Elitmus
Exam
Numerical Ability
Arithmetic
A natural number has exactly 10 divisors including 1 and itself. How many distinct prime factors can a number have...?
a. either 1 or 2
b. either 1 or 3
c. either 2 or 3
d. either 1, 2 or 3
Read Solution (Total 6)
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- ans is-either 1 or 2
Follow the link-http://www.m4maths.com/8864-a-natural-number-has-exactly-10-divisors-including-1-and-itself-how-many-distinct-prime-factors-can-this-natural-number.html - 7 years agoHelpfull: Yes(5) No(0)
- either 2 or 3 because 1 is not a prime number.
- 7 years agoHelpfull: Yes(1) No(2)
- ans is a.) either 1 or 2
B'z 1 is not prime no so we have only 9 divisor.
if we want to find the # of divisors of number so we take x^p1,y^p2 .... and so on
where x and y is prime factor and p1 and p2 is power.
so number of divisors =(p1+1) *(p2+1)
(p1+1) *(p2+1)=9
so we have only 3*3 means we have only 2 divisors or may be if p1=8 its means we have only one factors - 7 years agoHelpfull: Yes(1) No(0)
- either 2 or 3
eg; 10000 - 7 years agoHelpfull: Yes(0) No(2)
- either 1 or 2
- 7 years agoHelpfull: Yes(0) No(0)
- either 1or 2
- 7 years agoHelpfull: Yes(0) No(0)
Elitmus Other Question
If 1/a+1/b+1/c=1/(a+b+c) where a+b+c not equals=0, abc not equals=0, what is the value of (a+b) (b+c) (A+c)...??
a. Equals 0
b.Greater than 0
c.Less than 0
d. Can't be determined
If v, w,x,y,z are non negative integers, each less than 11, then how many distinct combinations of (v,w,x,y,z) satisfy v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001..??
a.0
b.1
c.2
d.3