# 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:
148 comments Page 1 of 15.

How would we know that H.C.F. of 3360, 2240 and 5600 = 1120.
(1)

Krishna kumar said:   1 decade ago
Sorry I can't understand why we are taking H. C. F. Of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).

Upto my understanding
(4665 - 1305)=3360, (6905 - 4665)=2240 and (6905 - 1305)=5600
3360:2240:5600
1120:1120:1120

So, Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Hope so,if am wrong means correct me

1120*3=3360
1120*2=2240
1120*5=5600

the highest common factorial num is:1120 for all 3 numbers
so that here 1+1+2+0=4 we got.....

How we know by seeing the numbers the H.C.F ?

First do this(4665 - 1305) , (6905 - 4665) and (6905 - 1305). we ll get 3360,2240,5600. then divide by 10.. now
336=2*2*2*2*3*7*10
224=2*2*2*2*2*7*10
560=2*2*2*2*5*7*10

So take the common multiples.. now we got 2*2*2*2*7*10=1120( this 1120 is "N")

So, question asked here is sum of digits "N"??
"N" is 1120; sum of digits "N" is 1+1+2+0=4.
Tats it!;)
(1)

Thanx priya.

Thanks sarsu.