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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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If n is an integer greater than 10, then the expression (n^2 - 2n)(n + 1)(n - 1) MUST be divisible by which of the following?

I. 4
II. 6
III. 18

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

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Originally posted by mkrishnabdrr on 01 Jul 2017, 01:30.
Last edited by Bunuel on 01 Jul 2017, 01:49, edited 1 time in total.
Edited the question.
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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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The expression ($$n^2$$-2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by $$24(2^3*3)$$
Rule : The product of any n consecutive integers will be always divisible by n!

Hence, the expression must be divisible by $$4(2^2)$$ and $$6(2*3)$$ but not $$18(2*3^2)$$
The answer is Option C(i and ii only)
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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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pushpitkc wrote:
The expression ($$n^2$$-2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by $$24(2^3*3)$$
Rule : The product of any n consecutive integers will be always divisible by n!

Hence, the expression must be divisible by $$4(2^2)$$ and $$6(2*3)$$ but not $$18(2*3^2)$$
The answer is Option C(i and ii only)

pushpitkc can you please share a link related to shared rule please
TIA
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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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ehsan090 wrote:
pushpitkc wrote:
The expression ($$n^2$$-2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by $$24(2^3*3)$$
Rule : The product of any n consecutive integers will be always divisible by n!

Hence, the expression must be divisible by $$4(2^2)$$ and $$6(2*3)$$ but not $$18(2*3^2)$$
The answer is Option C(i and ii only)

pushpitkc can you please share a link related to shared rule please
TIA

ehsan090 , the rule is under Consecutive Integers, here:

https://gmatclub.com/forum/math-number-theory-88376.html
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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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I took two values, i.e. 11 and 12 and applied them to the given formula. Only 4 and 6 satisfied both. While 18 satisfied only when n=11.
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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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mkrishnabdrr wrote:
If n is an integer greater than 10, then the expression (n^2 - 2n)(n + 1)(n - 1) MUST be divisible by which of the following?

I. 4
II. 6
III. 18

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

-----ASIDE---------------------
There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.
---------------------------------

The trick here is the see that (n² - 2n)(n + 1)(n - 1) is actually the product of 4 consecutive integers
Since we can factor (n² - 2n) as n(n - 2), we get: (n² - 2n)(n + 1)(n - 1) = (n)(n - 2)(n + 1)(n - 1)

Now rearrange the terms to get: (n - 2)(n - 1)(n)(n + 1)
Notice that n-2, n-1, n and n+1 represent 4 consecutive integers

According to the above rule, the product must be divisible by 4, 3, 2 and 1

So, the product is definitely divisible by 4

Since we know that the product is divisible by 2 AND 3, we know that it's divisible by 6

What about 18? Let's test a values of n.
If n = 13, then the product (n - 2)(n - 1)(n)(n + 1) becomes (11)(12)(13)(14)
Is that divisible by 18? It's hard to tell. Let's find the prime factorization of each value in the product.
(11)(12)(13)(14) = (11)(2)(2)(3)(13)(2)(7)
We can see that this is NOT divisible by 18 (since the product does NOT include two 3's and one 2)

Cheers,
Brent
_________________ Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +   [#permalink] 12 May 2019, 13:00
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