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Factor problem

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by anantbhatia » Fri Sep 10, 2010 10:36 pm
The following qn is from a set, most probably previous GMAT qns:-

If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what
is the value of r?
(1) 2 is not a factor of n.
(2) 3 is not a factor of n.

OA C
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Source: — Data Sufficiency |

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by Rahul@gurome » Sat Sep 11, 2010 12:27 am
Solution:
Consider (1) alone.
Let n = 5.
So (n-1)(n+1) = 4 * 6 = 24.
24 on being divided by 24 gives a remainder of 0.
Next let n = 9.
So (n-1)(n+1) = 8*10 = 80.
80 gives a remainder of 8 on being divided by 24.
Since nothing definite can be said, (1) alone is not sufficient.
Next consider (2) alone.
Take the examples of n = 5 and n = 4.
When n = 5, remainder is 0 on dividing (n-1)(n+1) by 24.
When n = 4, (n-1)(n+1) = 3*5 = 15 and remainder is 15.
Again nothing definite said be said from (2) alone.

Next combine both the statements together and check.
Since both 2 and 3 do not divide n, n is of the form 6k+1 or 6k+5, k being an integer.
Let n = 6k+1.
So (n-1)(n+1) = (6K+1-1)(6k+1+1) = 6k(6k+2) = 12k(3k+1).
If k is even, then 12k is a multiple of 24 and hence 12k(3k+1) is divisible by 24. Or remainder r is 0.
If k is odd 3k+1 is even . Or 12(3k+1) is divisible by 24 and so 12k(3k+1) is also divisible by 24. So remainder r is 0.
Next let n = 6k+5.
Or (n-1)(n+1) = (6k+5-1)(6k+5+1) = (6k+4)(6k+6) = 12(k+1)(3k+2).
If k is even (3k+2) is even or 12(3k+2) is divisible by 24 and so 12(k+1)(3k+2) is also divisible by 24.
If k is odd, (k+1) is even and 12(k+1) is divisible by 24. Or 12(k+1)(3k+2) is also divisible by 24
In any case remainder r is zero.

Or both statements together are sufficient and answer is (C).
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by pradeepkaushal9518 » Sat Sep 11, 2010 1:34 am
n is positive means n>0 r is remainder when (n-1)(n+1) is divided by 24 r=?

1. 2 is not a factor of n

so possible n= 3,5,7,9 etc

n=3 2*4=8 is not divisble

fr n=5 4*6=24 divided by 24 r=0

n=7 6*8=48 divided by 24 r=0
n=9 8*10 r= 8 so

this is not suff A and D out

2. 3 is not a factor of n

n= 4, 7,8,10 etc

n=7 6*8=48 r=0

n=8 7*9=63 r=15 so r is not fixed so insuff B out left C and E

combine 2 and 3 are not factor of n

n=4,5,7,11 etc

for all these value r=0 so suff

hence C
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by anantbhatia » Sat Sep 11, 2010 1:36 am
Thanks Pradeep.. you made it simpler! :)
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by Ian Stewart » Sat Sep 11, 2010 5:20 am
anantbhatia wrote:The following qn is from a set, most probably previous GMAT qns:-

If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what
is the value of r?
(1) 2 is not a factor of n.
(2) 3 is not a factor of n.

OA C
It's a GMATPrep question. You can also look at it as follows: n-1, n, n+1 are three consecutive integers.

From Statement 1, n is odd, so n-1 and n+1 are consecutive even numbers. Since even numbers alternate between multiples of 2 and multiples of 4, one of n-1 or n+1 is divisible by 2, and one is divisible by 4. So (n-1)(n+1) is divisible by 8.

From Statement 2, n is not divisible by 3, and among any three consecutive integers you always have exactly one multiple of 3, so either n-1 or n+1 is divisible by 3, and (n-1)(n+1) is divisible by 3.

So combining the Statements we know (n-1)(n+1) is divisible by 3 and by 8, and therefore by 24.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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