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 14 of 16.
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
Saravanan said:
1 decade ago
How will be taken sum of n digit(1+1+2+0)=4?
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 ?
Nagesh said:
1 decade ago
Yogesh thanks
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).
YUVARAJ said:
1 decade ago
Hey please explain why we take hcf of 4665-1305, 6905-4665 ?
Keerthi said:
1 decade ago
We have to find greatest num, hcf highest common factor.
So we did.
So we did.
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?
Shakti said:
1 decade ago
Thanks! saraswati.
Souravikiran said:
1 decade ago
Hi yogeash can you explain me the basic logic of hcf and lcm. How did we take numbers after factorising and what about co prime and the logic of selecting common factors. Please.
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