# 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:

150 comments Page 4 of 15.
Anom said:
4 years ago

Great explanation, Thanks @M. Harish.

Vishnu Sai said:
4 years ago

I have a question.

Can anyone please explain why they are taking the HCF of (4665-1305), (6905-4665) and (6905-1305) i.e. the HCF of 3360, 2240 and 5600.

Also in the question it states the remainder should be same, it does not state the remainder should only be zero.

So I wish some of the other clarifies my doubts.

Can anyone please explain why they are taking the HCF of (4665-1305), (6905-4665) and (6905-1305) i.e. the HCF of 3360, 2240 and 5600.

Also in the question it states the remainder should be same, it does not state the remainder should only be zero.

So I wish some of the other clarifies my doubts.

K.manoj said:
5 years ago

Thanks @Saraswati.

GIRISH SINGH BISHT said:
5 years ago

Thanks for explaining @Priya.

Harini said:
5 years ago

@Anchit.

We should subtract remainder. i.e 12-2=10.

We should subtract remainder. i.e 12-2=10.

Bimal said:
5 years ago

Let N be the greatest number that will divide 17, 41 and 77, leaving the same remainder in each case. Then the sum of the digits in N is:

It is easy to understand this with a simple example, and the logic behind this is so obvious.

Take number 12 (you can get any number you desire and practice it!).

Dividend = Divisor X Quotient + remainder (when the remainder is a constant!).

12 X 1 + 5 = 17.

12 X 3 + 5 = 41.

12 X 6 + 5 = 77.

Now, take the differences of 17, 41 and 77 in order considering the smallest, the middle and the largest numbers.

(41-17) = 24; (77-17) = 60; (77-41) =36.

Now take 24, 60, and 36 and find the HCF or HCD of these numbers: it is equal to 12.

Look at the "Divisor" used at the beginning. So, the divisor of that arrangement is equal to the HCF of 24, 60 and 36. When the remainder is constant, we can observe this pattern with any set of numbers arranged in this way.

The sum of digits of N = 1+2 =3.

It is easy to understand this with a simple example, and the logic behind this is so obvious.

Take number 12 (you can get any number you desire and practice it!).

Dividend = Divisor X Quotient + remainder (when the remainder is a constant!).

12 X 1 + 5 = 17.

12 X 3 + 5 = 41.

12 X 6 + 5 = 77.

Now, take the differences of 17, 41 and 77 in order considering the smallest, the middle and the largest numbers.

(41-17) = 24; (77-17) = 60; (77-41) =36.

Now take 24, 60, and 36 and find the HCF or HCD of these numbers: it is equal to 12.

Look at the "Divisor" used at the beginning. So, the divisor of that arrangement is equal to the HCF of 24, 60 and 36. When the remainder is constant, we can observe this pattern with any set of numbers arranged in this way.

The sum of digits of N = 1+2 =3.

(2)

Sachin said:
5 years ago

Thanks @Saraswati.

Kotresh said:
5 years ago

Good explanation, Thanks @Yogesh.

Provith said:
5 years ago

Thanks all for explaining the answer.

(1)

Akhil said:
5 years ago

Let consider 1305 as N * some value( say X) + remainder.

Similarly N*Y + r = 4665.

N*Z + r =6905,

For removing or subtract each other we get;

N* (X-Y) , N(Y-z), N*(Z-x).

After doing this the number may be changed. But the required HCF (ie N) remains the same. So we can find N. This is the Simplest method for eliminating the remainder.

Similarly N*Y + r = 4665.

N*Z + r =6905,

For removing or subtract each other we get;

N* (X-Y) , N(Y-z), N*(Z-x).

After doing this the number may be changed. But the required HCF (ie N) remains the same. So we can find N. This is the Simplest method for eliminating the remainder.

(1)

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