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Upon encountering an explanation to a problem in the MGMAT Number Properties, one question came to my mind: is number 0 divided by 24 (or any other number) can be said be divisible?

The problem is stated as below:

"If x^3 - x = p, and x is odd, is p divisible by 24?"

The explanation says that since p = (x-1)x(x+1), and x is odd, and (x-1) and (x+1) are consecutive multiple of 2, the statement is divisible by 24.

However, if (x-1) is 0, then, p equals to 0. In order for the statement to be a definite yes, 0/any number is considered divisible.

Does anyone know what is the 'divisibility rule' for zero to be divided by a number?

Thanks!
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if x^3-x=p, and x is odd, is p divisible by 24? Yes, p is divisible by 24.

x^3-x=x(x^2-1)=x(x+1)(x-1)=(x-1)x(x+1)

Thus; x^3-x is nothing but the product of 3 consecutive integers.

And we know x is odd; Means; x-1= even and x+1=even

In any set of 3 consecutive numbers; where there are two evens, we must have at least 3 2's as its prime factor. Also, there is always at least 1 3 as its prime factor. Thus, "x^3-x" is always divisible by 24 if x=odd.

does p have 2.2.2.3 in its prime box? if x=1, 1.1.1 - 1 = 0. 0/24 = no. if x = 3, 3.3.3 - 3 = 24 --> 24/24=1.

Hence, we can't say whether p is divisible by 24!!! ?? correct ??

0 *is* divisible by 24 (if you divide 0 by any positive integer, you get 0, which is an integer, so 0 is divisible by every positive integer). So in both the numerical examples you generated, you find that x^3 - x is divisible by 24. You'll always find that to be true, as fluke has explained above.
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The reason you use the formula is so that you don't have to waste time testing sets of numbers. I've seen problems where you'd have to test numbers for several minutes before you ran into one that contradicted the others, but with the formula, it's a matter of seconds.

instead of doing it the way you did (which for a newbie like me is a little complicated) was my way ok?

Quote:

x.x.x - x = p (means p is even).

does p have 2.2.2.3 in its prime box? if x=1, 1.1.1 - 1 = 0. 0/24 = yes. if x = 3, 3.3.3 - 3 = 24 --> 24/24=1.

therefore p is divisible by 24.

This is also a good way. It uses plugging in numbers to prove something.

Only problem with PIN is that SOMETIMES it is true for some cases and false for others. There is a chance that you miss to test those exceptional cases.

Well!!! In this case particularly, it is true as the expression is good for any odd "x". I advise you to learn the other method as well. It will come handy.
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Thus; x^3-x is nothing but the product of 3 consecutive integers.

And we know x is odd; Means; x-1= even and x+1=even

how do you go from the left to right? x(x+1)(x-1) = (x-1)x(x+1) Can we do that with any expression like this?

How does the above tell you that its 3 consecutive numbers? I thought that x(x+1)(x-1) simply means x = -1,0,1

Quote:

In any set of 3 consecutive numbers; where there are two evens, we must have at least 3 2's as its prime factor. Also, there is always at least 1 3 as its prime factor.

You mean when the 3 consecutive #'s are multiplied? 0,1,2 = 0?
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Thus; x^3-x is nothing but the product of 3 consecutive integers.

And we know x is odd; Means; x-1= even and x+1=even

how do you go from the left to right? x(x+1)(x-1) = (x-1)x(x+1) Can we do that with any expression like this?

How does the above tell you that its 3 consecutive numbers? I thought that x(x+1)(x-1) simply means x = -1,0,1

Does it matter whether you write:

-2*3*6 OR 3*-2*6 OR 6*3*-2

Likewise: x(x-1)(x-2) If x=odd means x=integer x-1=integer as integer(+-)integer=integer

Also; if x=odd x-1=Even and 1 less than x x+1=even and 1 more than x

So, (x-1)x(x+1) can be 0*1*2 for x=1 OR 1000*1001*1002 for x=1001. ***********************************

Please go through MGMAT Number Properties guide and practice few questions to assimilate this concept. You got to have Even/Odd properties and consecutive number properties at your fingertips to appreciate the solution.
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Consecutive Integers and Divisibility [#permalink]

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29 Jun 2011, 21:58

Guys,

These type of problems are really becoming a pain for me. can you please tell me what to do and where I am going wrong?

Question: If x^3 - x=p, and x is ODD, is p divisible by 24?

My solution: Factor x to give me x(x-1)(x+1) = p. So when x is ODD then (x-1) (x+1) will be even (ODD+-ODD=Even). That means p is even.

Now 24 = 2^3 * 3. Until here I am fine. And after this I get stuck on most of these types of problems. What to look for after this to get an answer?
_________________

Re: Consecutive Integers and Divisibility [#permalink]

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30 Jun 2011, 01:04

this much seems perfectly fine. you'll have to eliminate answers after this step...thats all you can do...
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Re: Consecutive Integers and Divisibility [#permalink]

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30 Jun 2011, 02:33

1

This post received KUDOS

enigma123 wrote:

Guys,

These type of problems are really becoming a pain for me. can you please tell me what to do and where I am going wrong?

Question: If x^3 - x=p, and x is ODD, is p divisible by 24?

My solution: Factor x to give me x(x-1)(x+1) = p. So when x is ODD then (x-1) (x+1) will be even (ODD+-ODD=Even). That means p is even.

Now 24 = 2^3 * 3. Until here I am fine. And after this I get stuck on most of these types of problems. What to look for after this to get an answer?

There, there. Okay, you're off to a good start: factoring and breaking down expressions and numbers is an excellent habit. But the "playing around" with numbers cannot stop there. You need to take this a step further and think about two important points:

First, if x is an integer, (x - 1), x and (x + 1) are by definition consecutive integers. Among three consecutive integers, one of them must be a multiple of 3.

Second, if x is odd, (x - 1) and (x + 1) are both even. And since every second even number is a multiple of 4, one of the two has to be a multiple of 4. And since 4 times 2 = 8, a multiple of 4 times a multiple of 2 has to be a multiple of 8.

So, x(x - 1)(x + 1) has to have a multiple of 3 somewhere in there, and it has to have 2^3 multiplied in there as well. And since 3 and 2 are both prime and neither can be overlapped with prime factors, we conclude that p must be a multiple of (2^3) * 3.

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