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integer greater than 6

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integer greater than 6

by money9111 » Mon Feb 01, 2010 8:17 pm
Hello everyone... So I did some studying tonight from the OG and here is a problem I had trouble setting up - therefore I couldn't solve it... please help!

If n is an integer greater than 6, which of the following must be divisible by 3?

a. n(n+1)(n-4)
b. n(n+2)(n-1)
c. n(n+3)(n-5)
d. n(n+4)(n-2)
e. n(n+5)(n-6)

i started plugging in numbers and I know this is incorrect...

OA A
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by thephoenix » Mon Feb 01, 2010 8:28 pm
i did it like this

6 _ _ 3*x _ _ 3*x _ _ 3*x

the two dash reperesent two numbers
now if u select the number at first dash then that number - 4 will give a number which is div by three---->n-4
if u select the number at second dash then that no. + 1 will give u a number which is div by 3---->n+1
hence n(n+1)(n-4)
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by sars72 » Mon Feb 01, 2010 8:32 pm
n is greater than 6

for the product of 3 numbers to be divisible by 3, one of the three numbers should be a multiple of 3

So, let us go about trying to find the answer
we are looking for an answer that will defiinitely contain a multiple of 3 as one of the factors
i'll start off with choice B, to explain why it is wrong
b. n(n+2)(n-1)
the result of (n+2) is 3 greater than (n-1) i.e. (n+2) = 3 + (n-1), since we do not know the value of n, we cannot say if
n(n+2)(n-1) is divisible by 3. Eliminated!
a. n(n+1)(n-4)
adding 3 to (n-4) --> 3 + (n-4) = n-1

--> n(n+1)(n-1)
Notice that these 3 are consecutive integers. Since one of the 3 have to be a multiple of 3, the product is therefore divisible by three and thus A is that answer
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by money9111 » Mon Feb 01, 2010 8:37 pm
oh thanks for the explanation Sars... i did not even think about looking at the equation like that... as groups related to each other. I mean I knew they were related to each other... but the way you put it made me think of it a different way.

Thanks!
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