Elitmus
Exam
Numerical Ability
Number System
The maximum number of factors below 600 are?
Read Solution (Total 7)
-
- ans: 24
600= 2^3∗3∗5^2
number of factor (3+1) (1+1) (2+1)=24 - 6 years agoHelpfull: Yes(16) No(1)
- If x=am∗bn∗cp...
where a, b, c, ... are the prime factors of x, then the number of factors of x is
(m+1)(n+1)(p+1)...
This is because any factor can be made by selecting 0 to m numbers of a (in m+1 ways), 0 to n numbers of b (in n+1 ways) and so on.
Since 600=23∗3∗52
Total number of factors of 600 = (3+1)(1+1)(2+1) = 24 - 6 years agoHelpfull: Yes(7) No(2)
- it is not about factors of 600
question is wrong - 5 years agoHelpfull: Yes(1) No(1)
- it's 23 since the factors should be less than 600.if the answer is 24 then 600 would be included which is wrong
- 5 years agoHelpfull: Yes(1) No(1)
- Factor of 600
(2^3)*(3^1)*(5^2)
(3+1)*(1+1)*(2+1)=4*2*3=24 - 6 years agoHelpfull: Yes(0) No(0)
- 24
the factors of 600 are
(2^3)(5^2)(3)
then total factor will be
4*3*2=24 - 5 years agoHelpfull: Yes(0) No(0)
- Ans =24 ;its not about to include 600 or not only take 480 =2^5 x 3 x 5
Number of divisor= (6) x 2 x 2 = 24 - 3 Months agoHelpfull: Yes(0) No(0)
Elitmus Other Question