Aptitude - Problems on H.C.F and L.C.M - Discussion

Discussion :: 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:

[A]. 4
[B]. 5
[C]. 6
[D]. 8

Answer: Option A

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


Sunny said: (Sep 6, 2010)  
How would we know that H.C.F. of 3360, 2240 and 5600 = 1120.

Krishna Kumar said: (Sep 11, 2010)  
Sorry I can't understand why we are taking H. C. F. Of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).

Priya said: (Dec 9, 2010)  
Upto my understanding
(4665 - 1305)=3360, (6905 - 4665)=2240 and (6905 - 1305)=5600
3360:2240:5600
1120:1120:1120

So, Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Hope so,if am wrong means correct me

Prashanthi said: (Dec 13, 2010)  
1120*3=3360
1120*2=2240
1120*5=5600

the highest common factorial num is:1120 for all 3 numbers
so that here 1+1+2+0=4 we got.....

Swarna said: (Dec 14, 2010)  
How we know by seeing the numbers the H.C.F ?

Saraswati said: (Dec 24, 2010)  
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!;)

Neha said: (Jan 11, 2011)  
Thanx priya.

Jinx said: (Jan 23, 2011)  
Thanks sarsu.

Vishal said: (Feb 1, 2011)  
Thanks saraswati.

Swati said: (Feb 17, 2011)  
Why we are taking the HCF of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).

Plzzz explain.

Zara said: (Apr 28, 2011)  
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4, how does it come and What N indicates?

Shanmugar said: (Apr 29, 2011)  
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.

Yogesh said: (Jun 30, 2011)  
HCF of 2240(smallest); 3360(middle); 5600(largest no)

         2240* ) 3360 ( 1
2240
------
1120** )2240* ( 2
2240
-------
0000
-------

1120**)5600 ( 5
5600
------
0000
------


Hence hcf ===1120

Savitha said: (Jul 5, 2011)  
Good explanation yogesh:).

Ramya said: (Jul 14, 2011)  
Thankx priya.

Chirag said: (Aug 7, 2011)  
Saraswati your explanation is superb.

Ashwin said: (Aug 27, 2011)  
Why are they finding H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) ?

Please explain.

Kumar said: (Sep 22, 2011)  
Thanks yogesh.

Souravikiran said: (Sep 24, 2011)  
Hi yogeash can you explain me the basic logic of hcf and lcm. How did we take numbers after factorising and what about co prime and the logic of selecting common factors. Please.

Shakti said: (Sep 25, 2011)  
Thanks! saraswati.

Nikhil said: (Sep 28, 2011)  
Hey anyone here can pls expain me why we take hcf of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).

Please explain if someone knows?

Keerthi said: (Oct 10, 2011)  
We have to find greatest num, hcf highest common factor.

So we did.

Yuvaraj said: (Oct 13, 2011)  
Hey please explain why we take hcf of 4665-1305, 6905-4665 ?

Sri said: (Oct 19, 2011)  
In question itself they gave N is greatest no for that only we take hcf.

Other name of hcf is greatest common divisor (GCD).

Nagesh said: (Oct 27, 2011)  
Yogesh thanks

Madhu said: (Jan 7, 2012)  
Using successive division to find HCF of 1305,4665 and 6905 is 5
How it is 1120 ?

Saravanan said: (Feb 17, 2012)  
How will be taken sum of n digit(1+1+2+0)=4?

Piyush said: (Feb 19, 2012)  
Hi saravanan..
in question it asks for "sum of digits in N" not "sum of n digits". As N=1120 so sum of digits i.e 1+1+2+0=4

Smilly Siva said: (Mar 3, 2012)  
Hey anyone here can please expain me why we take hcf of (4665 - 1305) , (6905 - 4665) and (6905 - 1305). ? please give explaination.

Sat said: (Apr 13, 2012)  
Yogesh is pretty much right.

