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
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There's a nice rule says:
The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1So, 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.
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The trick here is the see that (n² - 2n)(n + 1)(n - 1) is actually
the product of 4 consecutive integersSince 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 4Since we know that the product is divisible by 2 AND 3, we know that it's
divisible by 6What 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)
Answer: C
Cheers,
Brent
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