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

What is it exactly saying, when it says, "leaving the same remainder"?

We are ultimately finding the HCF, right? Then what is it defining by "Same remainder"?

Anyone, please explain me this in detail.

@All. Here's my solution.

Num 1305, 4665, 6905.
Difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
Now,
3360/2240 = 1120,
5600/1120 = 0.
HCF 3360, 2240 & 5560 is 1120.
n = 1 + 1 + 2 + 0 = 4;
So, n = 4.

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.

Step 1: Differences between the numbers.

4665−1305 = 3360.
6905−4665 = 2240.
6905−1305 = 5600.

Step 2: Find the GCD of these differences
common prime factors are 2^5 . 5. 7 = 1120.

Step 3: Sum of the digits of 𝑁
1 + 1 + 2 + 0 = 4.

@All.

So most of you people asked how (4665- 1305) (4665-6905) ( 6905- 1305).

So, By using (b-a) (b-c) (c-a) formula.
Given. a= 1305 , b=4665 c= 6905.

As explained in the question,

1305 = N(p) + r --->(1)
4665 = N(q) + r --->(2)
6905 = N(s) + r --->(3)

Now solve (2) - (1)=> 3360 = N (q-p) ---> (4)
and solve (3) - (2)=> 2240 = N (s-p) ---> (5)

Factorizing,
3360 = 1120 * 3 ---> (6)
2240 = 1120 * 2 ---> (7)

Comparing (4) and (5) with (6) and (7), we can clearly say that N=1120.
sum of digits in N = 1+1+2+0 = 4.

N = HCF of (4665-1305),(6905-4665) and (6905-1305).
= HCF of 3360,2240 and 5600.
= 2*2*2*2*2*5*7
= 1120.
Sum of digits in N = (1+1+2+0) = 4.

When these numbers are divided by 5 then the remainder is zero right. Then the Answer should be 5 right. Please explain to me.

Thanks for explaining @Saraswati.

@Anomi

Here, we need to find the greatest number. So 5 is not the greatest number.