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 5 of 16.
Helena said:
1 decade ago
The clue is in the question. A digit by definition is any number between 0 & 9.
N = 1120.
The digits of 1120 are the digits, 1, 1, 2, 0.
There are four of them.
Hope that helps.
N = 1120.
The digits of 1120 are the digits, 1, 1, 2, 0.
There are four of them.
Hope that helps.
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.
Himank said:
10 years ago
@Shrinivasmutagar.
HCF is calculated by seeing the common factors in all the numbers.
For Ex: 2, 2, 2, 2, 7 and 10 is present in all the numbers.
And on * you will get 1120.
HCF is calculated by seeing the common factors in all the numbers.
For Ex: 2, 2, 2, 2, 7 and 10 is present in all the numbers.
And on * you will get 1120.
Guru prasad said:
8 years ago
Subtract the numbers such as,
4665-1305 = 3360 and 6905 - 4665 = 2240 then subtracting the obtained results.
3360-2240 = 1120. Therefore 1 + 1 + 2 + 0 = 4.
4665-1305 = 3360 and 6905 - 4665 = 2240 then subtracting the obtained results.
3360-2240 = 1120. Therefore 1 + 1 + 2 + 0 = 4.
Aswini said:
4 years ago
@All.
So most of you people asked how (4665- 1305) (4665-6905) ( 6905- 1305).
So, By using (b-a) (b-c) (c-a) formula.
Given. a= 1305 , b=4665 c= 6905.
So most of you people asked how (4665- 1305) (4665-6905) ( 6905- 1305).
So, By using (b-a) (b-c) (c-a) formula.
Given. a= 1305 , b=4665 c= 6905.
(5)
Anita Chettri said:
5 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.
(1)
Vihaan said:
5 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)
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
Anita Chettri said:
5 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.
(4)
Rajeswari R said:
1 decade ago
@Madhu. They didn't mention exactly n divide these 3 numbers. They said when dividing it will leaves remainder please consider that part.
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