Nagarjuna D said: (May 14, 2012)  
I can't understand why we have to subtract one among other?

Rajeswari R said: (Jun 9, 2012)  
@Madhu. They didn't mention exactly n divide these 3 numbers. They said when dividing it will leaves remainder please consider that part.

Alisha (Alice) said: (Jun 27, 2012)  
HCF (Highest Common Factor) is also known as GCD (Greater Common Divisor).

And, there are various ways through which one can find HCF of the given numbers. Among these, one of the methods is well explained by Yogesh and another by Saraswati.

Sai Krishnan K said: (Jul 19, 2012)  
Why it is not taken 5. It also gives the same remainder 0.

M.Harish said: (Jul 24, 2012)  
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.

Sanatan said: (Aug 5, 2012)  
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.

Ramana said: (Aug 18, 2012)  
Thanks saraswathi. I get answer for this types of problems by following your explanation. Thanks a lot.

Anchit said: (Aug 18, 2012)  
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).. :)

Rangan said: (Aug 21, 2012)  
@Ramana : The Greatest common divisor (or the HCF) is not the solutiom needed here, it is rather the SUM OF THE DIGITS of THE HCF.


The Remainder being the same for each case, it is clear that subtracting the lesser no. from the greater ones renders the remainder zero. Also the HCF of te no.s left thereafter just wait to be known, because it perfectly divides the no.s and thus gets the same remainder in each case.

Gargi said: (Aug 28, 2012)  
5 is not correct answer. As question is for sum of the greatest devider. And greatest number which devides all the given numbers is : 1120 so, digit sum of the same is 4.

Correct answer is: 4.

Helena said: (Feb 9, 2013)  
The clue is in the question. A digit by definition is any number between 0 & 9.

N = 1120.
The digits of 1120 are the digits, 1, 1, 2, 0.
There are four of them.

Hope that helps.

Sahid said: (Jun 28, 2013)  
8 will divide each number and leave reminder 1, and 8 is greatest among all of the option.

Manish said: (Aug 28, 2013)  
People try dividing each number by 8 and you will get the same remainder.

Adarsha Gowda Mandya said: (Sep 25, 2013)  
Hello @Sarswathi, why are you taking 10 to multiply number after the subtraction. Could you please tell me?

Gopalakrishnan said: (Oct 24, 2013)  
@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.

Suleman said: (Dec 4, 2013)  
People try dividing each number by 8 and you will get the same remainder.

Ravi Jain said: (Dec 13, 2013)  
EASY ONE NO. ARE 6905, 4665, 1305

SO [(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.

NOW 1+1+2+0 = 4.

Chandra said: (Jan 12, 2014)  
[(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.

@Ravi can you explain how you did this step?

Srinivas said: (Jan 18, 2014)  
Question :

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 ____ .

Solution :

First we will find out N :

Use the formula :

Dividend = Divisor * Quotient + Remainder.

Given information :

1305 = N * x + Remainder ...equation 1.
4665 = N * y + Remainder ...equation 2.
6905 = N * z + Remainder ...equation 3.

Remainder in all the above three equations are equal but not known. Hence we eliminate them by performing subtraction as follows :

4665 - 1305 = N * (y-x) ...equation 2 - equation 1.
6905 - 4665 = N * (z-y) ...equation 3 - equation 2.
6905 - 1305 = N * (z-x) ...equation 3 - equation 1.

i.e.

3360 = N * (y-x) ...equation 4.
2240 = N * (z-y) ...equation 5.
5600 = N * (z-x) ...equation 6.

Values of x, y, z and N are not known. We are not concerned about values of x, y and z. From the equations 4, 5 and 6 it is clear N is a factor of 3360, 2240 and 5600. As the question states N is the GREATEST number, we will find out Highest Common Factor(HCF) of 3360, 2240 and 5600 which will be our N.

Factors :

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.

Hence N = 1120.

We have to find SUM of digits of N which is : 1+1+2+0 = 4.

