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:

SBI-PO-main

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:

A. 4
B. 5
C. 6
D. 8

Answer: A

Solution:

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = (1 + 1 + 2 + 0) = 4

Author Profile

IBPS Preparation Team
IBPS Preparation Team
At IBPS Preparation,We provide pathway for any candidate who wishes to shine in the recruitment process of IBPS.

NICL AO

Leave a Reply

Your email address will not be published. Required fields are marked *

7 + 2 =


© Copyright 2017 - All Rights Reserved | DesignzRush (OPC) Pvt. Ltd.
Inline
Inline