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 4 of 16.
Princely said:
7 years ago
Can we try it in this way?
3+3+6+0= 12
2+2+4+0= 8,
5+6+0+0= 11,
12+8+11= 31,
3+1= 4.
3+3+6+0= 12
2+2+4+0= 8,
5+6+0+0= 11,
12+8+11= 31,
3+1= 4.
(1)
Akshitha said:
7 years ago
Thnks @Sanchay.
(1)
Ramachandra said:
7 years ago
Good explanation @Yogesh.
(1)
Saniya taj said:
7 years ago
Easy to understand. Thanks @Suraj.
(1)
Akhil said:
7 years ago
Let consider 1305 as N * some value( say X) + remainder.
Similarly N*Y + r = 4665.
N*Z + r =6905,
For removing or subtract each other we get;
N* (X-Y) , N(Y-z), N*(Z-x).
After doing this the number may be changed. But the required HCF (ie N) remains the same. So we can find N. This is the Simplest method for eliminating the remainder.
Similarly N*Y + r = 4665.
N*Z + r =6905,
For removing or subtract each other we get;
N* (X-Y) , N(Y-z), N*(Z-x).
After doing this the number may be changed. But the required HCF (ie N) remains the same. So we can find N. This is the Simplest method for eliminating the remainder.
(1)
Provith said:
7 years ago
Thanks all for explaining the answer.
(1)
Jack said:
5 years ago
I can't understand why we are taking H.C.F. Of (4665 - 1305), (6905 - 4665) and (6905 - 1305).
Anyone please explain in detail.
Anyone please explain in detail.
(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)
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)
Dev said:
5 years ago
The HCF of any numbers would necessarily have to be a factor of the difference between any pair of numbers.
Here it is asked that find the number which when divides 4665, 6905 and 1305 give the same remainder.
So, to find this, we can find the HCF of the difference between these three numbers.
This number will be the greatest number which when divides 4665, 6905 and 305 gives the same remainder.
Here it is asked that find the number which when divides 4665, 6905 and 1305 give the same remainder.
So, to find this, we can find the HCF of the difference between these three numbers.
This number will be the greatest number which when divides 4665, 6905 and 305 gives the same remainder.
(1)
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