Hence 4 is the answer(option A).

I took help of comments of @M.Harish, @Anchit and @Saraswati to provide the solution. A big thanks to them. The solution is meant to ease out the confusion regarding the question. I have tried to explain in detail hence its lengthy. Its recommended people solve the problem with as many shortcuts as possible in competitive exams where time is a constraint.

Sujit said: (Jun 13, 2014)  
Why to take the difference between no.s? I don't get it.

Mehak said: (Jun 19, 2014)  
Why we have to add 1+1+2+0? I'm not getting it. Please someone explain it.

Bidya Sagar Behera said: (Jun 30, 2014)  
I can't understand because of taking H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305).

Swaminathan Vembu said: (Jun 30, 2014)  
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.

Swaminathan said: (Jul 1, 2014)  
Impossible to calculate the GCD without successive difference:

a. 6905 - 4665.
b. 4665 - 1305.
c. 6905 - 1305.

Interestingly when you replace the values used in the above method to calculate the successive difference as illustrated below :

a. 6905 - 1305 (c above).
b. 6905 - 4665 (a above).
and
c. 4665 - 1305 (above it would replace the statement => 4665 - 1305).

It will convincingly still give rise to a G.C.D = 1120.

Geetika said: (Jul 15, 2014)  
Please anyone tell me the concept of 1120?

Nandhini said: (Jul 18, 2014)  
Please can you explain how to calculate HCF and explain?

Pallavi said: (Jul 24, 2014)  
Please can you explain how to calculate HCF ?

Sasank said: (Sep 8, 2014)  
Here taking difference is mentioned because we need to find h.c.f.

Here 1305=pq1+r;
4665=pq2+r;
6905=pq3+r;.here p is divident, q=quotient, r=reminder.

We want to find p so,
Totally 3 equations so,
P(q2-q1) = 3360.
p(q3-q2) = 2240.
p(q3-q1) = 5600 now there are 3 p's we need p that is h.c.f of (3360, 2240, 5600) here finding hcf by taking difference it will make easy now p = 1120 and (1+1+2+0) = 4.

Hope you all understood.

Rohit said: (Oct 1, 2014)  
How to find HCF fast in simple method?

Vikram said: (Dec 1, 2014)  
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.

Suraj said: (Dec 10, 2014)  
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.

J.Manisha said: (Jan 27, 2015)  
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: (Jun 20, 2015)  
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.

Pavan said: (Jul 6, 2015)  
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.

Shrinivasmutagar said: (Jul 23, 2015)  
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?

Renu said: (Aug 3, 2015)  
Hey can anyone tell how to find HCF fast for big numbers?

Himank said: (Sep 13, 2015)  
@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.

Mukesh Priye said: (Sep 15, 2015)  
Common no. of each row is 22227 & 10 hence multiple of that Product equal to 1120.

Rekha Kharode said: (Dec 19, 2015)  
Please explain if someone knows why select only HCF?

Neha said: (Jan 10, 2016)  
How 1120?

Munsur said: (Mar 15, 2016)  
Will anyone explain the logic behind the Division Method of Finding HCF or GCD?

Niki said: (May 17, 2016)  
Thank you @Srinivas.

Prakash said: (May 25, 2016)  
Let r be the remainder then the numbers (1305 - r) , (4665 - r) and (6905 - r) are exactly divisible by the required greatest number N.

We know that if two numbers are divisible by a certain number, then their difference is also divisible by that number.

Hence, the numbers.

(4665 - r) - (1305 - r) = (4665 - 1305),
(6905 - r) - (4665 - r) = (6905 - 4665),
(6905 - r) - (1305 - r) = (6905 - 1305).

They are divisible by the required number.

Therefore, the greatest number N = HCF of (4665 - 1305), (6905 - 4665) and (6905 - 1305).

Dev said: (Jun 22, 2016)  
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.

