Remainder

[manik11]

If x and n are integers such that x = 12 + 22 + 32 + . . . + n^2, what is the remainder when x is divided by 5?

(1) n is an even integer with a units digit less than 6.
(2) The units digit of n is 2.

OA : D

[MartyMurray]

Does n^2 also have 2 as its units digit? In that case neither of the statements would work. In fact there is no integer the square of which ends in 2. So I guess the answer to my question is no.

Statement 1:

Case 1: If n = 10, then n^2 = 100.

I guess using that we can maybe, I guess, calculate x by adding 12 ... 92 + 100

We don't care about the 10's, because 10 is always a multiple of 5. The 2's add up to 18.

So the remainder when x is divided by 5 is 3.

Case 2: If n = 12, n^2 = 144.

So x = 12 ... 142 + 144.

Once again the 10's don't matter. In this case we get 14 2's + 4 = 28 + 4 = 32 The remainder when divided by 5 is 2.

So Statement 1 is insufficient.

The OA says differently. LOL

I think the OA should be I for INSANE.

Time to ditch this unclear, ridiculous question.

[Brent@GMATPrepNow]

I believe (based on the OA) that the question should read as follows:

If x and n are integers such that x = 1² + 2² + 3² + . . . + n², what is the remainder when x is divided by 5?

(1) n is an even integer with a units digit less than 6.
(2) The units digit of n is 2.

OA : D

Target question: What is the remainder when x is divided by 5?

Given: x = 1² + 2² + 3² + . . . + n²
Before we examine each statement, let's TEST some values of n to see if there's a PATTERN.
n = 1: x = 1² = 1. So, when x is divided by 5, the remainder is 1
n = 2: x = 1² + 2² = 5. So, when x is divided by 5, the remainder is 0
n = 3: x = 1² + 2² + 3² = 14. So, when x is divided by 5, the remainder is 4
n = 4: x = 1² + 2² + 3² + 4² = 30. So, when x is divided by 5, the remainder is 0
n = 5: x = 55. So, when x is divided by 5, the remainder is 0
n = 6: x = 91. So, when x is divided by 5, the remainder is 1
n = 7: x = 140. So, when x is divided by 5, the remainder is 0
n = 8: x = 204. So, when x is divided by 5, the remainder is 4
n = 9: x = 285. So, when x is divided by 5, the remainder is 0
n = 10: x = 385. So, when x is divided by 5, the remainder is 0
n = 11: x = 506. So, when x is divided by 5, the remainder is 1
n = 12: x = 650. So, when x is divided by 5, the remainder is 0
n = 13: x = 819. So, when x is divided by 5, the remainder is 4
n = 14: x = 1015. So, when x is divided by 5, the remainder is 0
.
.
.
Aha, so the pattern repeats itself every 5 instances.
So, the remainder when n = 3 is the same as the remainder when n = 8, 13, 18, 23, etc
Likewise, the remainder when n = 4 is the same as the remainder when n = 9, 14, 19, 24, etc

Statement 1: n is an even integer with a units digit less than 6.
In other words, the units digit of n is 0, 2 or 4
From our calculations above, we can conclude that when we divide x by 5, the remainder will be 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The units digit of n is 2.
From our calculations above, we can conclude that when we divide x by 5, the remainder will be 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

[MartyMurray]

I believe (based on the OA) that the question should read as follows:

If x and n are integers such that x = 1² + 2² + 3² + . . . + n², what is the remainder when x is divided by 5?

(1) n is an even integer with a units digit less than 6.
(2) The units digit of n is 2.

OA : D

Nice. Now THAT I kinda wish I had seen.