How many factors of 72^2 are multiples of 6?

Question

A. 35
B. 25
C. 30
D. 24
E. 12

exc4libur · Accepted Answer

prime factorization of  (72)^2=(36•2)^2=(6^2•2)^2=(2^23^2•2)^2=2^63^4 
num factors of an integer is the product of the "power+1" of its prime factors:
 72^2=2^63^4…6=2•3…m(6)=72^2=(2•3)(2^53^3)…num.factors=(2^53^3)=(5+1)(3+1)=24

Answer (D)

chetan2u · Answer

Let us factorization 72^2..
 72^2=(2^33^2)^2=2^63^4 
Number of factors = (6+1)(4+1)=7*5=35
Now 6 is 2*3, so just multiple of just 2 and 3 are (6+1) and (4+1), so total 7+5=12..
But in each case, we have added factor 1 twice, once with 2 and once with 3..
So 12-1=11...
Number of factors that are multiples of 6 are 35-11=24..

D

07tiger · Answer

Thank you. I was struggling to wrap my head around this one.

SWAPNILP · Answer

Hi chetan 2u

kindly explain this concept 
"But in each case, we have added factor 1 twice, once with 2 and once with 3..So 12-1=11..." I am not able to understand

ashu2107 · Answer

Hi  Bunuel ,
Could you please explain the concept and solution for this question.

chetan2u · Answer

Hi 
what are the factors of 2^6..
Total factors are (6+1) and they are  1,2,2^2,2^3,2^4,2^5,2^6 
What about factors of 3^4...they are (4+1) and they are  1,3,3^2,3^3,3^4 
So total are 12 but 1 has got repeated twice so cancel out one of them, hence 12-1=11

Hope it helps you.

imants12 · Answer

My 2 cents please...
There are four distinct numbers with powers of three here, ie.,3^1,3^2,3^3 & 3^4. Now there are six distinct numbers with power of two, ie.,2^1,2^2,2^3,2^4,2^5 & 2^6. Only these combinations produce multiples of 6 : 4×6 = 24

Posted from my mobile device

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How many factors of 72^2 are multiples of 6?

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How many factors of 72^2 are multiples of 6?

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