Elitmus
Exam
Numerical Ability
Number System
What is the number of odd factors of 49000000?
Ans.21
Read Solution (Total 6)
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- For these types of questions:
First resolve given number into powers of prime factors(i.e 2*2*5**5.....)
Then:
to find number of even factors: ignore 2^0 and find product of (prime factor's powers+1)
to find number of odd factors: ignore all 2 powers except 2^0 and find product of (prime factor's power+1)
for 49000000= 7^2*2^6*5^6
ignore all 2powers except 2^0 i.e now 7^2*2^0*5^6
num of factors (as said above)= (2+1)(0+1)(6+1)=7*1*3=21 - 10 years agoHelpfull: Yes(43) No(0)
- 49000000 factors are 7^2 * 5^6 * 2^6
we need only odd factors
so only consider 7 & 5 combinations
is (2+1)*(6+1) = 21
(short cut of finding factor of any combination) - 10 years agoHelpfull: Yes(5) No(0)
- the factors of 49000000 is 7^2*5^6*2^6
so to have odd factors the ans would be given as
(2+1(6+1)=21
bcz to hav odd factors odd*odd=odd
therefore 2^6 is neglected - 10 years agoHelpfull: Yes(2) No(0)
- 49000000=7^2*2^6*5^6
total number of factors=(2+1)*(6+1)*(6+1)=7*7*3
now total number of even factors=6(occurance of 2)*(2+1)*(6+1);
hence odd factors=7*7*3-6*3*7=7*3(7-6)=21 ans. - 10 years agoHelpfull: Yes(1) No(0)
- 49000000
49*1000000
7^2*10^6
(2+1)*(6+1)=21 - 10 years agoHelpfull: Yes(1) No(0)
- I think ans will be 20
49000000=7*7*2*5*2*5*2*5*2*5*2*5*2*5
so odd factors are
5
5*5
5*5*5
5*5*5*5
5*5*5*5*5
5*5*5*5*5*5
5*7
5*5*7
5*5*5*7
5*5*5*5*7
5*5*5*5*5*7
5*5*5*5*5*5*7
5*7*7
5*5*7*7
5*5*5*7*7
5*5*5*5*7*7
5*5*5*5*5*7*7
5*5*5*5*5*5*7*7
7
7*7 - 10 years agoHelpfull: Yes(0) No(9)
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