How many positive odd integers are factors of 120^2?

Question

A. 4
B. 5
C. 9
D. 31
E. 32

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paragw · Answer

120^2 = 2^6 × 3^2 × 5^2.
Odd factors cannot include factor 2, so ignore all the powers of 2 and use only 3^2 × 5^2.
Number of odd factors = (2 + 1)(2 + 1) = 3 × 3 = 9
So the correct option is  C

PrafulLata · Answer

Answer is C

120^2 = (2^3 * 3 * 5)^2
 = 2^6 * 3^2 * 5^2
So, total number of factors = (6+1)*(2+1)*(2+1) = 63

But question is asking positive odd factors. So there should no be any multiple of 2.

Odd factors = 2^0 * 3^2 * 5^2
 = 1 * (2+1) * ( 2+1)
 = 1 * 3 * 3
 = 9

Dereno · Answer

Let’s first prime factorize the given number.

120 = 4*3*2*5 = (2^3 *3*5)

But, the number given is 120^2 =  ^2

= 2^6 * 3^2 * 5^2

The total number of factors = (6+1)*(2+1)*(2+1) = 7*3*3 = 63

we need to find the number of ODD factors. This can be done in two ways :

Method 1   :

2^6 has 7 factors. They are. (2^0, 2^1, 2^2 , 2^3, 2^4 , 2^5, 2^6) . Remember 1 is a factor of all numbers.

Similarly 3^2 has 3 factors. They are (3^0, 3^1, 3^2).

and, 5^2 has also 3 factors. They are (5^0, 5^1, 5^2).

Total even factors = any number multiplied by 2 or the power of 2, will give us even number.

Amongst the 7 factors of 2,  EXCEPT 2^0 , all other powers of 2 will yield a even number.

so, now we are left with 6 factors of 2.

2 ^ (greater than 0) when multiplied with 3^0 or 5^0 will yield a even number only.

Thus, number of even factors = 6*3*3 = 54 factors.

Odd factors = Total - number of even factors

= 63 - 54

=  9 factors

Option C

Method 2:

Number of ODD factors

= removing all powers of 2 which will give us a even power * powers of 3 * powers of 5

if we remove all powers or 2 which will give us even number, we will be left with 2^0 alone.

= 1* 3 * 3

=  9 factors.

Option C

rohitz98 · Answer

Given 120^2
Factorizing 2^6 x 5^2 x 3^2
Hence for odd factors considering only powers of odd numbers = (2+1) x (2+1) = 9  (Option C)

007Phoenix007 · Answer

120^2 = (4×5×6)^2
= (2^3×3×5)^2
So for finding odd integers we need to eliminate the multiples of 2
Therefore 
120^2= 2^0×3^2×5^2 (for odd factors)
Therefore no. of odd factors
= 1×3×3
=9

ExpertsGlobal5 · Answer

Video Explanation:

https://www.youtube.com/watch?v=C4mK5XvyAiU

How many positive odd integers are factors of 120^2?

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