# 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:

Answer: Option

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:

152 comments Page 2 of 16.
Ashu said:
2 years ago

Thanks for explaining @Saraswati.

(4)

Komal said:
3 years ago

Thank you for explaining the answer @Greeshma.

(2)

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.

(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)

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.

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.

(5)

Anomi said:
3 years ago

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

(4)

Elissa said:
3 years ago

I understood well, thanks to all for explaining.

(1)

Deeksha said:
3 years ago

Good, Thanks you @Greeshma.

K.snigdha shree said:
3 years ago

Thanks all for epxlaining.

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

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

(8)

Subhasis said:
3 years ago

Thanks @Saraswati.

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