Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : 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?
Answer: Option
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.
Discussion:
12 comments Page 1 of 2.
Dipasha Gautam said:
10 months ago
Very helpful. Thanks all for explaining the solution.
Abdul Gani said:
4 years ago
Excellent, Thanks @Sairam.
AGk Durga said:
6 years ago
Thank you for explaining @Vijay.
Shiv said:
7 years ago
@Sairam.
Well explained.
The easiest way to find the factors and then Divisors.
Well explained.
The easiest way to find the factors and then Divisors.
(1)
Vijay said:
7 years ago
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.
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.
Rupesh Kashyap said:
7 years ago
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;
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;
Mydul islam said:
8 years ago
176 is correct. I too agree.
Chota rehaman said:
9 years ago
Yes you are right @Teoh.
(1)
Teoh said:
1 decade ago
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?
(2)
Tamilazhagan said:
1 decade ago
Any easiest way for finding no of divisors?
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