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If the remainder is 1, when integer n(n>1) is divided by 3,

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Saraforgre
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If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   30 Jul 2020, 15:20
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n is an integer greater than 1. If the remainder is 1, when n is divided by 3, then (n² + n - 2) must be divisible by which 2-digit number?

A. 12
B. 14
C. 16
D. 18
E. 20
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D
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sukrut96
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Re: If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   30 Jul 2020, 22:59
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When n is divided by 3, the remainder is 1.
So lets say n is 7, when when divided by 3 gives remainder as 1.
Now, according to the equation, (n^2+n-2)
49 + 7 - 2 = 54
Only 18 can divide 54 completely. Thus, answer is D :)
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Re: If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   03 Aug 2020, 04:45
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In general if we have integers x,y,m,and r such that

x/y=m+(r/y) then r is the remainder and m is the whole number part of the division
e.g. 4/3=1+1/3 here x=4, y=3, m=1, r=1


Let n and m be integers

n/3=m+(1/3)

m is the whole number part of the division and 1 in (1/3) represents the remainder.

multiple 3 on both sides

n=3m+1

therefore

n^2-n-2= (n+2)(n-1)= ((3m+1)+2)((3m+1)-1) =(3m+3)(3m) =3(m+1)3m= 9(m+1)(m)

Finally since m and m+1 are consecutive integers one of them must be even therefore we can factor out a 2 from it giving us

=9*2*((m+1)m)/2
=18*((m+1)(m))/2

Since 18 is a factor of n^2-n-2. Then n^2-n-2 is divisible by 18.

Final Answer: D
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Re: If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   04 Aug 2020, 12:23
Saraforgre wrote:
n is an integer greater than 1. If the remainder is 1, when n is divided by 3, then (n² + n - 2) must be divisible by which double-digit number?

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


If n/3 yields a remainder of 1, then n must be one greater than a multiple of 3. Since the problem specifies that the given equation MUST be divisible by one of the answers choices, we know that we can choose any value that satisfies the criteria.

Let n = 4. (4)^2 + 4 - 2 = 18, so D is clearly the correct answer.
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Re: If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   28 Oct 2021, 19:25
GreenlightTestPrep sir, please explain this question in your approach. I am a fan of your problem-solving technique!
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If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   29 Oct 2021, 06:00
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Saraforgre wrote:
n is an integer greater than 1. If the remainder is 1, when n is divided by 3, then (n² + n - 2) must be divisible by which 2-digit number?

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



Approach #1: Test a number
For many Integer Properties questions, the fastest and easiest approach is to simply test a value that satisfies the given information.
We're told that, when n is divided by 3, the remainder is 1.
So it could be the case that n = 4

If n = 4, then n² + n - 2 = 4² + 4 - 2 = 18
Check the answer choices. . . . . . answer choice D is the only option that is a divisor of 18
Answer: D


Approach #2: Apply some algebra
When n is divided by 3, the remainder is 1
This means n is 1 greater than some multiple of 3.
In other words: n = 3k + 1 (for some integer k)

n² + n - 2 must be divisible by which 2-digit number
Since n = 3k + 1, we can determine the value of n² + n - 2 by replacing n with 3k + 1.
We get: n² + n - 2 = (3k + 1)² + (3k + 1) - 2
= 9k² + 6k + 1 + 3k + 1 - 2
= 9k² + 9k
= 9(k² + k)
= 9(k)(k + 1)

At this point, we can see that 9(k)(k + 1) it's definitely divisible by 9.
Also recognize that k and k+1 are consecutive integers, which means one of them is ODD and one of them is EVEN.
This mean (k)(k + 1) is definitely divisible by 2.
Since 9(k)(k + 1) is divisible by 9 and by 2, we can be certain that it is divisible by 18

Answer: D

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Re: If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink] New post   07 Feb 2023, 17:46
Hello from the GRE Prep Club BumpBot!

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Re: If the remainder is 1, when integer n(n>1) is divided by 3, [#permalink]
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