Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then the sum of digits of N is:

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Solution

The correct option is A
4

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.

Suggest Corrections

Similar questions

Q. 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

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:

Q. Let N be the greatest number that will divide

1305, 4665 a n d 6905

leaving the same remainder in each case. The sum of the digits of N is

Q. Question: 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

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

Aren't we supposed to find the HCF (1305, 4665, 6905) rather than that of the HCF(4665 - 1305), (6905 - 4665) and (6905 - 1305)?

Let Me be the greatest number that will divide 1305,4665 & 6905, leaving the same remainder ineach case. Sum of digits of N is ---------

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Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then the sum of digits of N is:

The correct option is A 4 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.

The correct option is A
4

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.