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ritula
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lylya4
I don't think you can simply calculate some first prime number to get the conclusion

Among 3 consecutive integer n - 1, n, n + 1
n is not factor of 3 => n - 1 or n + 1 is divisible by 3
=> (n-1)(n+1) divisible by 3 (1)

n is not factor of 2 => n - 1 or n + 1 is divisible by 2
between n - 1 and n + 1, if n - 1 is divisible by 2, n + 1 is divisible by 4, if n - 1 is divisible by 4, n + 1 is divisible by 2

=> (n-1)(n+1) is divisible by 8. (2)

from (1) & (2), (n-1)(n+1) is divisible by 12, r = 0

I agree with you. However, during actual test, if I can find a pattern with couple of numbers, I will probably go with that approach in case I do not come up with an immediate algebraic approach.
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well in fact, from (1) and (2) we can concur that n is prime with the exception of 2 and 3.
i.e n=1,5,7....
answer is C.
(if n=1 and product is 0, remainder is still 0)
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yezz
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from 1

(n-1),n,(n+1) are 3 consecs where (n-1) and (n+1) are even.

when

(n-1)(n+1) are devided by 2^3 * 3 the remainder could be anything.......insuff

from2

if 3 is not a factor of n then sure either (n-1) or (n+1) is. however.........insuff

both

in the 3 cosnecs one of them is even multiple of 3 , second is even and one of the 2 evens has one more factor of 2 in it, as such

the 2 evens are in the form = 2*2*2*3*x, where x is any +ve intiger

thus remainder is 0..........suff
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I think any of the above process is correct. Based on whether one is good at forming calculation with numbers or at forming quick algebric equations, one can take any of the approach.



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