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OG12 #170: If n is a positive integer....

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OG12 #170: If n is a positive integer....

by Fractal » Fri Sep 17, 2010 4:18 am
If n is a positive integer, is n^3 - n divisible by 4?

(1) n = 2k + 1, where k is an integer
(2) n^2 + n is divisible by 6

Correct answer is A

My question is more general, but i can illustrate it with the above exercise:

For example, if i realised that in statement (1), n must be an odd integer, but i didn't realise that i can transform the expression n^3 - n to n(n-1)(n+1), how can i solve this exercise then? or how do i know which expression i have to transform? does there exist any tips?

i hope my question is clear ;-)

thx
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by John@Knewton » Fri Sep 17, 2010 5:30 am
Realizing that n^3 - n = n(n + 1)(n - 1) allows you to see that n^3 - n must have the same parity (oddness/evenness) as n. You can do this without factoring, also:

n^3 will have the same parity as n because if n is odd, then n^3 = n*n*n is the product of three odd numbers, so n^3 will also be odd; similarly, if n is even, n^3 will be even. That means n^3 - n is the difference of two odd numbers (if n is odd) or the difference of two even numbers (if n is even); either way, the difference is even since the difference of two odd numbers or two even numbers is always even.
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by Fractal » Fri Sep 17, 2010 6:59 am
that's all fine, but wouldn't help me to answer the question, because i don't just have to know if n is odd or even, but if n is divisible by 4 ....
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by John@Knewton » Fri Sep 17, 2010 7:09 am
Sorry for being incomplete: it does help because when considering Statement 1, you can see that n is odd, as you noted. Since n^3 - n must then also be odd, it can't be divisible by 4, or by any other even factor. For the purposes of evaluating Statement 1, the prompt might as well say "divisible by 14," "divisible by 212," or divisible by any even number. While it is not true that EVERY even number is divisible by 4, it is true that NO odd number is divisible by 4, so knowing that n^3 - n is odd is enough to deem the statement insufficient.
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by John@Knewton » Fri Sep 17, 2010 7:18 am
***To deem the statement sufficient. Please forgive the typo.
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by Fractal » Fri Sep 17, 2010 7:18 am
john@knewton wrote:Sorry for being incomplete: it does help because when considering Statement 1, you can see that n is odd, as you noted. Since n^3 - n must then also be odd, it can't be divisible by 4, or by any other even factor. For the purposes of evaluating Statement 1, the prompt might as well say "divisible by 14," "divisible by 212," or divisible by any even number. While it is not true that EVERY even number is divisible by 4, it is true that NO odd number is divisible by 4, so knowing that n^3 - n is odd is enough to deem the statement insufficient.
you say that if n is odd, n^3 - n must then also be odd. but take 3 as an example: 3^3 - 3 = 24 --> even, not odd!

odd - odd = even

am i wrong? this is why i meant that it doesn't help to answer the exercise!
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by Ian Stewart » Fri Sep 17, 2010 7:26 am
john@knewton wrote:Realizing that n^3 - n = n(n + 1)(n - 1) allows you to see that n^3 - n must have the same parity (oddness/evenness) as n.
n^3 - n is always even. It does not have the same 'parity' as n.
john@knewton wrote: it does help because when considering Statement 1, you can see that n is odd, as you noted. Since n^3 - n must then also be odd, it can't be divisible by 4, or by any other even factor.
Again, if n is odd, n^3 - n is even, not odd; n^3 - n = (n-1)(n)(n+1) is the product of three consecutive integers, so must be even.
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by debmalya_dutta » Fri Sep 17, 2010 7:45 am
Fractal wrote:If n is a positive integer, is n^3 - n divisible by 4?

(1) n = 2k + 1, where k is an integer
(2) n^2 + n is divisible by 6

Correct answer is A

My question is more general, but i can illustrate it with the above exercise:

For example, if i realised that in statement (1), n must be an odd integer, but i didn't realise that i can transform the expression n^3 - n to n(n-1)(n+1), how can i solve this exercise then? or how do i know which expression i have to transform? does there exist any tips?

i hope my question is clear ;-)

thx
(n-1).n.(n+1)
Statement 1 : this says n is odd . That means n+1 is even and n-1 is even. So product of 2 even numbers is always divisible by 4
hence sufficient

Statement 2
this says n.(n+1) = 6 . so what this means is product of 2 consecutive numbers = 6
the numbers can only be n=2 and n+1= 3..That in turn implies that n-1 is 1...
hence we , know that (n-1).n.(n+1) = 6 which is not divisible by 4....

but now if n.(n+1) = 12 ....same concepts
(n-1).n.(n+1) = 2.3.4 - this however is divisible by 4
Hence 2 is insufficient
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by John@Knewton » Fri Sep 17, 2010 7:55 am
You are correct, Fractal. I was rushing through writing this and started writing about a few different questions. My full attention is here now. To answer your question of doing the problem without factoring, there is at least a way of doing it with LESS factoring. You can factor out an n from n^3 - n, to get n(n^2 - 1). THEN you can use the fact that n is odd to see that every square of an odd number is one more than a multiple of 4. Try testing cases, 1^2 = 1, and 1 -1 is 0, a multiple of 4; 3^2 = 9, one more than a multiple of 4, 5^2 is 25, one more than a multiple of 4.

Still, though, the ideal approach is recognizing the difference of squares, but you can be reasonably confident if you don't recognize that.
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