Aptitude - Problems on H.C.F and L.C.M - Discussion

The clue is in the question. A digit by definition is any number between 0 & 9.

N = 1120.
The digits of 1120 are the digits, 1, 1, 2, 0.
There are four of them.

Hope that helps.

8 will divide each number and leave reminder 1, and 8 is greatest among all of the option.

People try dividing each number by 8 and you will get the same remainder.

Hello @Sarswathi, why are you taking 10 to multiply number after the subtraction. Could you please tell me?

@All.

Is very simple to find the answer!

In question clearly given N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. So we dividing the each number by giving option as 4 5 6 8 and we get same remainder as option A) 4.

People try dividing each number by 8 and you will get the same remainder.

EASY ONE NO. ARE 6905, 4665, 1305

SO [(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.

NOW 1+1+2+0 = 4.

[(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.

@Ravi can you explain how you did this step?

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 :

First we will find out N :

Use the formula :

Dividend = Divisor * Quotient + Remainder.

Given information :

1305 = N * x + Remainder ...equation 1.
4665 = N * y + Remainder ...equation 2.
6905 = N * z + Remainder ...equation 3.

Remainder in all the above three equations are equal but not known. Hence we eliminate them by performing subtraction as follows :

4665 - 1305 = N * (y-x) ...equation 2 - equation 1.
6905 - 4665 = N * (z-y) ...equation 3 - equation 2.
6905 - 1305 = N * (z-x) ...equation 3 - equation 1.

i.e.

3360 = N * (y-x) ...equation 4.
2240 = N * (z-y) ...equation 5.
5600 = N * (z-x) ...equation 6.

Values of x, y, z and N are not known. We are not concerned about values of x, y and z. From the equations 4, 5 and 6 it is clear N is a factor of 3360, 2240 and 5600. As the question states N is the GREATEST number, we will find out Highest Common Factor(HCF) of 3360, 2240 and 5600 which will be our N.

Factors :

3360 = 2*2*2*2*3*7*10.
2240 = 2*2*2*2*2*7*10.
5600 = 2*2*2*2*5*7*10.
HCF = 2*2*2*2*7*10 = 1120.

Hence N = 1120.

We have to find SUM of digits of N which is : 1+1+2+0 = 4.

Hence 4 is the answer(option A).

I took help of comments of @M.Harish, @Anchit and @Saraswati to provide the solution. A big thanks to them. The solution is meant to ease out the confusion regarding the question. I have tried to explain in detail hence its lengthy. Its recommended people solve the problem with as many shortcuts as possible in competitive exams where time is a constraint.

Why to take the difference between no.s? I don't get it.