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divisibility & primes

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divisibility & primes [#permalink] New post 25 Jun 2011, 10:56
if x^3-x=p, and x is odd, is p divisible by 24?


the book gives a complicated approach. can't we just plugin numbers to see?

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 = no.
if x = 3, 3.3.3 - 3 = 24 --> 24/24=1.

Hence, we can't say whether p is divisible by 24!!! ?? correct ??
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Re: divisibility & primes [#permalink] New post 25 Jun 2011, 11:30
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386390 wrote:
if x^3-x=p, and x is odd, is p divisible by 24?


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.

0,1,2
2,3,4
4,5,6
6,7,8
8,9,10
10,11,12
12,13,14
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Re: divisibility & primes [#permalink] New post 25 Jun 2011, 11:35
Expert's post
386390 wrote:

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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Re: divisibility & primes [#permalink] New post 25 Jun 2011, 15:43
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.
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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:14
IanStewart wrote:
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 0 is a multiple of every number? But not a factor! I guess x/0=undefined! Right?
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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:19
386390 wrote:
IanStewart wrote:
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 0 is a multiple of every number? But not a factor! I guess x/0=undefined! Right?


Right. "0" can not be a factor of ANY integer.
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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:23
So fluke,

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.
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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:31
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386390 wrote:
So fluke,

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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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:39
fluke wrote:
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



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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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:48
386390 wrote:
fluke wrote:
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



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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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 07:53
makes sense.....


thanks. i guess i gotta go over the num properties book once more.
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Re: divisibility & primes [#permalink] New post 27 Jun 2011, 08:31
Nice explanations..in light of the above..Yes
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Consecutive Integers and Divisibility [#permalink] New post 29 Jun 2011, 20:58
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? :cry:
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Re: Consecutive Integers and Divisibility [#permalink] New post 30 Jun 2011, 00:04
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] New post 30 Jun 2011, 01:06
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? :cry:


Can you please post all the options? Also, i want to know why you have taken 24 = 2^3 * 3. I dint understand this step..

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Re: Consecutive Integers and Divisibility [#permalink] New post 30 Jun 2011, 01:33
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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? :cry:


There, there. :wink: 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.

That help?
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Re: Consecutive Integers and Divisibility [#permalink] New post 02 Jul 2011, 11:22
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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Re: divisibility & primes [#permalink] New post 02 Jul 2011, 11:45
Thanks Fluke and rustrypolymath.
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Re: Consecutive Integers and Divisibility [#permalink] New post 03 Jul 2011, 23:46
enigma123 wrote:
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?


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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is p divisible by 24? [#permalink] New post 16 Apr 2012, 09:57
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.
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is p divisible by 24?   [#permalink] 16 Apr 2012, 09:57

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