E"
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If 1 book was left when a pile of books was stacked in rows of 5
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02 Feb 2015, 21:41
Responding to a pm: Quote: If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
(1) When the books were stacked in rows of 7, 1 book was left
(2) When the books were stacked in rows of 8, 4 books were left
How I can know that there are addition numbers more than 36 that left remaining 1 when divided by 5 and 7 and left
remaining 2 when divided by 8. I mean how I can get 316 by fast way and under 2 minutes.
Explanation: Here, you cannot get the number of books even with the information given in both the statements. This is so because there will be infinite numbers satisfying these conditions. Before you read the explanation, you should check out these posts discussing divisibility and remainders to really understand the explanation. http://www.veritasprep.com/blog/2011/04 ... unraveled/http://www.veritasprep.com/blog/2011/04 ... yapplied/http://www.veritasprep.com/blog/2011/05 ... emainders/http://www.veritasprep.com/blog/2011/05 ... spartii/Say, total number of books is N. If 1 book was left when a pile of books was stacked in rows of 5, N = 5a + 1 (1) When the books were stacked in rows of 7, 1 book was left N = 7b + 1 Using the two expressions for N given above, we know that N must be of the form 5*7c+1 = 35c + 1 (2) When the books were stacked in rows of 8, 4 books were left N = 8d + 4 Now consider these two: N = 35c + 1 N = 8d + 4 If c = 1, N = 36. 36 is of the form 8d + 4 if d = 4. Hence the smallest value that satisfies all three conditions is 36. There will be many more such values given by 35*8e + 36 = 280e + 36 So N can be 36 or 316 (if e = 1) or 586 (if e = 2) etc. Note that you don't really need to do these calculations. You know that you will get a value for N and if you get one value, you will get infinite values. Hence answer must be (E).
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Re: If 1 book was left when a pile of books was stacked in rows
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04 Apr 2018, 17:50
noboru wrote: If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
(1) When the books were stacked in rows of 7, 1 book was left
(2) When the books were stacked in rows of 8, 4 books were left We can create the equation: Number of books = 5Q + 1 where Q is a positive integer So the numbers of books can be: 6, 11, 16, 21, 26, 31, 36, … Statement One Alone: When the books were stacked in rows of 7, 1 book was left. So the numbers of books can be: 8, 15, 22, 29, 36, … We see that we could have 36 books. We can also add the LCM of 5 and 7, which is 35, to 36 to obtain 71 books. Both 36 books and 71 books will have 1 book left when they were stacked in rows of 5 or 7. Statement one alone is not sufficient to answer the question. Statement Two Alone: When the books were stacked in rows of 8, 4 books were left. So the numbers of books can be: 12, 20, 28, 36, … We see that we could have 36 books. We can also add the LCM of 5 and 8, which is 40, to 36 to obtain 76 books. Both 36 books and 76 books will have 1 book left when they were stacked in rows of 5 and 4 books left when they were stacked in rows of 8. Statement two alone is not sufficient to answer the question. Statements One and Two Together: Again, we see that we could have 36 books. We can also add the LCM of 5, 7 and 8, which is 280, to 36 to obtain 316 books. Both 36 books and 316 books will have 1 book left when they were stacked in rows of 5 or 7, and 4 books left when they were stacked in rows of 8. Both statements together are not sufficient to answer the question. Answer: E
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