Please help!!!!

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Please help!!!!

by eustudent » Mon Jun 01, 2009 9:00 pm
If N is a positive integer and R is the reminder when (N-1)(N+1) is divided by 24,w hat is the value of R?

1) N is not divisible by 2

2) N is not divisible by 3

Answer is C

Ok, considering both statements together, N could be 7, 13, 19, etc.
Then R will have several values, therefore E should be the right answer.

I believe, this problem is mistaken because we cannot find the value of R with the information given.

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by mike22629 » Tue Jun 02, 2009 4:47 am
You are right that N can be those values, but you have not completely thought out the problem.

Lets say n = 7

n+1=8 and n-1 = 6
8*6 = 48

R = 0

N = 13
N+1 = 14 and n - 1 = 12

12*14 = 168

R = 0

So, no matter what the remainder is 0

Hint: You do not have to multiply all the possibilities out, all you have to do is find factors equaling 24 to make it divisible by 24

So:

12 * 14 = 2*2*3*7*2 - Since 2*2*2*3 = 24, this product is divisible by 24

You can do this with any possibility and always find factors that will equal 24.

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by mike22629 » Tue Jun 02, 2009 4:52 am
P.S.

N could also equal 1

n-1 = 0
n+1 = 2

0*2 = 0

R = 0

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by jba05d » Thu Jun 04, 2009 11:39 am
Another way to think about this problem is to consider what we know about consecutive numbers and prime factors. For example, the questions stem gives us a hint that we are dealing with a series of three consecutive digits (i.e. the dividend was originally N and the dividend is now (n-1)(n+1), therefore (n-1)(n)(n+1) is a set up three consecutive digits). We know that in a set of three consecutive digits that one of the digits must be divisible by the number 3, and that at least one of the other numbers is divisible by the number 2, depending on whether or not the first number is even or odd. Consequently we know that (N-1)(N+1) must have prime factors of at least 2 or 3 or 2, 2, and 3, depending on whether N is even or odd. The prime factors of 24 are 2X2X2X3.

The first statement tells us that N is not divisible by 2. Therefore, N is not even and N-1 and N+1 must both be even. Consequently, if N-1 is not 0 then (N-1)(N+1) has at least 2X2X2 it is prime box, that is n-1 must 1 must be even and N+1 is even and 2 higher then N-1. Furthermore, (N-1)(N+1) may have a three but we don't know yet. Therefore, Stmt 1 is insufficient. Statement two tells us that N is not divisible by three. Therefore, we know that either (N-1) or (N+2) must be divisible by three. However, since this statement tells us nothing about the sign of N we cannot determine what other factors are in the prime box of (N-1)(N+2).

Taken together, however, we are able to determine that N is odd and N is not divisible by 3. Therefore, if (N-1) is not zero then (N-1)(N+1) must contain 2X2X2X3 its in prime box. Since 24 has a 2X2X2X3 in its prime box the statements together are sufficient. If N-1 is zero the statements are still sufficient sine the remainder will still be zero.

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by Osirus@VeritasPrep » Thu Jun 04, 2009 12:01 pm
jba05d wrote:Another way to think about this problem is to consider what we know about consecutive numbers and prime factors. For example, the questions stem gives us a hint that we are dealing with a series of three consecutive digits (i.e. the dividend was originally N and the dividend is now (n-1)(n+1), therefore (n-1)(n)(n+1) is a set up three consecutive digits). We know that in a set of three consecutive digits that one of the digits must be divisible by the number 3, and that at least one of the other numbers is divisible by the number 2, depending on whether or not the first number is even or odd. Consequently we know that (N-1)(N+1) must have prime factors of at least 2 or 3 or 2, 2, and 3, depending on whether N is even or odd. The prime factors of 24 are 2X2X2X3.

The first statement tells us that N is not divisible by 2. Therefore, N is not even and N-1 and N+1 must both be even. Consequently, if N-1 is not 0 then (N-1)(N+1) has at least 2X2X2 it is prime box, that is n-1 must 1 must be even and N+1 is even and 2 higher then N-1. Furthermore, (N-1)(N+1) may have a three but we don't know yet. Therefore, Stmt 1 is insufficient. Statement two tells us that N is not divisible by three. Therefore, we know that either (N-1) or (N+2) must be divisible by three. However, since this statement tells us nothing about the sign of N we cannot determine what other factors are in the prime box of (N-1)(N+2).

