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(#M40161660) ELITMUS Numerical Ability question Keep an EYE

A natural number has exactly 10 divisers including one and itself how many distinct prime factors can these natural number have ?

Asked In Elitmus Amey Satish Marathe (8 years ago)

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pankaj bhardwaj  (8 years ago)

take any 2 digit number.........for ex. 20 ...we can see 20 have 6 divisors 1,2,4,5,10,20 . According to question..... number have exactly 10 divisors. So we have to choose another number which have 10 divisors .........again we have choose number 40......... we can see 40 have 8 divisors 1,2,4,5,8,10,20,40.......again we have to choose another number that is 80......we can see 80 have 10 divisors 1,2,4,5,8,10,16,20,40,80. So prime factor of number 80 is 2*2*2*2*5 and distinct prime factor of 80 is two that is 2 and 5.
answer: There are two distinct prime factor that is 2 and 5.
Tips : Number will be choose by doing double and calculate the divisors.

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Pradosh kumar  (8 years ago)

Let N be the number and has 10 divisors.
We know that N= a^p * b^q * c^r........
So here a,b,c,....... are the prime factors and
(p+1)(q+1)(r+1).....are the no. of factors.
So number N has 10 divisors or factors which means
N=9+1 or N=(4+1)(1+1) i.e. these 2 are the only cases.
So which means answer is
Either 1 or 2 solutions.

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SATISH KUMAR SINGH  (8 years ago)

If n is a composite number of form (a^p)*(b^q)*(c^r) where a , b , c are the prime factors of number n then number of prime factors will be given as (p+1)(q+1)(r+1)... Since here (p+1)(q+1)(r+1) = 10. Solving we get number n= (a^4)*(b) hence there can be only 2 distinct prime factors only.

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apoorva nayak  (8 years ago)

SATISH KUMAR SINGH ...sir please elaborate your answer.i cant understand.

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pritesh  (8 years ago)

this que is belong to like find the number of factors of a given no. for ex:-find the no of all factor of 40?
soln;-
40=2^3*5^1 //then no of all factor = (3+1)(1+1)=8
i.e 40 is divisible by total 8 different whole no. are 1,2,4,5,8,10,20,40.
So, the conclusion is no of divisor including 1 and itself= no of all factor of any whole no.
Now, for que given 10 divisor i.e total 10 factor.
and, the no. of factor =10 i.e 5*2 which is also write as (4+1)(1+1)
Lets take a no. 80 its facorization =2^4*5^1 //now see the power of the factor it is 4and 1.
then the total no of factor = (4+1)(1+1)=10=10 divisor .
now here the trick is no of braces() = no of distinct factor .
Hence for 10 divisor i.e 5*2 i.e (4+1)(1+1) i.e total 2 distinct factor(then see factor is prime or not)

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Amit  (8 years ago)

As given it has 10 divisors so it can be through
5*2 or 10*1
so there can be maxm. of 3 prime factors and min. 2 prime factors.

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SATISH KUMAR SINGH  (8 years ago)

Suppose a number n is 18, it can be expressed as (2^1)*(9^2) then total no. of divisors it can have including 1 and itself is (1+1)(2+1) = 6. Similarly the above question be solved as formerly solved

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SATISH KUMAR SINGH  (8 years ago)

Sorry there is correction 18 = 2*(3^2)

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shubham kumar  (8 years ago)

why not only 1 distinct prime number.suppose 512(1,2,4,8,16,32,64,128,256,512).
plese comment..

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naveen  (7 years ago)

The number of factors or divisors will be the product of all the powers after incremented by 1 like (a+1)(a+1).(b+1)(b+1).(c+1)(c+1)..and so on.

So, in the question we got 10, now 10 can be a result if either r a single power or product of 2 powers only and not more than that. So, here

Case 1: (a+1)(a+1)=10 Case 2: (a+1)(a+1)(b+1)(b+1) = 10.

so the answer is either 1 or 2.

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