Aptitude - Problems on H.C.F and L.C.M - Discussion

In general, the no of divisors can be found as follows:

Write the number as the product of powers of prime numbers

If the no can be written as (2 power a)*(3 power b)*.

Where 2,3,5,. should be strictly prime numbers.

Note: 1 should not be included because it is not a prime number.

Then the no of divisors are = (a+1)*(b+1).

For 99 = 3*3*11 = (3to the power 2)*(11 to the power 1).

No.of divisors of 99 = (2+1)*(1+1) = 6.

For 101 = (101 power 1).

No.of divisors of 101 = (1+1) = 2.

For 176 = (2 power 4)*(11 power 1).

No.of divisors of 176 = (4+1)*(1+1) = 10.

For 182 = (2 power 1)*(7 power 1)*(13 power 1).

No.of divisors of 182 = (1+1)*(1+1)+(1+1) = 8.

Most no of divisors for the above problem is 176.

Would a shorter answer would simply be to count the number of prime factors? Since divisors are generated from the prime factors, one can safely assume the number with the most prime factors has the most factors?

Find the prime factors:

176 =2^4 & 11.
so we can take from no of 2 Zoro or more and 11 Zero or more.
M^n * X^y ; {(n+1)*(y+1)}.
So in 176 case : {(4+1)*(1+1)}; =>5*2= 10;

Find the prime factors:

176 =2^4 & 11.
so we can take from no of 2 Zoro or more and 11 Zero or more.
M^n * X^y ; {(n+1)*(y+1)}.
So, in 176 case : {(4+1)*(1+1)}; =>5*2= 10.

@Sairam.

Well explained.

The easiest way to find the factors and then Divisors.

Very helpful. Thanks all for explaining the solution.

Any easiest way for finding no of divisors?

Please explain any short method.

Thank you for explaining @Vijay.

176 is correct. I too agree.