Taken together, however, we are able to determine that N is odd and N is not divisible by 3. Therefore, if (N-1) is not zero then (N-1)(N+1) must contain 2X2X2X3 its in prime box. Since 24 has a 2X2X2X3 in its prime box the statements together are sufficient. If N-1 is zero the statements are still sufficient sine the remainder will still be zero.
These are the only problems that I can't solve at least some of the time. Would you have been able to figure these problems out without MGMAT numbers properties book? Is there any web resource that would help or should I just break down and purchase that book?

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by lunarpower » Fri Jun 05, 2009 2:29 am
osirus0830 wrote:These are the only problems that I can't solve at least some of the time. Would you have been able to figure these problems out without MGMAT numbers properties book? Is there any web resource that would help or should I just break down and purchase that book?
on REMAINDER problems, such as these, you should almost always be able to solve the problem by SYSTEMATIC PLUGGING-IN OF NUMBERS.

since remainders are fundamentally based on repetition - i.e., remainders themselves repeat in cycles, over and over again, predictably and systematically - you should be able to discover PATTERNS in them, simply by trying enough examples.

just GENERATE EXHAUSTIVE LISTS, try them, and, there you go.
the only catch with this method is that this is not fast, so you have to DECIDE RIGHT AWAY if you want to use this method.

the idea is to try the lists until one of two things happens:
(a) you see an obvious PATTERN
(b) you get "INSUFFICIENT"

as soon as either of these things happens, you are finished.

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here's a taste of this method, applied here:

statement (1)

this means odd numbers. so, try odd numbers, and see what happens.

n = 1 --> (n + 1)(n - 1) = 0 --> remainder when 0 is divided by 24 = 0
n = 3 --> (n + 1)(n - 1) = 8 --> remainder when 8 is divided by 24 = 8

if you don't like dividing numbers smaller than 24 by 24, then try the next two or three:
n = 5 --> (n + 1)(n - 1) = 24 --> remainder when 24 is divided by 24 = 0
n = 7 --> (n + 1)(n - 1) = 48 --> remainder when 48 is divided by 24 = 0
n = 9 --> (n + 1)(n - 1) = 80 --> remainder when 80 is divided by 24 = not 0 (we don't care what it is, as long as it's different from the previous answers, since that's "insufficient")

insufficient.

statement (2)

n = 1 --> (n + 1)(n - 1) = 0 --> remainder when 0 is divided by 24 = 0
n = 2 --> (n + 1)(n - 1) = 3 --> remainder when 3 is divided by 24 = 3

if you don't like dividing numbers smaller than 24 by 24, then try the next two or three:
n = 5 --> (n + 1)(n - 1) = 24 --> remainder when 24 is divided by 24 = 0
n = 7 --> (n + 1)(n - 1) = 48 --> remainder when 48 is divided by 24 = 0
n = 8 --> (n + 1)(n - 1) = 63 --> remainder when 63 is divided by 24 = not 0 (we don't care what it is, as long as it's different from the previous answers, since that's "insufficient")

insufficient.

together:
n = 1 --> (n + 1)(n - 1) = 0 --> remainder when 0 is divided by 24 = 0
n = 5 --> (n + 1)(n - 1) = 24 --> remainder when this is divided by 24 = 0
n = 7 --> (n + 1)(n - 1) = 48 --> remainder when this is divided by 24 = 0
n = 11 --> (n + 1)(n - 1) = 120 --> remainder when this is divided by 24 = 0
n = 13 --> (n + 1)(n - 1) = 168 --> remainder when 48 is divided by 24 = 0
ok, the pattern here is pretty obvious.
it's always 0.
sufficient.
Ron has been teaching various standardized tests for 20 years.

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Re: Please help!!!!

by lunarpower » Fri Jun 05, 2009 2:32 am
to the original poster,
Ok, considering both statements together, N could be 7, 13, 19, etc.
Then R will have several values, therefore E should be the right answer.
(this was fallacious logic)

this just goes to show one thing of which you should be well aware: DON'T MAKE ASSUMPTIONS.

this poster assumed that, because there were lots of possibilities for R, there would likewise be lots of different possible remainders.

it's not so.

unless you are ABSOLUTELY SURE that there is some sort of RULE that justifies the assumption(s) you're making, don't make assumptions. just DO the work!
especially in the case of such things as computing remainders, which doesn't take long at all.
the last thing you want is to miss a problem because you didn't do something that takes only a matter of seconds.
Ron has been teaching various standardized tests for 20 years.

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by eustudent » Sat Aug 15, 2009 7:12 am
thank you Ron!