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 Post subject: GMAT Number Theory
PostPosted: Wed May 26, 2010 3:16 pm 
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Positive integer n is divisible by 5. What is the remainder when dividing (n + 1)(n + 2)(n + 8)(n + 9) by 5?
A. 0
B. 1
C. 2
D. 3
E. 4

(E) We can factor (n + 1)(n + 2)(n + 8)(n + 9), then analyze the summands and find a remainder. But there is no need in lengthy factoring if we see that each summand except 1 × 2 × 8 × 9 would contain n. If it is not clear to you, do some factoring in following way:
(n + 1)(n + 2)(n + 8)(n + 9) = n(n + 2)(n + 8)(n + 9) + 1 × (n + 2)(n + 8)(n + 9).
First summand is divisible by n so we can leave it the way it is. Let’s do some factoring for second summand:
1 × (n + 2)(n + 8)(n + 9) = 1 × n(n + 8)(n + 9) + 1 × 2 × (n + 8)(n + 9). Again the first summand is divisible by n so we leave it the way it is and repeat the process couple more times:
1 × 2 × (n + 8)(n + 9) = 1 × 2 × n(n + 9) + 1 × 2 × 8 × (n + 9)
1 × 2 × 8 × (n + 9) = 1 × 2 × 8 × n + 1 × 2 × 8 × 9

In result we get:
(n + 1)(n + 2)(n + 8)(n + 9) = n(n + 2)(n + 8)(n + 9) + 1 × n(n + 8)(n + 9) + 1 × 2 × n(n + 9) + 1 × 2 × 8 × n + 1 × 2 × 8 × 9

All the summands that contain n would be divisible by 5. Therefore the remainder when dividing (n + 1)(n + 2)(n + 8)(n + 9) by 5 is the same as the remainder when dividing 1 × 2 × 8 × 9 by 5. 1 × 2 × 8 × 9 = 144. Therefore the remainder is 4 and the answer is (E).

Plugging in a positive integer divisible by 5 is also an option, e.g. 5 itself. We can do that using an assumption that the remainder must be the same for any positive integer divisible by 5.[/color]
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Very lengthy explanation. I believe this can question can be solved easily by substitution. E.g. take n as 5 and find out what is the remainder. It takes less than 1 min to solve this question this way. After all GMAT is not about showing complex logic to solve a problem but about arriving at the right answer.


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 Post subject: Re: math: number theory, remainder
PostPosted: Wed May 26, 2010 3:18 pm 
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It is true. Using an assumption that any positive integer divisible by 5 must give the same remainder, we can plug in number 5 itself. However if you see right away that each summand except 1 × 2 × 8 × 9 would contain n then you calculate 1 × 2 × 8 × 9 instead of 6 × 7 × 13 × 14.

Besides, given explanation should help you to understand a problem in details and find a solution to similar questions, e.g.:

Positive integer n is divisible by 144. What is the remainder when dividing (n + 1)(n + 2)(n + 8)(n + 9) by 144?


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