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First of all thanks for editing my question in an appropiate manner.

Secondly is clear. I should infer a lot on this question. I always have the doubt that these kind of questions are too convoluted for the exam.


Anyway, thanks :(
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great solution buenel..!!!

I didt consider while solving that Rectangular box must have integer value for its L,B &H.

I went unnecessarily on jumping to fractional values..!!

Poor of me...
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Bunuel
Jean puts N identical cubes, the sides of which are 1 inch long, inside a rectangular box, each side of which is longer than 1 inch, such that the box is completely filled with no gaps and no cubes left over. What is N?

Notice that, since the volume of each cube is 1 inch^3, then the volume of N cubes (the volume of the rectangular box) is N inch^3. For example if there are 10 cubes, then the volume of the rectangular box (total volume of 10 cubes) is 10 inch^3. Next, we are told that the length, the width and the height of the rectangular box is longer than 1 inch and there are no gaps when all cubes are put in the box, so the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... Thus each dimension of the rectangular box must have at least one prime in it, so the volume (length*width*height) must be the product of at least 3 primes (not necessarily distinct primes).

(1) 56 < N < 63. N could be 57, 58, 59, 60, 61, or 62. Analyze each case:

57=3*19 --> just two primes. Discard.
58=2*29 --> just two primes. Discard.
59 --> prime itself. Discard.
60=2^2*3*5 --> the product of 4 primes. OK. For example, the the length, the width and the height of the cube cold be 2, by 6, by 5.
61 --> prime itself. Discard.
62=2*31 --> just two primes. Discard.

As we can see, N can only be 60. Sufficient.

(2) N is a multiple of 3. Multiple values of N are possible so that it to be a multiple of 3 AND the product of at least 3 primes, for example 27 or 60. Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunnel,

The question says "each side of which is longer than 1 inch" so the lengths of each sides can even be "2, 2, 2, or 3,3,3" may i know why you took the lenghts as "2,3,4"

Regards
Srinath
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Bunuel
Jean puts N identical cubes, the sides of which are 1 inch long, inside a rectangular box, each side of which is longer than 1 inch, such that the box is completely filled with no gaps and no cubes left over. What is N?

Notice that, since the volume of each cube is 1 inch^3, then the volume of N cubes (the volume of the rectangular box) is N inch^3. For example if there are 10 cubes, then the volume of the rectangular box (total volume of 10 cubes) is 10 inch^3. Next, we are told that the length, the width and the height of the rectangular box is longer than 1 inch and there are no gaps when all cubes are put in the box, so the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... Thus each dimension of the rectangular box must have at least one prime in it, so the volume (length*width*height) must be the product of at least 3 primes (not necessarily distinct primes).

(1) 56 < N < 63. N could be 57, 58, 59, 60, 61, or 62. Analyze each case:

57=3*19 --> just two primes. Discard.
58=2*29 --> just two primes. Discard.
59 --> prime itself. Discard.
60=2^2*3*5 --> the product of 4 primes. OK. For example, the the length, the width and the height of the cube cold be 2, by 6, by 5.
61 --> prime itself. Discard.
62=2*31 --> just two primes. Discard.

As we can see, N can only be 60. Sufficient.

(2) N is a multiple of 3. Multiple values of N are possible so that it to be a multiple of 3 AND the product of at least 3 primes, for example 27 or 60. Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunnel,

The question says "each side of which is longer than 1 inch" so the lengths of each sides can even be "2, 2, 2, or 3,3,3" may i know why you took the lenghts as "2,3,4"

Regards
Srinath

I said that "the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... " So, yes, each dimension can be 2 or 3, for example .
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Bunuel
kotela
Bunuel
Jean puts N identical cubes, the sides of which are 1 inch long, inside a rectangular box, each side of which is longer than 1 inch, such that the box is completely filled with no gaps and no cubes left over. What is N?

Notice that, since the volume of each cube is 1 inch^3, then the volume of N cubes (the volume of the rectangular box) is N inch^3. For example if there are 10 cubes, then the volume of the rectangular box (total volume of 10 cubes) is 10 inch^3. Next, we are told that the length, the width and the height of the rectangular box is longer than 1 inch and there are no gaps when all cubes are put in the box, so the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... Thus each dimension of the rectangular box must have at least one prime in it, so the volume (length*width*height) must be the product of at least 3 primes (not necessarily distinct primes).

(1) 56 < N < 63. N could be 57, 58, 59, 60, 61, or 62. Analyze each case:

57=3*19 --> just two primes. Discard.
58=2*29 --> just two primes. Discard.
59 --> prime itself. Discard.
60=2^2*3*5 --> the product of 4 primes. OK. For example, the the length, the width and the height of the cube cold be 2, by 6, by 5.
61 --> prime itself. Discard.
62=2*31 --> just two primes. Discard.

As we can see, N can only be 60. Sufficient.

(2) N is a multiple of 3. Multiple values of N are possible so that it to be a multiple of 3 AND the product of at least 3 primes, for example 27 or 60. Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunnel,

The question says "each side of which is longer than 1 inch" so the lengths of each sides can even be "2, 2, 2, or 3,3,3" may i know why you took the lenghts as "2,3,4"

Regards
Srinath

I said that "the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... " So, yes, each dimension can be 2 or 3, for example .

