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

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 4)
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:
4
5
6
8
Explanation:

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

Discussion:
150 comments Page 1 of 15.

Barath dasari said:   3 years ago
(1305 - 4665) = 3360.
(4665 - 6905 ) = 2240.
( 6905 - 1305) = 5600.

Then HCF Of 3360 - 3 * 1120,
HCF Of 2240 - 2 * 1120,
HCF Of 5600 - 5 * 1120,
Sum Of Digits in N = 1 + 1 + 2 + 0 = 4.

I hope it's easy to understand.
(33)

Sumit said:   2 years ago
If we only take two differences i.e. 6905 - 4665 = 2240, 4665 - 1305 = 2360.
HCF of 2240 and 2360 is 40.
4+0 = 4.
(19)

Krishnakanth said:   1 year ago
Given numbers 1305, 4665, 6905.

Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.

3360 factor it will give you 4,4,10,7 all divisible by the other 2 numbers.
So, multiple them and you will get 1120 add them because in question they said digits give you 4 i.e answer.
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Agondeze milly said:   1 year ago
How do you get 4? I am not getting it clearly.
(10)

Suraj Chaudhary said:   10 months ago
I can't understand this question properly.

(10)

Surya said:   8 months ago
Why should we subtract the Numbers with the smallest number and so HCF for that?
(10)

Jai said:   6 months ago
Given numbers 1305, 4665, 6905.

Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
now,
Divide 3360/2240=1120, and
Now divide 5600/1120=0.
so, the HCF of 3360, 2240 & 5560 is 1120.
n=1+1+2+0=4;
So, n=4
(10)

Greeshma said:   3 years ago
Given numbers 1305, 4665, 6905.

Now take difference;
4665-1305= 3360,
6905-4665=2240,
6905-1305=5600.

Take the smallest difference i.e 2240 find factors of it;
1-> 2240
2-> 1120
4-> 560
5-> 448
and so on;

Check from highest factors if any one of them leaves the same remainder for all given numbers
So, 1st one 2240 it can't divide 1305. So, go for the second one 1120.

1305Ã·1120 leaves 185 as the remainder
4665Ã·1120 leaves 185 as the remainder
6905Ã·1120 leaves 185 as the remainder.

As all given conditions fulfilled :)
We got it;
1120.
1+1+2+0= 4.
Edits are welcome
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Aswini said:   3 years ago
@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.
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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.
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