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Is n divisible by 8????

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Is n divisible by 8????

by bhumika.k.shah » Mon Mar 01, 2010 4:15 am
If x^3 - x = n and x is a positive integer greater than 1, is n divisible by 8?

(1) When 3x is divided by 2, there is a remainder.

(2) x = 4y + 1, where y is an integer.

I get that statement I is sufficient . But how is statement II sufficient ?

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by rohan_vus » Mon Mar 01, 2010 4:53 am
x^3 - x is x*(x-1)*(x+1) ---(1)

Now stmnt 2 says x = 4y+1, so plainly substitute this into eqn 1

x*(x-1)*(x+1) = (4y+1)*4y*(4y+2) ==> ((4y+1)*4y*2*(2y+1)==>8y*(4y+1)*(2y+1) ==> 8*(some integer value)

So stmnt 2 tells that the expression x*(x-1)*(x+1) is a multiple 8
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by scoobydooby » Mon Mar 01, 2010 7:51 am
product of 2 consecutive even numbers is always divisible by 8

1) =>x is odd=> (x+1) and (x-1) are even, so divisible by 8

2) n= 4y*(4y+1)*(4y+2)=> even*odd*even, so divisible by 8

hence, D
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