Comments on: HCF Question

By: wael.salman

wael.salman — Wed, 18 Jun 2008 08:58:54 +0000

Vikas aggarwal

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

Can you explain that please??

By: SD

SD — Tue, 10 Jun 2008 12:42:59 +0000

yeah ans is 4

By: Atul

Atul — Mon, 02 Jun 2008 10:16:38 +0000

By: urefriendforever

urefriendforever — Thu, 29 May 2008 16:47:57 +0000

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

By: piyush

piyush — Tue, 27 May 2008 13:54:57 +0000

ANSEWR is 4 pair

By: Vikas aggarwal

Vikas aggarwal — Tue, 27 May 2008 11:26:36 +0000

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)

By: Varun Gupta

Varun Gupta — Sun, 25 May 2008 07:34:33 +0000

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

By: Varun Gupta

Varun Gupta — Sun, 25 May 2008 07:26:49 +0000

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.

By: pavan

pavan — Sat, 24 May 2008 04:38:23 +0000

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

The answer is 4 pairs – choice a

By: Math Crazy

Math Crazy — Fri, 23 May 2008 19:30:50 +0000

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