# 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 3 of 15.
Babu said:
3 years ago

Thank you all.

Kezang Wanhchuk said:
3 years ago

@All.

The solution is;

=> 1305/4= remiander is .250, 4664/4= .250 and 6905/4= .250

1305/5= no remainder to all.

=> 1305/6= .50, 4665/6= .50 and 6905/6= .833(remainder is diferent)

=> 1305/8= .125 so, remainder is less.

Whereas we are finding HCF so less remainder is not HCF.

The solution is;

=> 1305/4= remiander is .250, 4664/4= .250 and 6905/4= .250

1305/5= no remainder to all.

=> 1305/6= .50, 4665/6= .50 and 6905/6= .833(remainder is diferent)

=> 1305/8= .125 so, remainder is less.

Whereas we are finding HCF so less remainder is not HCF.

(1)

Dev said:
3 years ago

The HCF of any numbers would necessarily have to be a factor of the difference between any pair of numbers.

Here it is asked that find the number which when divides 4665, 6905 and 1305 give the same remainder.

So, to find this, we can find the HCF of the difference between these three numbers.

This number will be the greatest number which when divides 4665, 6905 and 305 gives the same remainder.

Here it is asked that find the number which when divides 4665, 6905 and 1305 give the same remainder.

So, to find this, we can find the HCF of the difference between these three numbers.

This number will be the greatest number which when divides 4665, 6905 and 305 gives the same remainder.

Anita Chettri said:
3 years ago

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.

= HCF of 3360,2240 and 5600.

= 2*2*2*2*2*5*7

= 1120.

Sum of digits in N = (1+1+2+0) = 4.

(3)

Anita Chettri said:
3 years ago

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.

= HCF of 3360,2240 and 5600,

= 2 * 2 * 2 * 2 * 2 * 5 * 7.

= 1120

Sum of digits in N = (1+1+2+0) = 4.

Haq ahmed said:
3 years ago

How we get the 1120 HCF of those number, can someone explain? please.

Vihaan said:
3 years ago

I can't understand why are we taking the numbers (4665 - 1305), (6905 - 4665) and (6905 - 1305)?

Can anyone please explain to me this in brief?

Can anyone please explain to me this in brief?

(1)

Jack said:
4 years ago

I can't understand why we are taking H.C.F. Of (4665 - 1305), (6905 - 4665) and (6905 - 1305).

Anyone please explain in detail.

Anyone please explain in detail.

(1)

Prince Khan said:
4 years ago

Thanks for the answer @Saraswati.

Jaya said:
4 years ago

Thanks @Saraswati.

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