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

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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 $N = (1 + 1 + 2 + 0) = 4$

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Similar questions

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

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:

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 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:

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 sum of the digits in N is

The correct option is A 4N= 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

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

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 $N = (1 + 1 + 2 + 0) = 4$