TIFR PHD CS & SS 2012
October 28, 2023KVS 30-12-2018 Part B
October 28, 2023NIELIT Scientist -B CS (22-07-17)
|
Question 15
|
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is
|
4
|
|
|
5
|
|
|
6
|
|
|
8
|
Question 15 Explanation:
For this type of questions we have to find gcd of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
So, = H.C.F. of 3360, 2240 and 5600 = 1120.
sum = 1+1+2+0 = 4.
So, = H.C.F. of 3360, 2240 and 5600 = 1120.
sum = 1+1+2+0 = 4.
Correct Answer: A
Question 15 Explanation:
For this type of questions we have to find gcd of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
So, = H.C.F. of 3360, 2240 and 5600 = 1120.
sum = 1+1+2+0 = 4.
So, = H.C.F. of 3360, 2240 and 5600 = 1120.
sum = 1+1+2+0 = 4.
Subscribe
Login
0 Comments