M4maths Previous Puzzles

29 Oct 2019 Even Number

How many factors of 42996970800 are even numbers?

OPtion

1) 750
2) 800
3) 640
4) 960
5) 580
6) 150
7) 125
8) 648
9) 240
10) None of these

Solution

42996970800 = 53^1 * 73^1 * 7^3 * 5^2 * 2^4 * 3^4

Any factor of this number should be of the form 53^a * 73^b * 7^c * 5^d * 2^e * 3^f.

For the factor to be an even number e should be non zero.

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

Total number of odd factors = 2*2*4*3*4*5 = 960

Option 4)

Correct Option: 4 Best Solution (2)

G.S.Kantharao   5 years AGO

42996970800=2^4*3^4*5^2*7^3*53*73
Total factors=(4+1)(4+1)(2+1)(3+1)(1+1)(1+1)=1200
Odd factors=5*3*4*2*2=240
Even factors=1200-240=960
My option is 4👈

Like? Yes (1) | No   

Varun   5 years AGO

Given = 2 2 2 2 3 3 3 3 5 5 7 7 53 73
So 960

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