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)

Problems on H.C.F and L.C.M

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:

158 comments Page 12 of 16.

Saravanan said: 1 decade ago

How will be taken sum of n digit(1+1+2+0)=4?

Geetika said: 1 decade ago

Please anyone tell me the concept of 1120?

Shirish said: 8 years ago

Thanks to all for the correct explanation.

Ity said: 9 years ago

What is the HCF of no's 1365, 4665, 6905?

Madhu said: 9 years ago

Thanks for excellent explanation @Yogesh.

Mounika said: 9 years ago

Thanks for Excellent explanation @Suraj.

Rohit said: 1 decade ago

How to find HCF fast in simple method?

Thanuja said: 3 days ago

1120 doesn't divides 3 numbers. Right?

Chirag said: 1 decade ago

Saraswati your explanation is superb.

Deepa said: 9 years ago

How (1 + 1 + 2 + 0)?

Please tell me.

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