M4maths Previous Puzzles

05 Apr 2018 Factors

How many factors of 16757044128 are even numbers?

OPtion

1) 360
2) 300
3) 120
4) 90
5) 150
6) 540
7) 180
8) 480
9) 125
10) None of these

Solution

16757044128= (2^5)*(3^4)*(11^2)*(23^2)*101

Any factor of this number should be of the form 23^a * 11^b * 3^c * 2^d * 101^e

For the factors of this number to be an even number d should be non 0

a can take values 0, 1, 2
b can take values 0, 1, 2
c can take values 0, 1, 2, 3, 4
d can take values 1, 2, 3, 4, 5
e can take value 0, 1

Total number of odd factors = 3*3*5*5*2= 450

Option 10)

Correct Option: 10 Best Solution (5)

Varun   7 years AGO

1675704413 = 2^5 * 3^4 * 11^2 * 23^2 *101
=> Total number of odd factors = (4+1)*(2+1)*(2+1)*(1+1) = 5*3*3*2 = 90

Total number of factors = 90*(5+1) = 540

Therefore number of odd factors = 540-90 = 450
= Option 10)

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PRABHAT KUMAR   7 years AGO

Factor=32*81*121*529*101
No of even factor=5*5*3*3*2=450
Option 10

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Akshay   7 years AGO

167557044128=(2^5)*(3^4)*(11^2)*(23^2)*(101)
total fators=(5+1-1)*(4+1)*(2+1)*(2+1)(1+1)
=5*5*3*3*2=450
we use -1
for 2^0

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ruby   7 years AGO

16757044128=2^5*3^4*11^2*23^2*101
factors shld b in d form of
3^a*11^b*23^c*101^d*2^e
values for a,b,c,d,e
a= 0,1,2,3,4
b=0,1,2
c=0,1,2
d=0,1
e=1,2,3,4,5
Total number of Even factors = 5*3*3*2*5 = 450
hence option 10

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Mukund Manohar   7 years AGO

16757044128=2^5*3^4*11^2*23^2*101^1
2 should have a non zero power.
So, No. of even factors = 5*5*3*3*2 = 450

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