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On page 28 of the 3rd edition number properties guide there is this sample question:

If k^3 is divisible by 240, what is the least possible value of interger k?
a) 12 b) 30 c)60 d)90 e) 120

The answer in the book is C. However in the explanation, the says that k has 2, 2, 3, and 5 in its prime box, so k must be a multiple of 60.

My question is if k must be a multiple of 60 then why is it the least possible value of interger k?

mizzzie wrote:On page 28 of the 3rd edition number properties guide there is this sample question:

If k^3 is divisible by 240, what is the least possible value of interger k?
a) 12 b) 30 c)60 d)90 e) 120

The answer in the book is C. However in the explanation, the says that k has 2, 2, 3, and 5 in its prime box, so k must be a multiple of 60.

My question is if k must be a multiple of 60 then why is it the least possible value of interger k?

Mizzie, since we know that K has all of the factors, 2,2, 3 and 5 the smallest number that can have these factors is 2x2x3x5 = 60.

Where did you get hold of your 3rd edition guide from?

Thanks for the reply praneeth. But I don't get why 12, 30, 90 or 120 would be more likely than 60 to be the integer?

I swapped the 3rd edition guides the past weekend at the Manhattan corporate office. I started on the Number Properties guide and this new edition covers more areas of the topic and has more practice questions at the end of each chapter.

mizzzie wrote:Thanks for the reply praneeth. But I don't get why 12, 30, 90 or 120 would be more likely than 60 to be the integer?

I swapped the 3rd edition guides the past weekend at the Manhattan corporate office. I started on the Number Properties guide and this new edition covers more areas of the topic and has more practice questions at the end of each chapter.

Start with the factors that we know of K^3, since we know that k^3 is divisible by 240, k^3 has the following prime factors (at a minimum): 2,2,2,2,3,5

If K^3 has at least four (2's) in its prime factorization k must have had at least two 2s in its prime factorization. (one would have only given three 2's)

If k^3 has a 3 in its prime factorization, k must have had at least one 3.

If k^3 has a 5 in its prime factorization, k must have had at least one 5.

Therefore we know that K must at least have the prime factors 2,2,3 and 5.

The smallest number that has all of these factors is the number that has just these 4 factors and nothing else, i.e. 60.

Hope this is helpful.

It makes more sense now. Thanks

Thanks for jumping in there, praneeth. Normally, this question would be more suited for the GMAT Math part of the forums, but I'm glad it got answered!

Best,