HCF Question

Patel
#1. May 23rd, 2008, at 2:10 PM.

I think the answer is C

pavan
#2. May 23rd, 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.

Rajib Banerjee
#3. May 23rd, 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????

Math Crazy
#4. May 24th, 2008, at 1:00 AM.

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

pavan
#5. May 24th, 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

Varun Gupta
#6. May 25th, 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.

Varun Gupta
#7. May 25th, 2008, at 1:04 PM.

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

Vikas aggarwal
#8. May 27th, 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)

piyush
#9. May 27th, 2008, at 7:24 PM.

ANSEWR is 4 pair

urefriendforever
#10. May 29th, 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

Atul
#11. June 2nd, 2008, at 3:46 PM.

4

SD
#12. June 10th, 2008, at 6:12 PM.

yeah ans is 4

wael.salman
#13. June 18th, 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??