# 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:

150 comments Page 2 of 15.
Dev said:
4 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.

Saraswati said:
1 decade ago

First do this(4665 - 1305) , (6905 - 4665) and (6905 - 1305). we ll get 3360,2240,5600. then divide by 10.. now

336=2*2*2*2*3*7*10

224=2*2*2*2*2*7*10

560=2*2*2*2*5*7*10

So take the common multiples.. now we got 2*2*2*2*7*10=1120( this 1120 is "N")

So, question asked here is sum of digits "N"??

"N" is 1120; sum of digits "N" is 1+1+2+0=4.

Tats it!;)

336=2*2*2*2*3*7*10

224=2*2*2*2*2*7*10

560=2*2*2*2*5*7*10

So take the common multiples.. now we got 2*2*2*2*7*10=1120( this 1120 is "N")

So, question asked here is sum of digits "N"??

"N" is 1120; sum of digits "N" is 1+1+2+0=4.

Tats it!;)

(2)

Ayush said:
9 years 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.

M.Harish said:
1 decade ago

1305, 4665, 6905 ----- These are the numbers.

In this question why we subtract the numbers because here it is said that they have same remainder.

So, 1305 = hcf * x + remainder.

4665 = hcf * y + remainder.

6905 = hcf * z + remainder.

Where x, y, z are respective quotients.

So. When we subtract those numbers, they become multiples of the hcf. Then we can directly calculate hcf.

In this question why we subtract the numbers because here it is said that they have same remainder.

So, 1305 = hcf * x + remainder.

4665 = hcf * y + remainder.

6905 = hcf * z + remainder.

Where x, y, z are respective quotients.

So. When we subtract those numbers, they become multiples of the hcf. Then we can directly calculate hcf.

Yogesh said:
1 decade ago

HCF of 2240(smallest); 3360(middle); 5600(largest no)

Hence hcf ===1120

2240* ) 3360 ( 1

2240

------

1120** )2240* ( 2

2240

-------

0000

-------

1120**)5600 ( 5

5600

------

0000

------

Hence hcf ===1120

Meghana said:
7 years ago

Indeed,

1. It is given in the question that the remainder is same and

2. we know that the greatest number is leaving the same number.

Thus we are subtracting two numbers to get value exclusive of remainder

Now, if A and B are divisible by say x them A-B is also divisible by x.

With this logic, we are first eliminating the common remainder and later finding the HCF.

1. It is given in the question that the remainder is same and

2. we know that the greatest number is leaving the same number.

Thus we are subtracting two numbers to get value exclusive of remainder

Now, if A and B are divisible by say x them A-B is also divisible by x.

With this logic, we are first eliminating the common remainder and later finding the HCF.

Dev said:
8 years ago

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder is,

Answer LCM of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3.

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

My question how to find out K value in this question.

Answer LCM of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3.

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

My question how to find out K value in this question.

Vikram said:
10 years 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.

Akhil said:
5 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)

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).. :)

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