M23-29

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
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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]

mahuya78

Basically 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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chetan2u

You 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 .

Hello moderators,
Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahna
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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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)

I think this is a high-quality question.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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