rma26 wrote:
Bunuel wrote:
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.