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:
158 comments Page 7 of 16.
Nikhil said:
1 decade ago
Hey anyone here can pls expain me why we take hcf of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).
Please explain if someone knows?
Please explain if someone knows?
Keerthi said:
1 decade ago
We have to find greatest num, hcf highest common factor.
So we did.
So we did.
YUVARAJ said:
1 decade ago
Hey please explain why we take hcf of 4665-1305, 6905-4665 ?
Sri said:
1 decade ago
In question itself they gave N is greatest no for that only we take hcf.
Other name of hcf is greatest common divisor (GCD).
Other name of hcf is greatest common divisor (GCD).
Nagesh said:
1 decade ago
Yogesh thanks
Madhu said:
1 decade ago
Using successive division to find HCF of 1305,4665 and 6905 is 5
How it is 1120 ?
How it is 1120 ?
Saravanan said:
1 decade ago
How will be taken sum of n digit(1+1+2+0)=4?
Piyush said:
1 decade ago
Hi saravanan..
in question it asks for "sum of digits in N" not "sum of n digits". As N=1120 so sum of digits i.e 1+1+2+0=4
in question it asks for "sum of digits in N" not "sum of n digits". As N=1120 so sum of digits i.e 1+1+2+0=4
Smilly siva said:
1 decade ago
Hey anyone here can please expain me why we take hcf of (4665 - 1305) , (6905 - 4665) and (6905 - 1305). ? please give explaination.
Sat said:
1 decade ago
Yogesh is pretty much right.
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