- dear friends

this is a straight CAT 2002 pblm.

7^6n-6^6n, where n is an integer >0 is divisible by

1. 13

2. 127

3. 559

4. all of these

answer given is 1. (i.e) 13.

my doubt is

1,x^n - a^n is exactly divisible by (x+a) when n is even

2,x^n - a^n is exactly divisible ny (x-a) irrespective of n

by this funda 1, we get two answers

1. 1st case (7^3)^2n-(6^3)^2n in this case it will be 7^3+6^3

which is 559.

2. 2nd case (7)^6n-(6)^6n in this case it will be 7+6 which is

13.

by funda 2, we get two answers

1. 1st case (7^2)^3n-(6^2)^3n in this case it is x-a so it is

13

2. 2nd case (7^3)^2n-(6^3)^2n in this case it is x-a so 7*7*7 -

6*6*6 which is 127

so the answer should be all of these.

but in the cat booklet they have taken into considetation only one

aspect. so how to go about it??

pblm 2

a watch ticks 90 times in 95 seconds and another watch ticks 315

times in 323 seconds. if both the watches are started together,

how many times will they tick together in the first hour?

ans is 101.

kly help me out

thanx and regards

subbu

___________________________________________________

Medicine meets Marketing; Dr. Swati Weds Jayaram.

Rediff Matchmaker strikes another interesting match !!

Visit http://rediff.com/matchmaker?2 - hi subbuwell for the first problem, i have arrived at the same answer, i.e. all the options marked are correct. so perhaps there was a typing mistake in the cat booklet answers!!!for your second problem solution will be as follows:The first watch ticks once every 95/90 seconds, and the second ticks once every 323/315 secondsso they will tick together at the LCM of 95/90 and 315/323 i.e. 323/9 secondsTherefore, every hour or every 3600 seconds they will tick together 3600/ (323/9) times or 100 times ( ignoring the remainder 100/323). But since they were started together, we add one more to the number of times they tick together hence answer is 101Hope this would have helpedregardsmadan
wrote:*subbu narayanaswamy <samarpan_subbu@...>*`dear friends`

this is a straight CAT 2002 pblm.

7^6n-6^6n, where n is an integer >0 is divisible by

1. 13

2. 127

3. 559

4. all of these

answer given is 1. (i.e) 13.

my doubt is

1,x^n - a^n is exactly divisible by (x+a) when n is even

2,x^n - a^n is exactly divisible ny (x-a) irrespective of n

by this funda 1, we get two answers

1. 1st case (7^3)^2n-(6^3)^2n in this case it will be 7^3+6^3

which is 559.

2. 2nd case (7)^6n-(6)^6n in this case it will be 7+6 which is

13.

by funda 2, we get two answers

1. 1st case (7^2)^3n-(6^2)^3n in this case it is x-a so it is

13

2. 2nd case (7^3)^2n-(6^3)^2n in this case it is x-a so 7*7*7 -

6*6*6 which is 127

so the answer should be all of these.

but in the cat booklet they have taken into considetation only one

aspect. so how to go about it??

pblm 2

a watch ticks 90 times in 95 seconds and another watch ticks 315

times in 323 seconds. if both the watches are started together,

how many times will they tick together in the first hour?

ans is 101.

kly help me out

thanx and regards

subbu

___________________________________________________

Medicine meets Marketing; Dr. Swati Weds Jayaram.

Rediff Matchmaker strikes another interesting match !!

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