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 3 of 16.
Anchit said:
1 decade ago
We are taking hcf of (4665 - 1305) , (6905 - 4665) and (6905 - 1305)..because in this way the remainder is cancelled...
take an example..
12 and 7..both give remainder 2 when divided by 5.
now when we substact (12-7) the differance 2 is cancelled ...
hence we get a number that is completely divisible i.e a factor of 5..(or the hcf).. :)
take an example..
12 and 7..both give remainder 2 when divided by 5.
now when we substact (12-7) the differance 2 is cancelled ...
hence we get a number that is completely divisible i.e a factor of 5..(or the hcf).. :)
Swaminathan Vembu said:
1 decade ago
I calculated the HCF for the 3 numbers and got an ans of 40.
This would also mean that the sum of the digits is 4 = correct answer.
1120 = 2*2*2*2*2*2*5*7.
3360 = 2*2*2*2*2*3*5*7.
5600 = 2*2*2*3*3*3*5*5.
This method does not warrant/validate the condition that the GCD needs to be calculated via successive difference.
This would also mean that the sum of the digits is 4 = correct answer.
1120 = 2*2*2*2*2*2*5*7.
3360 = 2*2*2*2*2*3*5*7.
5600 = 2*2*2*3*3*3*5*5.
This method does not warrant/validate the condition that the GCD needs to be calculated via successive difference.
Vishnu Sai said:
6 years ago
I have a question.
Can anyone please explain why they are taking the HCF of (4665-1305), (6905-4665) and (6905-1305) i.e. the HCF of 3360, 2240 and 5600.
Also in the question it states the remainder should be same, it does not state the remainder should only be zero.
So I wish some of the other clarifies my doubts.
Can anyone please explain why they are taking the HCF of (4665-1305), (6905-4665) and (6905-1305) i.e. the HCF of 3360, 2240 and 5600.
Also in the question it states the remainder should be same, it does not state the remainder should only be zero.
So I wish some of the other clarifies my doubts.
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.
Krishnakanth said:
3 years ago
Given numbers 1305, 4665, 6905.
Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
3360 factor it will give you 4,4,10,7 all divisible by the other 2 numbers.
So, multiple them and you will get 1120 add them because in question they said digits give you 4 i.e answer.
Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
3360 factor it will give you 4,4,10,7 all divisible by the other 2 numbers.
So, multiple them and you will get 1120 add them because in question they said digits give you 4 i.e answer.
(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)
Kezang Wanhchuk said:
5 years ago
@All.
The solution is;
=> 1305/4= remiander is .250, 4664/4= .250 and 6905/4= .250
1305/5= no remainder to all.
=> 1305/6= .50, 4665/6= .50 and 6905/6= .833(remainder is diferent)
=> 1305/8= .125 so, remainder is less.
Whereas we are finding HCF so less remainder is not HCF.
The solution is;
=> 1305/4= remiander is .250, 4664/4= .250 and 6905/4= .250
1305/5= no remainder to all.
=> 1305/6= .50, 4665/6= .50 and 6905/6= .833(remainder is diferent)
=> 1305/8= .125 so, remainder is less.
Whereas we are finding HCF so less remainder is not HCF.
(1)
Sanatan said:
1 decade ago
The answer is actually 5 and is pretty simple. You can see by dividing the given numbers by 5 we get the same remainder zero as asked in the question.
As we have to find the max number which yields the same remainder for the given numbers. 4 is not the correct answer here.
As we have to find the max number which yields the same remainder for the given numbers. 4 is not the correct answer here.
Gopalakrishnan said:
1 decade ago
@All.
Is very simple to find the answer!
In question clearly given N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. So we dividing the each number by giving option as 4 5 6 8 and we get same remainder as option A) 4.
Is very simple to find the answer!
In question clearly given N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. So we dividing the each number by giving option as 4 5 6 8 and we get same remainder as option A) 4.
Shanmugar said:
1 decade ago
Let they find the hcf of 3 numbers is that. 1350, 4665, 6905.
Why are they finding the hcf of 3360, 2240, 5600. And one more thing in the question they have that N divides this numbers with same reminder they didn't mention that "N exactly divides" or reminder is zero.
Why are they finding the hcf of 3360, 2240, 5600. And one more thing in the question they have that N divides this numbers with same reminder they didn't mention that "N exactly divides" or reminder is zero.
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