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.
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