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 10 of 16.
Mukesh priye said:
10 years ago
Common no. of each row is 22227 & 10 hence multiple of that Product equal to 1120.
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.
Renu said:
1 decade ago
Hey can anyone tell how to find HCF fast for big 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?
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.
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.
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.
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)
Vikram said:
1 decade ago
Why do we need to take the difference and then calculate HCF here? As per my understanding in the question to find HCF for given three numbers and the reminder should be same for all three numbers.
In that case why can't I divide all the numbers by 5 so that the reminder will be zero and I will get the HCF also. Where my idea gets slips. Please help.
In that case why can't I divide all the numbers by 5 so that the reminder will be zero and I will get the HCF also. Where my idea gets slips. Please help.
Rohit said:
1 decade ago
How to find HCF fast in simple method?
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