Anoop Singh said: (Jul 13, 2016)  
Anyone explain how the sum of digits in N = (1 + 1 + 2 + 0) = 4?

Ity said: (Jul 20, 2016)  
What is the HCF of no's 1365, 4665, 6905?

Komathy said: (Jul 30, 2016)  
I Can't understand the N value, please explain the sum of digits N = (1 + 1 + 2 + 0) = 4?

Sudhanshu said: (Aug 13, 2016)  
Will you explain the below?

3360:2240:5600.

1120:1120:1120.

How we got this 1120 in common.

Becky said: (Aug 18, 2016)  
Thanks @Saraswati.

Deepa said: (Aug 19, 2016)  
How (1 + 1 + 2 + 0)?

Please tell me.

Madhu said: (Aug 24, 2016)  
Thanks for excellent explanation @Yogesh.

Moni Kumari said: (Aug 27, 2016)  
Thanks for this complete explanation @M. Harish and @Srinivas.

Debita said: (Sep 21, 2016)  
@Srinivas.

Thanks, this is the perfect explanation like a true mathematician.

Bunny said: (Sep 23, 2016)  
Why should we subtract before finding out the HCF? can anybody tell me?

Faiz said: (Oct 1, 2016)  
1305, 4665, 6905 -----> These are the numbers.

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

So, 1305 = hcf * x + remainder -----> (1),

4665 = hcf * y + remainder -----> (2),

6905 = hcf * z + remainder -----> (3),

Where x, y, z are respective quotients.
Now we take (2) - (1).
3360 =hcf (y-x) -----> (4),
From (3) - (2) we have
2240 = hcf (z - y) -----> (5),
From(3) - (1) we have
5600 = hcf (z - x) -----> (6),
Now we can see from the equation (4), (5), (6) hcf of the three numbers is the hcf of the three numbers. So the hcf of 6905, 4665 and 1305 = hcf of the three numbers 5600, 3360, 2240 = 1120.

Shreya said: (Nov 3, 2016)  
Thank you @Harish.

Tushar said: (Dec 9, 2016)  
@All.

Whenever we see highest, greatest, largest, etc -> this kind words in question, then take HCF.

And whenever we see smallest, lowest, etc -> this kind of words in question, then take LCM.

Mounika said: (Dec 20, 2016)  
Thanks for Excellent explanation @Suraj.

Mounika said: (Dec 20, 2016)  
Thank you all for explaining it.

Supriya said: (Dec 28, 2016)  
Thanks @Priya.

Hanumant said: (Feb 7, 2017)  
Thank you all for explaining it.

Shirish said: (Feb 18, 2017)  
Thanks to all for the correct explanation.

Diya said: (May 24, 2017)  
Thanks @Prakash.

Amit Yadav said: (Jun 18, 2017)  
Any no. Can be represented as of the form D*q+r right. so we want remainder same in each case. So
1305=N*q1+r
4665=N*q2+r
6905=N*q3+r
Now, now subtracting 2 eq from 1 then ..3 to 2 then 3 to 1 we get N*(q2-q1)=3360.
Similarly, we will get the rest ones. now here, we just want the greatest no. Which divides all this num3360 2240 nd 5600 ....so simply take the hcf off all. We will get N=1120.

I hope this helps you.

Dilz said: (Jun 22, 2017)  
Subtraction of these numbers cancels remainder but how?

Sumit said: (Aug 12, 2017)  
For
4*2+3=11
4*8+3=35
4*11+3=47
Here you all can see that taking same remainder h.c.f seems 4.
H.C.F. OF (35-11),(47-11) and (47-35)=12.
So taking this method answers can differ.

Jayant Verma said: (Oct 2, 2017)  
8*163 = 1304 thus 1 remainder
583*8 = 4664 thus 1 remainder
863*8 = 6904 thus 1 remainder
So the answer should be 8.

Meghana said: (Oct 6, 2017)  
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.

