HCF Question

The sum of 2 positive numbers is 1215. HCF is 81. How many pairs of such integers are possible?

a. 4

b. 6

c. 7

d. 8

e. 14

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1. wael.salman says:

   June 18, 2008 at 2:28 pm

   Vikas aggarwal

   Why you are looking for unique pairs which doesn’t have any common factor other than 1.

   Can you explain that please??

   Reply
2. SD says:

   June 10, 2008 at 6:12 pm

   yeah ans is 4

   Reply
3. Atul says:

   June 2, 2008 at 3:46 pm

   4

   Reply
4. urefriendforever says:

   May 29, 2008 at 10:17 pm

   Answer is 4, consider this

   1215 / 81 = 15

   so 81 + 81 X 14 = 1215 , hcf is 81
   81 X 2 + 81 X 13 = 1215 , hcf is 81
   81 X 3 + 81 X 12 = 1215 , hcf is 81 X 3 because 81 X 12 = 81 X 3 X 4
   keep going like this till 81 X 6 and you will get 4 pairs

   Reply
5. piyush says:

   May 27, 2008 at 7:24 pm

   ANSEWR is 4 pair

   Reply
6. Vikas aggarwal says:

   May 27, 2008 at 4:56 pm

   A.

   1215=81(x+y)
   x+y=15

   Now we have to just look for unique pairs which doesnt have any common factor other than 1. such pairs are (1,14), (2,13), (4,11) & (7,8)

   Reply
7. Varun Gupta says:

   May 25, 2008 at 1:04 pm

   I also agree with Rajib, i just missed on it. Anyways ans is (A)

   Reply
8. Varun Gupta says:

   May 25, 2008 at 12:56 pm

   Answer is C for sure. Its explanation is as follows:
   Let 1st No be A = 81*a (a= any natural no)
   Let 2nd No be B = 1215-81*a (As sum of 2 no’s A+B=1215)
   PUT a=1, A=81 and B=1134
   a=2, A=162, B = 1053
   a=3, A= 243, B = 972
   a=4, A = 324, B= 891
   a=5, A = 405, B = 810
   a=6, A= 486, B= 729
   a=7, A= 567, B = 648
   After a = 7, we start getting the same values again. So this could be seen here that we have 7 pairs.

   Reply
9. pavan says:

   May 24, 2008 at 10:08 am

   I agree with Rajib…
   I just overlooked at it… Thanks for that Rajib

   The answer is 4 pairs – choice a

   Reply
10. Math Crazy says:

    May 24, 2008 at 1:00 am

    Zero is a multiple of every integer so the answer is 7.

    Reply
11. Rajib Banerjee says:

    May 23, 2008 at 9:27 pm

    Answer is “4″ or the option 1.
    From the pairs shown by Pavan, only {81,1134}, {162,1053}, {324,891} and {648,567} has HCF of 81 and sum equal to 1215. Prove me wrong????

    Reply
12. pavan says:

    May 23, 2008 at 3:41 pm

    Ans is b – 6 pairs –
    {81,1134}, {162,1053}, {243,972}, {324,891},{729,486} and {648,567}

    HCF is 81… and sum of 2 no’s is 1215
    Consider the multiples of 81till less than 1215
    Therefore we have {81,162,243,324,405,567,648,729,810,891,972,1053,1134,1215}
    Among the set, consider the no’s which can give the sum 1215
    we have – {81,1134}, {162,1053}, {243,972}, {324,891}, {405,810}, {729,486} and {648,567} giving a sum of 1215…
    Among all the above pairs, Pair {405,810} which gives HCF of 135
    and rest all gives 81 as HCF.

    Therefore we have 6 pairs.

    Reply
13. Patel says:

    May 23, 2008 at 2:10 pm

    I think the answer is C

    Reply