Question

Please answer this question

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

Answer 1

Let the numbers be represented as

1305 = Na + d ---------------------(1)

where q is the remainder

Also ,

4665 = Nb + d ---------------------(2)

6905 = Nc + d ---------------------(3)

Subtracting eq(1) from eq(2) we get

Nb - Na = 3360

N(b-a) = 3360

for N to be maximum (b-a) is to be minimum

Also , subtracting eq(1) from eq(3) we get

Nc - Na = 5600

N(c-a) = 5600

Also subtracting eq(2) from eq(3) we get

Nc - Nb = 2240

N(c-a) = 2240

Now we need to find HCF of (3360 , 5600 , 2240) , that is 80 which is equal to value of N

Thus N is 1120

Sum of digits of N is 1+1+2+0 = 4 Ans