If 1 book was left when a pile of books was stacked in rows

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.

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

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"

Responding to a pm:


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.


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).

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

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