How many factors of 1247465610 are odd numbers?

M4maths Previous Puzzles

30 Nov 2017 ODD Number

How many factors of 1247465610 are odd numbers?

OPtion

1) 24
2) 36
3) 120
4) 60
5) 48
6) 75
7) 180
8) 16
9) 90
10) None of these

Solution

1247465610 = 73^2 * 17^2 * 5^1 * 3^4 * 2^1

Any factor of this number should be of the form 73^a * 17^b * 5^c * 3^d * 2^e

For the factors of this number to be an odd number e should be 0

a can take values 0, 1, 2
b can take values 0, 1, 2
c can take values 0, 1
d can take values 0, 1, 2, 3, 4
and e can take values 0


Total number of odd factors = 3*3*2*5= 90

Option 9)

Correct Option: 9 Best Solution (6)

Dr.Dipin Singh   7 years AGO

1247465610= 2*5*(3^4)*(17^2)*(73^2)
Hence odd factors are 2*5*3*3=90

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Rajaboina Radhika   7 years AGO

1247465610=2*3*3*3*3*5*17*17*73*73
total no of factors=(1+1)*(4+1)(1+1)*(2+1)*(2+1)=180
toatl no of odd factors =(4+1)(1+1)(2+1)(2+1)=5*2*3*3=90
ans is option)9

Like? Yes (2) | No (1)  

Aerpula Swarnalatha   7 years AGO

Factors of 1247465610=2*3^4*5*17^2*73^2
Odd factors are 3,5,17, and 73
No.of Odd factors=(4+1)*(1+1)*(2+1)*(2+1)
=90
Hence the option is (9)

Like? Yes (2) | No (1)  

Varun   7 years AGO

1247465610 = 3^4 * 5 * 17^2 * 73^2 * 2
Odd factors are 3,5,17,73 that are repeated 4,1,2,2 times respectively,
So number of odd factors from them = (4+1)*(1+1)*(2+1)*(2+1) = Product of 5, 2, 3, 3, 3 = 90
= Option 9

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SAYAN CHATTERJEE   7 years AGO

1247465610= 5^1*2^1*17^2*3^4*73^2
So, no. of odd factors are (1+1)*(2+1)*(4+1)*(2+1) = 90.
Option 9.

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Prasanna   7 years AGO

1247465610=2*3*3*3*3*5*17*17*73*73
total no of factors=(1+1)(4+1)(1+1)(2+1)(2+1)=180
total no of odd factors=(4+1)(1+1)(2+1)(2+1)=5*2*3*3=90
hence ans is option:9

Like? Yes (1) | No (2)