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:
4
5
6
8
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).. :)

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.

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.

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.

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.
(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.
(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.
(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.

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.

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.


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