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.
SURAJ said:
1 decade ago
The easiest way to find this answer is subtracting until you get repeated value.
(4465-1305 = 3360.
6905-4665 = 2240.
6905-1305 = 5600).
(3360-2240 = 1120.
5600-2240 = 3360.
5600-3360 = 2240).
(2240-1120 = 1120.
3360-1120 = 2240.
3360-2240 = 1120).
So, the ans is 1120.
1+1+2+0 = 4.
(4465-1305 = 3360.
6905-4665 = 2240.
6905-1305 = 5600).
(3360-2240 = 1120.
5600-2240 = 3360.
5600-3360 = 2240).
(2240-1120 = 1120.
3360-1120 = 2240.
3360-2240 = 1120).
So, the ans is 1120.
1+1+2+0 = 4.
(1)
J.manisha said:
1 decade ago
Sorry I can't understand this process. Please kindly give correct explanation. How will take this "N" value N = 1+1+2+0 = 4.
Ayush said:
1 decade ago
If two number gives same remainder when divided by a number (say N) then their diff will be perfectly divisible by that number (N).
Example 23 and 49 gives same remainder 1 when divided by two.
So the difference 49-23 = 26 is perfectly divisible by 2.
So our question suffices to finding the HCF of the differences. (Since we need the greatest number).
Hope now it is clear for all.
Example 23 and 49 gives same remainder 1 when divided by two.
So the difference 49-23 = 26 is perfectly divisible by 2.
So our question suffices to finding the HCF of the differences. (Since we need the greatest number).
Hope now it is clear for all.
Pavan said:
1 decade ago
The reason they're taking the HCF of the difference of the numbers is because it's said that the numbers give equal remainder on being divided. So number n%p=q [% denotes remainder operation] and n1%p=q.
So (n1-n) %p=0. We can say that the number (p) is a factor of the difference of the two numbers.
So (n1-n) %p=0. We can say that the number (p) is a factor of the difference of the two numbers.
Shrinivasmutagar said:
1 decade ago
3360 = 2*2*2*2*3*7*10.
2240 = 2*2*2*2*2*7*10.
5600 = 2*2*2*2*5*7*10.
HCF = 2*2*2*2*7*10 = 1120.
How this is done can any one explain?
2240 = 2*2*2*2*2*7*10.
5600 = 2*2*2*2*5*7*10.
HCF = 2*2*2*2*7*10 = 1120.
How this is done can any one explain?
Renu said:
1 decade ago
Hey can anyone tell how to find HCF fast for big numbers?
Himank said:
1 decade 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.
Mukesh priye said:
1 decade ago
Common no. of each row is 22227 & 10 hence multiple of that Product equal to 1120.
Rekha kharode said:
10 years ago
Please explain if someone knows why select only HCF?
Neha said:
10 years ago
How 1120?
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