###### TIFR PHD CS & SS 2012

October 28, 2023###### KVS 30-12-2018 Part B

October 28, 2023# NIELIT 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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