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

Here taking difference is mentioned because we need to find h.c.f.

Here 1305=pq1+r;
4665=pq2+r;
6905=pq3+r;.here p is divident, q=quotient, r=reminder.

We want to find p so,
Totally 3 equations so,
P(q2-q1) = 3360.
p(q3-q2) = 2240.
p(q3-q1) = 5600 now there are 3 p's we need p that is h.c.f of (3360, 2240, 5600) here finding hcf by taking difference it will make easy now p = 1120 and (1+1+2+0) = 4.

Hope you all understood.

Please can you explain how to calculate HCF ?

Please can you explain how to calculate HCF and explain?

Please anyone tell me the concept of 1120?

Impossible to calculate the GCD without successive difference:

a. 6905 - 4665.
b. 4665 - 1305.
c. 6905 - 1305.

Interestingly when you replace the values used in the above method to calculate the successive difference as illustrated below :

a. 6905 - 1305 (c above).
b. 6905 - 4665 (a above).
and
c. 4665 - 1305 (above it would replace the statement => 4665 - 1305).

It will convincingly still give rise to a G.C.D = 1120.

I calculated the HCF for the 3 numbers and got an ans of 40.
This would also mean that the sum of the digits is 4 = correct answer.

1120 = 2*2*2*2*2*2*5*7.
3360 = 2*2*2*2*2*3*5*7.
5600 = 2*2*2*3*3*3*5*5.

This method does not warrant/validate the condition that the GCD needs to be calculated via successive difference.

I can't understand because of taking H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305).

Why we have to add 1+1+2+0? I'm not getting it. Please someone explain it.

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

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.