n is an integer greater than 1. If the remainder is 1, when n is divid

Question

n is an integer greater than 1. If the remainder is 1, when n is divided by 3, then (n^2 + n - 2) must be divisible by which of he following double-digit number?

A. 12
B. 14
C. 16
D. 18
E. 20

BrentGMATPrepNow · Accepted Answer

Another approach....

If the remainder is 1, when n is divided by 3, then  n is 1 greater than some multiple of 3 
So, we can say   n = 3k + 1   for some integer k.

We want to find the value of n² + n - 2
Replace n with   3k + 1   to get: n² + n - 2 = (  3k + 1  )² + (  3k + 1  ) - 2
Simplify to get: 9k² + 9k
Factor:   9  (k² + k)
So, the correct answer must be a multiple of   9

Answer: D

Cheers,
Brent

Gauravji21 · Answer

n= 4, 7 ,10...
n2+n-2 
16+4-2 = 18
49+7-2 = 54
100+10-2= 108

all are divisible by 18..

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Capakki · Answer

If the remainder is 1 when a number is divided by 3 then the number must b 1 more than every multiple of 3. 
So possible values: 1,4,7,10....
All will leave remainder 1.

Lets check with the minimum value from above range:
If number (n) = 1
Then, n^2+n-2= 0
Not in option

Lets check next number: 4
n^2+n-2= 18

Upon checking the options, 18 is the answer. 
Hence D

Posted from my mobile device

IanStewart · Answer

The fastest way to do this problem is to notice n can equal 4, and just plug n=4 into n^2 + n - 2 to get 18. Only D can be right. That's how I'd solve the problem.

If one wanted to see why that's true: n^2 + n - 2 = (n + 2)(n - 1). Now if the remainder is 1 when n is divided by 3, that means n is 1 greater than a multiple of 3. So n-1 will be divisible by 3, and so will be (n-1)+3 = n+2. So (n+2)(n-1) is a product of two multiples of 3, and must be divisible by 3^2. It's also the product of one even and one odd number, so it must further be divisible by 2, and thus by (2)(3^2) = 18.

I'm not sure why the question uses the phrase "double-digit number." That is not mathematical language (the wording would always be "two-digit number"), and there's no reason for the question to even mention how many digits the right answer has.

Akp880 · Answer

n is an integer greater than 1. If the remainder is 1, when n is divided by 3, then (n^2 + n - 2) must be divisible by which of he following double-digit number?

A. 12
B. 14
C. 16
D. 18
E. 20

Explanation:

let n=3k+1, (where k is an integer >0 as given n>1)
Now  n^2+n-2 = (n+2)(n-1)
=(3k+1+2)(3k+1-1)
=3(k+1)3(k)
= 9k(k+1) 
since K is an integer >0 thus multipliction of  k*(k+1) will be always divisible by 2

Hence  (n^2 + n - 2) must be divisible by 18

Thus D is the correct answer.

BrushMyQuant · Answer

If the remainder is 1, when n is divided by 3

Theory: Dividend = Divisor*Quotient + Remainder

n -> Dividend
3 -> Divisor
a -> Quotient (Assume)
1 -> Remainders
=> n = 3*a + 1 = 3a + 1

For simplicity, lets take a = 1
=> n = 3*1 + 1 = 4

( n^2  + n - 2) must be divisible by which of he following double-digit number

n^2  + n - 2 =  4^2  + 4 - 2 = 16 + 2 = 18
=>  n^2  + n - 2 will be divisible by 18

So,  Answer will be D 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

bumpbot · Answer

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n is an integer greater than 1. If the remainder is 1, when n is divid

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n is an integer greater than 1. If the remainder is 1, when n is divid

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