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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)
Problems on H.C.F and L.C.M
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 15 of 16.
Kumar said:   1 decade ago
Thanks yogesh.
Ashwin said:   1 decade ago
Why are they finding H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) ?

Please explain.
Chirag said:   1 decade ago
Saraswati your explanation is superb.
Ramya said:   1 decade ago
Thankx priya.
Savitha said:   1 decade ago
Good explanation yogesh:).
Yogesh said:   1 decade ago
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
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.
Zara said:   1 decade ago
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4, how does it come and What N indicates?
Swati said:   1 decade ago
Why we are taking the HCF of (4665 - 1305) , (6905 - 4665) and (6905 - 1305).

Plzzz explain.
Vishal said:   1 decade ago
Thanks saraswati.
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