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

### Discussion :: Problems on H.C.F and L.C.M - General Questions (Q.No.27)

27.

Which of the following has the most number of divisors?

 [A]. 99 [B]. 101 [C]. 176 [D]. 182

Explanation:

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

 Shweta said: (Apr 10, 2012) Please explain any short method.

 Sairam said: (Jul 3, 2013) 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.

 Tamilazhagan said: (Oct 9, 2014) Any easiest way for finding no of divisors?

 Teoh said: (May 22, 2015) 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?

 Chota Rehaman said: (Jan 27, 2017) Yes you are right @Teoh.

 Mydul Islam said: (Jan 6, 2018) 176 is correct. I too agree.

 Rupesh Kashyap said: (Jul 30, 2018) 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;

 Vijay said: (Aug 1, 2018) 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.

 Shiv said: (Dec 8, 2018) @Sairam. Well explained. The easiest way to find the factors and then Divisors.

 Agk Durga said: (Jan 23, 2020) Thank you for explaining @Vijay.