Meher Priyanka said: (Oct 6, 2017)  
N(q2-q1)=(4665-1305)=3360
N(q3-q2)=(6905-4665)=2240
N(q3-q1)=(6905-1305)=5600

N(q1+q2+q3)=3360+2240+5600=11200

N(1+1+2+0+0)=4.

Guru Prasad said: (Nov 9, 2017)  
Subtract the numbers such as,

4665-1305 = 3360 and 6905 - 4665 = 2240 then subtracting the obtained results.

3360-2240 = 1120. Therefore 1 + 1 + 2 + 0 = 4.

Sibaram said: (Dec 28, 2017)  
What if the question is for finding smallest such number? Is it same like we have to find lcm of difference between numbers?

Samir Alvani said: (Dec 31, 2017)  
When you divide 1305, 4665 and 6905 by 1120 you will get the same remainder 185 so N is 1120 so the sum of the digits is 1+1+2+0=4.

Gopika said: (Jan 11, 2018)  
@Saraswati.

I can't understand why you are multiplying 10?

Please explain it.

Aman Srivastava said: (Jan 30, 2018)  
Thanks @Yogesh.

Kashi said: (Feb 18, 2018)  
Thanks @Prakash.

Princely said: (May 10, 2018)  
Can we try it in this way?

3+3+6+0= 12
2+2+4+0= 8,
5+6+0+0= 11,
12+8+11= 31,
3+1= 4.

Sanchay said: (May 24, 2018)  
We subtracted 1305 from 4665, 4665 from 6905 and 1305 from 6905 because we needed n as a common factor which can be obtained this way.

Let us consider 1305 as q1
4665 as q2
And 6905 as we
Then,

4665-1305 = N(q2 -q1).
6905-4665 = N(q3-q2 ).
6905-1305 = N(q3-q2 ).

Now we get here N as the factor of all 3 digits and we need to find the greatest factor so, take out the HCF of all three numbers i.e 1305, 4665, 6905 = 1120.

And this is your answer if your question asks no. Of digits in the number simply add the digits i.e 1+1+2+0=4.

Hope this helped.

Akshitha said: (Jun 4, 2018)  
Thnks @Sanchay.

Ramachandra said: (Jun 8, 2018)  
Good explanation @Yogesh.

Saniya Taj said: (Jul 11, 2018)  
Easy to understand. Thanks @Suraj.

Akhil said: (Dec 27, 2018)  
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.

Provith said: (Jan 11, 2019)  
Thanks all for explaining the answer.

Kotresh said: (Feb 21, 2019)  
Good explanation, Thanks @Yogesh.

Sachin said: (Feb 23, 2019)  
Thanks @Saraswati.

Bimal said: (Mar 1, 2019)  
Let N be the greatest number that will divide 17, 41 and 77, leaving the same remainder in each case. Then the sum of the digits in N is:

It is easy to understand this with a simple example, and the logic behind this is so obvious.
Take number 12 (you can get any number you desire and practice it!).

Dividend = Divisor X Quotient + remainder (when the remainder is a constant!).

12 X 1 + 5 = 17.
12 X 3 + 5 = 41.
12 X 6 + 5 = 77.

Now, take the differences of 17, 41 and 77 in order considering the smallest, the middle and the largest numbers.

(41-17) = 24; (77-17) = 60; (77-41) =36.

Now take 24, 60, and 36 and find the HCF or HCD of these numbers: it is equal to 12.
Look at the "Divisor" used at the beginning. So, the divisor of that arrangement is equal to the HCF of 24, 60 and 36. When the remainder is constant, we can observe this pattern with any set of numbers arranged in this way.

The sum of digits of N = 1+2 =3.

Harini said: (Jun 22, 2019)  
@Anchit.

We should subtract remainder. i.e 12-2=10.

Girish Singh Bisht said: (Jul 7, 2019)  
Thanks for explaining @Priya.

K.Manoj said: (Jul 19, 2019)  
Thanks @Saraswati.

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