Divisibility: Advanced Strategy pg. 124

[omalkin20]

My question comes from Strategy Guide 1, pg. 124. The answer is not clear to me for the questions, "If k^3 is divisible by 240, what is the least possible value of integar k?". I picked (B) 30 because 30 contains the factors 2,3,5, which 240 also contains. How do we know there should be 2 2s in the prime box and not just 1? Thanks, appreciate a clearer explanation.

[mithunsam]

240 = 2^4 * 3 * 5

For k^3 to be a multiple of 240, k^3 should contain four 2s, one 3 and one 5.

if you take k = 2*3*5, then k^3 will be 2^3 * 3^3 * 5^3, which is not a multiple of 240 (still short of one 2).

[If you want to test it... 2^3 * 3^3 * 5^3 = (2^3 * 3 * 5) * 3^2 * 5^2 = (120) * 3^2 * 5^2]

Since we are short of 2, we have to increase the power of 2 in k. Now try k = 2^2 * 3 * 4

Now, k^3 = (2^2 * 3 * 4)^3 = 2^6 * 3^3 * 4^3, which satisfies the minimum criteria of four 2s, one 3 and one 5.

[If you want to try it... 2^6 * 3^3 * 4^3 = (2^4 * 3 * 5) * 2^2 * 3^2 * 5^2 = (240) * 2^2 * 3^2 * 5^2]

[jnelson0612]

mithunsam, once again, great job! Thanks!