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16 Sep 2014, 01:19
Kudos
Bookmarks
If a stack of books was arranged into rows of 5 and only 1 book was left over, how many total books were there in the stack?
(1) When the books were arranged into rows of 7, there was 1 book left over.
(2) When the books were arranged into rows of 8, there were 4 books left over.
(1) When the books were arranged into rows of 7, there was 1 book left over.
(2) When the books were arranged into rows of 8, there were 4 books left over.
16 Sep 2014, 01:19
Kudos
Bookmarks
Official Solution:
If a stack of books was arranged into rows of 5 and only 1 book was left over, how many total books were there in the stack?
The condition where a stack of books is arranged into rows of 5 and only 1 book is left over can be represented as \(total = 5q + 1\). So, the total number of books is 1 more than a multiple of 5, thus the total could be 1, 6, 11, 16, 21, 26, 31, 36, and so on.
(1) When the books were arranged into rows of 7, there was 1 book left over.
The above can be represented as \(total = 7a + 1\). Hence, the total number of books is 1 more than a multiple of 7, so the total could be 1, 8, 15, 22, 36, and so on.
We can derive a general formula for the total (of a type \(total=dx+r\), where \(d\) is the divisor and \(r\) is the remainder) based on the two formulas given: \(total = 5q + 1\) (1, 6, 11, 16, 21, 26, 31, 36, and so on) and \(total = 7a + 1\) (1, 8, 15, 22, 36, and so on).
The divisor \(d\) would be the least common multiple of the two divisors 5 and 7, hence \(d=35\) and the remainder \(r\) would be the first common integer in the two patterns, hence \(r=1\).
So, the general formula based on both pieces of information is \(total = 35x+1\). Thus, the total number of books is 1 more than a multiple of 35, so the total could be 1, 36, 71, 106, and so on.
Not sufficient.
(2) When the books were arranged into rows of 8, there were 4 books left over.
The above can be represented as \(total = 8b + 4\). Hence, the total number of books is 4 more than a multiple of 8, and it could be 4, 12, 20, 28, 36, and so on.
Similarly, we can derive one formula based on \(total = 5q + 1\) (1, 6, 11, 16, 21, 26, 31, 36, ...) and \(total = 8b + 4\) (4, 12, 20, 28, 36, .. ).
The divisor would be the least common multiple of the two divisors 5 and 8, hence 40, and the remainder would be the first common integer in the two patterns, hence 36.
So, the general formula based on both pieces of information is \(total = 40y+36\). Therefore, the total number of books is 36 more than a multiple of 40, and so the total could be 36, 76, 116, and so on.
Not sufficient.
(1)+(2) From (1), we have \(total = 35x+1\) (1, 36, 71, 106, ...) and from (2), we have \(total = 40y+36\) (36, 76, 116, ...). Therefore, our combined formula is \(total = 280z + 36\). So, the total number of books could be 36, 316, 596, and so on. Not sufficient.
The method above uses algebra to help us solve this problem. But it's also possible to solve it without any formula. Since the total number of books isn't limited, there will be infinitely many numbers which simultaneously satisfy all three conditions: being 1 more than a multiple of 5; being 1 more than a multiple of 7; and being 4 more than a multiple of 8. As shown above, there are infinitely many such numbers: 36, 316, 596, ...
Answer: E
If a stack of books was arranged into rows of 5 and only 1 book was left over, how many total books were there in the stack?
The condition where a stack of books is arranged into rows of 5 and only 1 book is left over can be represented as \(total = 5q + 1\). So, the total number of books is 1 more than a multiple of 5, thus the total could be 1, 6, 11, 16, 21, 26, 31, 36, and so on.
(1) When the books were arranged into rows of 7, there was 1 book left over.
The above can be represented as \(total = 7a + 1\). Hence, the total number of books is 1 more than a multiple of 7, so the total could be 1, 8, 15, 22, 36, and so on.
We can derive a general formula for the total (of a type \(total=dx+r\), where \(d\) is the divisor and \(r\) is the remainder) based on the two formulas given: \(total = 5q + 1\) (1, 6, 11, 16, 21, 26, 31, 36, and so on) and \(total = 7a + 1\) (1, 8, 15, 22, 36, and so on).
The divisor \(d\) would be the least common multiple of the two divisors 5 and 7, hence \(d=35\) and the remainder \(r\) would be the first common integer in the two patterns, hence \(r=1\).
So, the general formula based on both pieces of information is \(total = 35x+1\). Thus, the total number of books is 1 more than a multiple of 35, so the total could be 1, 36, 71, 106, and so on.
Not sufficient.
(2) When the books were arranged into rows of 8, there were 4 books left over.
The above can be represented as \(total = 8b + 4\). Hence, the total number of books is 4 more than a multiple of 8, and it could be 4, 12, 20, 28, 36, and so on.
Similarly, we can derive one formula based on \(total = 5q + 1\) (1, 6, 11, 16, 21, 26, 31, 36, ...) and \(total = 8b + 4\) (4, 12, 20, 28, 36, .. ).
The divisor would be the least common multiple of the two divisors 5 and 8, hence 40, and the remainder would be the first common integer in the two patterns, hence 36.
So, the general formula based on both pieces of information is \(total = 40y+36\). Therefore, the total number of books is 36 more than a multiple of 40, and so the total could be 36, 76, 116, and so on.
Not sufficient.
(1)+(2) From (1), we have \(total = 35x+1\) (1, 36, 71, 106, ...) and from (2), we have \(total = 40y+36\) (36, 76, 116, ...). Therefore, our combined formula is \(total = 280z + 36\). So, the total number of books could be 36, 316, 596, and so on. Not sufficient.
The method above uses algebra to help us solve this problem. But it's also possible to solve it without any formula. Since the total number of books isn't limited, there will be infinitely many numbers which simultaneously satisfy all three conditions: being 1 more than a multiple of 5; being 1 more than a multiple of 7; and being 4 more than a multiple of 8. As shown above, there are infinitely many such numbers: 36, 316, 596, ...
Answer: E
27 May 2015, 22:40
Kudos
Bookmarks
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]
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]
