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 2 of 2.
Sairam said:
1 decade ago
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.
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.
(9)
Shweta said:
1 decade ago
Please explain any short method.
(1)
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