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If n is a positive Integer and r is the remainder when [#permalink]
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10 Dec 2004, 07:30
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If n is a positive Integer and r is the remainder when (n1)(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



Director
Joined: 03 Nov 2004
Posts: 853

I will go with â€˜Eâ€™ on this one.
From statement 1: n is odd.
From statement 2: The sum of the digits of n is not divisible by 3
Both put together: n can be 5, 7, 11, 13, so depending on n the remainder r varies.



Director
Joined: 31 Aug 2004
Posts: 606

Dookie style !!
1st statement : n is obviously odd which means (n1)(n+1) is a multiple of 4  not enough
2nd statement : 3 is not a factor of n so (n1)(n+1) is a multiple of 3
2 combined : multiple of 12 so remainder could be either 0 or 12.
I pick E



Senior Manager
Joined: 19 May 2004
Posts: 291

Gotcha again !
Wanna give it another try ?



Manager
Joined: 28 Aug 2004
Posts: 205

C.
n would be a prime above 3. Remainder is 0.



Senior Manager
Joined: 19 May 2004
Posts: 291

N doesn't have to be a prime. It could be.. 85 for example.



Manager
Joined: 07 Nov 2004
Posts: 89
Location: London

C
[1] n is odd and that both (n+1) and (n1) are divisible by 2
[2] Either (n1) or (n+1) is divisible by 3
consequently (n1)(n+1) is divisible by 12. Since n has to be greater than
24 for (n1)(n+1) to leave a remainder when divided by 24 it is therfore divisible by 12 and at least its "first" multiple, that is, 24.
so both statements together are sufficient.
The remainder r = 0.



Manager
Joined: 28 Aug 2004
Posts: 205

yes, I tried n^21 for the first few (prime) numbers... but it could work for others too.



CIO
Joined: 09 Mar 2003
Posts: 463

oxon wrote: C
[1] n is odd and that both (n+1) and (n1) are divisible by 2 [2] Either (n1) or (n+1) is divisible by 3
consequently (n1)(n+1) is divisible by 12. Since n has to be greater than 24 for (n1)(n+1) to leave a remainder when divided by 24 it is therfore divisible by 12 and at least its "first" multiple, that is, 24.
so both statements together are sufficient.
The remainder r = 0.
I think it's C too, but I have a different explanation.
Look at what they're saying. n isn't divisible by 2 or 3. That means n is an odd number that's not divisible by 3.
That means it could be:
5
7
11
13
17
19
23
25
and on and on. In all cases, the numbers above and below each of these will either be divisible by 4 or by 6. It just works out that way. So when we multiply them together, the answer will be divisible by 24.
Even if we start with 5, this will be true.
Great Q dookie!



Senior Manager
Joined: 19 May 2004
Posts: 291

Great, C it is.
1st statement : n is odd. (n1)(n+1) is a multiple of 8
2nd statement : 3 is not a factor of n so (n1)(n+1) is a multiple of 3
2 combined : multiple of 24. Remainder = 0.



Manager
Joined: 07 Nov 2004
Posts: 89
Location: London

ian777  great stuff!
what I meant in my post is that if you pick n<24 (n<23 in fact) (n1)(n+1)
will be divisible by both 4 and 3. If you pick n>23 then (n1)(n+1) will be divisible by at least one multiple of 12 hence 24.
Dookie  You rock!



Director
Joined: 03 Nov 2004
Posts: 853

Dookie wrote: Great, C it is.
1st statement : n is odd. (n1)(n+1) is a multiple of 8
2nd statement : 3 is not a factor of n so (n1)(n+1) is a multiple of 3
2 combined : multiple of 24. Remainder = 0.
Dookie, Can you explain how from statement 1 you got that (n1)(n+1) is a multiple of 8



CIO
Joined: 09 Mar 2003
Posts: 463

rthothad wrote: Dookie wrote: Great, C it is.
1st statement : n is odd. (n1)(n+1) is a multiple of 8
2nd statement : 3 is not a factor of n so (n1)(n+1) is a multiple of 3
2 combined : multiple of 24. Remainder = 0. Dookie, Can you explain how from statement 1 you got that (n1)(n+1) is a multiple of 8
Since n is odd, n1 and n+1 are both even, and, in fact, they are consecutive even number. Consecutive even numbers alternate between being multiples of 4 and not being. So in any two, one will be divisible by 2 only, and the other will be divisible by 4. So multiplying them together gives a number with 3 two's in it, or a multiple of 8.



Director
Joined: 31 Aug 2004
Posts: 606

Great question Dookie, as usual !
In fact simply picking numbers provided the answer... my mistake...
Than you Ian for your insight.



Director
Joined: 03 Nov 2004
Posts: 853

Thanks ian7777 for your explanation



Director
Joined: 19 Nov 2004
Posts: 556
Location: SF Bay Area, USA

This is a very good one.
1) If 2 is not a factor of n then n could be 1,3,5,7... (set of odd numbers)
(n1)(n+1) ranges from 0, 8, 24 ...
We can't get the same remainder if (n1)(n+1) is divided by 24
2) If 3 is not a factor of n then n could be 1,2,4, 5...
(n1)(n+1) ranges from 0, 3, 15 ...
We can't get the same remainder if (n1)(n+1) is divided by 24
When both statements are combined
n could be 1 5 7 11 ..
All these numbers or odd. Any odd number is preceeded and followed by an even number. With n starting from 5, it turns out that (n1)(n2) will have the factor 3*2*2*2 = 24 (proof for this might be complicated; by trial). So remainder when divided by 24 will be 0.
For n =1, (n1)(n+1) = 0, so remainder is still 0.
C works Ok.
This question seems a bit hard to appear on a GMAT and to be solved in 2 minutes, because we have to extrapolate the result by substitution w/o a clear proof.
You never know!










