Civil Engineering Mechanical Engineering Chemical Engineering Networking Database Questions Computer Science Basic Electronics Digital Electronics Electronic Devices Circuit Simulation Electrical Enigneering Engineering Mechanics Technical Drawing

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

@ : Home > Aptitude > Problems on H.C.F and L.C.M > General Questions - Discussion

"Weakness of attitude becomes weakness of character."
- Albert Einstein
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

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: (Mon, Sep 6, 2010 11:31:46 AM) How would we know that H.C.F. of 3360, 2240 and 5600 = 1120.

 Krishna Kumar said: (Sat, Sep 11, 2010 01:47:02 PM) Sorry I can't understand why we are taking H. C. F. Of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).

 Priya said: (Thu, Dec 9, 2010 02:00:50 AM) 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: (Mon, Dec 13, 2010 01:49:35 PM) 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: (Tue, Dec 14, 2010 03:05:35 AM) How we know by seeing the numbers the H.C.F ?

 Saraswati said: (Fri, Dec 24, 2010 03:49:58 AM) 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: (Tue, Jan 11, 2011 10:48:07 AM) Thanx priya.

 Jinx said: (Sun, Jan 23, 2011 05:59:25 AM) Thanks sarsu.

 Vishal said: (Tue, Feb 1, 2011 10:36:42 AM) Thanks saraswati.

 Swati said: (Thu, Feb 17, 2011 08:37:50 AM) Why we are taking the HCF of (4665 - 1305) , (6905 - 4665) and (6905 - 1305). Plzzz explain.

 Zara said: (Thu, Apr 28, 2011 12:35:26 AM) Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4, how does it come and What N indicates?

 Shanmugar said: (Fri, Apr 29, 2011 05:21:47 AM) 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: (Thu, Jun 30, 2011 11:28:39 AM) 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: (Tue, Jul 5, 2011 03:17:45 AM) Good explanation yogesh:).

 Ramya said: (Thu, Jul 14, 2011 11:23:11 AM) Thankx priya.

 Chirag said: (Sun, Aug 7, 2011 03:07:02 PM) Saraswati your explanation is superb.

 Ashwin said: (Sat, Aug 27, 2011 08:18:38 AM) Why are they finding H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) ? Please explain.

 Kumar said: (Thu, Sep 22, 2011 08:07:47 PM) Thanks yogesh.

 Souravikiran said: (Sat, Sep 24, 2011 12:16:19 AM) 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: (Sun, Sep 25, 2011 03:37:32 PM) Thanks! saraswati.

 Nikhil said: (Wed, Sep 28, 2011 02:19:37 PM) 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: (Mon, Oct 10, 2011 01:10:36 PM) We have to find greatest num, hcf highest common factor. So we did.

 Yuvaraj said: (Thu, Oct 13, 2011 10:54:32 AM) Hey please explain why we take hcf of 4665-1305, 6905-4665 ?

 Sri said: (Wed, Oct 19, 2011 10:56:42 PM) 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: (Thu, Oct 27, 2011 06:55:32 PM) Yogesh thanks

 Madhu said: (Sat, Jan 7, 2012 08:11:26 PM) Using successive division to find HCF of 1305,4665 and 6905 is 5 How it is 1120 ?

 Saravanan said: (Fri, Feb 17, 2012 09:32:21 PM) How will be taken sum of n digit(1+1+2+0)=4?

 Piyush said: (Sun, Feb 19, 2012 10:16:32 PM) 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: (Sat, Mar 3, 2012 03:20:56 PM) Hey anyone here can please expain me why we take hcf of (4665 - 1305) , (6905 - 4665) and (6905 - 1305). ? please give explaination.

 Sat said: (Fri, Apr 13, 2012 02:43:41 AM) Yogesh is pretty much right.

 Nagarjuna D said: (Mon, May 14, 2012 05:36:31 PM) I can't understand why we have to subtract one among other?

 Rajeswari R said: (Sat, Jun 9, 2012 10:51:11 PM) @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: (Wed, Jun 27, 2012 12:43:38 PM) 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: (Thu, Jul 19, 2012 02:07:15 PM) Why it is not taken 5. It also gives the same remainder 0.

 M.Harish said: (Tue, Jul 24, 2012 08:32:16 PM) 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: (Sun, Aug 5, 2012 03:16:12 PM) 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: (Sat, Aug 18, 2012 11:23:58 AM) Thanks saraswathi. I get answer for this types of problems by following your explanation. Thanks a lot.

 Anchit said: (Sat, Aug 18, 2012 01:48:53 PM) 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: (Tue, Aug 21, 2012 04:57:58 PM) @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: (Tue, Aug 28, 2012 08:53:02 PM) 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: (Sat, Feb 9, 2013 12:02:15 AM) 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.