Divisibility and Primes - Advanced Strategy pg 124 4th ed

[kapalzazzner]

Hi,

The question is:
If K^(3) is divisible by 240, what is the least possible value of integer k?

The multiple choice answers are 12, 30, 60, 90, and 120.

When I read the explanation of why the answer is C (60), it seems to me that they are solving for what is the MOST possible value of integer K versus the LEAST, which the question asks for. Factoring down 240 gives one 2^(4) * 3 * 5. The only answer choice that does not contain the factor 5 is 12, so I thought it would be 12. Am I reading this incorrectly? What is another way to think about this problem? Stepping through the book's explanation was not helpful.

Given the explanation, it would seem 60 and 120 are the MOST probable values for integer K (containing the factors 2^(2), 3, and 5).

Regards,
Andrew

[vikrant.bapat]

The question asks for the smallest number whose cube is divisible by 240. , so to be stated more directly, what is smallest k, if k*k*k/240 = integer. Factoring 240 would definitely make our job easier. After factoring, we need to find out the smallest number which would cancel out ALL the factors of 240 (2*2*2*3*2*5)

If we consider 12, 12*12*12 => this would not be divisible by 240, since there is no factor in 12^(3) which would be divisible by 5.

if we consider 30, (30*30*30)/(2*2*2*3*2*5) => now, in this case, we would be able to take care of 2*5, 3 would be canceled, and 2*2 would be canceled, leaving us with 15*15/2 (Try it out)

With 60, we would be able to cancel out the whole denominator(240) easily. Hence 60 wins!

Hope that helps!

[Ben Ku]

Hi,

The question is:
If K^(3) is divisible by 240, what is the least possible value of integer k?

The multiple choice answers are 12, 30, 60, 90, and 120.

When I read the explanation of why the answer is C (60), it seems to me that they are solving for what is the MOST possible value of integer K versus the LEAST, which the question asks for. Factoring down 240 gives one 2^(4) * 3 * 5. The only answer choice that does not contain the factor 5 is 12, so I thought it would be 12. Am I reading this incorrectly? What is another way to think about this problem? Stepping through the book's explanation was not helpful.

Given the explanation, it would seem 60 and 120 are the MOST probable values for integer K (containing the factors 2^(2), 3, and 5).

Regards,
Andrew

Let's examine general ideas about cubes. If x = 3*5, then x^3 = 3*3*3*5*5*5 or 3^3 * 5^3. Note that there are three of each prime factor.

We can go backward. If x^3 is divisible by 2*2*2*3, we can notice that to complete the factors, x^3 must also be divisible by 2*2*2*3*3*3, so x is divisible by 2*3.

If k^3 is divisible by 240, which is 2*2*2*2*3*5 or 2^3 * 2 * 3 * 5, then it is also divisible by 2^3 * 2^3 * 3^3 * 5^2. Therefore, k is divisible by 2*2*3*5 = 60.

[kapalzazzner]

Ok - so I misread the question is the bottom line. They are asking for the SMALLEST possible value versus the MOST UNLIKELY value. I read the phrase LEAST POSSIBLE as MOST UNLIKELY.

-Andrew

[tim]

Yep, super important to read the questions carefully. :) Actually, if you turn this into a prime boxes question it will be a lot easier and you don't run nearly as great a risk of misinterpreting the question..

-Tim

[khushbumerchant]

I have gone through the explanation & have understood.

Can someone help me with an example of similar problem, to understand the concept better?

Thanks in advance!

[tim]

Sure; just change the k^3 to k^2 in the original problem and try it that way! Or change the 240 to a different number. You'll find that you can get a lot of extra practice on some math problems by changing part of the problem and redoing the problem.