Thanks Bunnel got it....
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Bunuel
Jean puts N identical cubes, the sides of which are 1 inch long, inside a rectangular box, each side of which is longer than 1 inch, such that the box is completely filled with no gaps and no cubes left over. What is N?

Notice that, since the volume of each cube is 1 inch^3, then the volume of N cubes (the volume of the rectangular box) is N inch^3. For example if there are 10 cubes, then the volume of the rectangular box (total volume of 10 cubes) is 10 inch^3. Next, we are told that the length, the width and the height of the rectangular box is longer than 1 inch and there are no gaps when all cubes are put in the box, so the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... Thus each dimension of the rectangular box must have at least one prime in it, so the volume (length*width*height) must be the product of at least 3 primes (not necessarily distinct primes).

(1) 56 < N < 63. N could be 57, 58, 59, 60, 61, or 62. Analyze each case:

57=3*19 --> just two primes. Discard.
58=2*29 --> just two primes. Discard.
59 --> prime itself. Discard.
60=2^2*3*5 --> the product of 4 primes. OK. For example, the the length, the width and the height of the cube cold be 2, by 6, by 5.
61 --> prime itself. Discard.
62=2*31 --> just two primes. Discard.

As we can see, N can only be 60. Sufficient.

(2) N is a multiple of 3. Multiple values of N are possible so that it to be a multiple of 3 AND the product of at least 3 primes, for example 27 or 60. Not sufficient.

Answer: A.

Hope it's clear.

I don't understand the highlighted part. Help me plz! How this part can be deduced??
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rma26
Bunuel
Jean puts N identical cubes, the sides of which are 1 inch long, inside a rectangular box, each side of which is longer than 1 inch, such that the box is completely filled with no gaps and no cubes left over. What is N?

Notice that, since the volume of each cube is 1 inch^3, then the volume of N cubes (the volume of the rectangular box) is N inch^3. For example if there are 10 cubes, then the volume of the rectangular box (total volume of 10 cubes) is 10 inch^3. Next, we are told that the length, the width and the height of the rectangular box is longer than 1 inch and there are no gaps when all cubes are put in the box, so the length, the width and the height of the rectangular box are integers more than one: 2, 3, 4, ... Thus each dimension of the rectangular box must have at least one prime in it, so the volume (length*width*height) must be the product of at least 3 primes (not necessarily distinct primes).

(1) 56 < N < 63. N could be 57, 58, 59, 60, 61, or 62. Analyze each case:

57=3*19 --> just two primes. Discard.
58=2*29 --> just two primes. Discard.
59 --> prime itself. Discard.
60=2^2*3*5 --> the product of 4 primes. OK. For example, the the length, the width and the height of the cube cold be 2, by 6, by 5.
61 --> prime itself. Discard.
62=2*31 --> just two primes. Discard.

As we can see, N can only be 60. Sufficient.

(2) N is a multiple of 3. Multiple values of N are possible so that it to be a multiple of 3 AND the product of at least 3 primes, for example 27 or 60. Not sufficient.

Answer: A.

Hope it's clear.

I don't understand the highlighted part. Help me plz! How this part can be deduced??

The question stem states that the sides of the rectangular box are greater than 1. This means the sides can be 2, or 3, or 4, etc.
So when we calculate the volume, it'll be 2*2*2, or 3*3*3, or 4*4*4, etc.
=> the volume of the rectangular box will be a multiple of 3 prime numbers. They could be the same prime number, or distinct prime numbers.

1) 56 < N < 63
All sides are greater than 1, so any value of N with 1 as a multiple will be discarded.
57 = 1*3*19 -> OUT
58 = 1*2*29 -> OUT
59 = 1*59 -> OUT
60 = \(2^2\)*3*5 -> Keep
61 = 1*61 -> OUT
62 = 1*2*31 -> OUT

N = 60. Sufficient.

2) N = 3a
N could be 27 = \(3^3\), or 60 = \(2^2\)*3*5
2 values, hence, insufficient.

A is the answer.
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Didn't get it! :-(

Oh! May be as 2 is the smallest prime, he said so.
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Didn't get it! :-(

Oh! May be as 2 is the smallest prime, he said so.

Primes are numbers that are divisible by 1, and itself.
And all numbers are made up of prime numbers.

4 = \(2^2\) -> made up of 1 prime number 2.
12 = \(2^2\)*3 -> made up of 2 prime numbers 2 and 3.

The question stem states that the rectangular box has side lengths >1.

Volume of box = l*b*h
=> l,b,h > 1
So any value l,b,h take will consist of prime numbers.

Does this help?
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Yes! I thought so ,as I wrote in my last post. Thank you!

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Why exactly primes? It can be anything...
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Let the sides of the Rectangle be L,B,H. These sides have to be integers since it is given that 1" cubes fill the rectangular box completely. ( This also implies that the Volume of rectangular Box = N, but this would not help much).

And each side is greater than 1",

Take Statement 1
Check for all possible values of N between 56 and 63.
L x B x H = 57 can be factored in = 1x 57 x 1 NP
3x 19x 1 NP
take others like the same as well and you will find only 60 is the one that satisfies the condition given in Question

Take statement 2
There can be many values which are multiples of 3 with sides greater than 1"
For Eg - 30 - 3 5 2
24 - 3 2 4
and others

So answer is A
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