If 2^5, 3^3, and 13^2 are all factors of the product of 936 and w wher

Question

If 2^5, 3^3, and 13^2 are all factors of the product of 936 and w where w is a positive integer, what is the smallest possible value of w?

A. 26 
B. 39 
C. 42 
D. 65 
E. 156

stonecold · Answer

Here 156 has three two's
 two three's 
and one 13
rest of them must be in w
so w= 13*3*4 = 156

Smash E

PhilSad92 · Answer

E) is the right answer, but don't you mean 936 in the first sentence?

936= 2^3 *3^2 *13^1

So what you need is 2 more 2's, 1 more 3 and 1 more 13.
This gives us 2^2 *3 *13=156 E)

Abhishek009 · Answer

Yup definitely its a TYPO

936 =  2^3 x  3^2 x  13^1

riyan · Answer

2^5, 3^3, and 13^2 are all factors of the product of 936 and w
936=2^33^213
So, we left with 2^2*3*13=156
Hence answer is E

MathRevolution · Answer

936 =  2^3 * 3^2  * 13

Factors are:  2^5, 3^3, 13^2

Therefore 936 * w =  2^5 * 3^3 * 13^2

=>  2^3 * 3^2  * 13 * w =  2^5 * 3^3 * 13^2

=> w =  2^2  * 3 * 13 = 156

Answer E

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