Hi, 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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divisibility & primes

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enigma123

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02 Jul 2011, 12:22

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Thanks Rustypolymath. Here is my doubt buddy:

You said 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.

Let's say x=1, then consecutive integers are 0,1,2. Then the above doesn't apply? or am I reading something wrong?

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02 Jul 2011, 12:45

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Thanks Fluke and rustrypolymath.

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04 Jul 2011, 00:46

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enigma123

Yes, the above does still apply. (0)(1)(2) = 0, a multiple of 24. 0 is a multiple of 4--indeed, 0 is a multiple of every integer.

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16 Apr 2012, 10:57

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I have a question on the below problem.

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

And the answer is yes. It is divisible by 24.

the reason being the above can be simplified into (x-1)(x)(x+1) which are consecutive integers. so (x-1) & (x+1) are even integers. and so the the total product should have factors 2*3*4.

Now, if the problem is exactly as given above, should we also not consider the below scenarios.

X-1 could be zero which is also an even integer. So p = 0. But again considering that zero is also divisible by 24, is this why the answer is correct. How are such questions to be Handled. Any inputs on how to consider the last 'zero' scenario please.

olwan

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16 Apr 2012, 11:53

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We can just pick odd numbers and work directly on this question

I picked
3^3-3=24 Divisible
11^3-11=1320/24 Divisible

Bunuel

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16 Apr 2012, 12:15

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mohankumarbd

\(x^3-x=(x-1)*x*(x+1)\).

Since \(x=odd\) then \(x-1\) and \(x+1\) are consecutive even integers. Now, the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2).

Next, \((x-1)*x*(x+1)\) is also the product of three consecutive integers. Out of three consecutive integers one is always divisible by 3, so \((x-1)*x*(x+1)\) is divisible by 3 too.

Which means that \((x-1)(x)(x+1)\) is divisible by both 3 and 8, so by 3*8=24.

As for zero: zero is a divisible by every integer, except zero itself. So, if \(p=0\)then it's divisible by 24 as well as by all other integers but zero itself.

Hope it's clear.

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16 Apr 2012, 20:35

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Bunuel,

I was clear on the '24 is a factor of p' part of the question.

My only question is around below scenario.

Based on the conditions set in the question, one of the possible scenarios could be
(x-1) = 0
x = 1
(x+1) = 2

I am trying to figure if it is right to consider this scenario always, as long as the conditions set in the question allows for it.

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17 Apr 2012, 01:46

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mohankumarbd

I'm not sure understood your question.

After some point we have that (x-1)x(x+1) is divisible by both 3 and 8, so by 3*8=24, which means that we already answered the question and we don't need to consider ANY additional scenarios at all.

If you ask whether 0 is divisible by 24 then the answer is YES (see my previous post).

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20 Apr 2012, 10:24

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Interesting question, thanks for the in-depth analysis!

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