Let N be the largest integer that divides 1305, 4665, and 6905 without leaving the same residual in each case. Then in N, the total of the digits is:
A. 4
B. 5
C. 6
D. 8
Answer:A
Let N be the largest integer that divides 1305, 4665, and 6905 without leaving the same residual in each case. Then in N, the total of the digits is:
A. 4 B. 5 C. 6 D. 8
Explanation: N = HCF of (4665 - 1305), (6905 - 4665), and (6905 - 4665). (6905 - 1305),3360, 2240, and 5600 H.C.F. = 1120,In N, the sum of the digits is ( 1 + 1 + 2 + 0 ) = 4.