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Checking the options:
If the remainder is 0, then n^2-8=33a+0, so n^2=33a+8
If the remainder is 15, then n^2-8=33a+15, so n^2=33a+23
If the remainder is 21, then n^2-8=33a+21, so n^2=33a+29

To check if n^2 can be equal to 33a+8 for n and a being integers, the easiest way is to check the remainder of both terms when they are divided by 3.

Remainder of n^2 divided by 3 has three possibilities:
- if n is a multiple of 3 (3b) then n^2=9b^2 and divided by 3 has 0 as remainder.
- if n is a multiple of 3 plus 1 (3b+1) then n^2=(9b^2+6b+1) and divided by 3 has 1 as remainder.
- if n is a multiple of 3 minus 1 (3b-1) then n^2=(9b^2-6b+1) and divided by 3 has 1 as remainder.

Remainder of 33a+8 divided by 3 is 2
So 0 is not a valid option.

Remainder of 33a+23 divided by 3 is 2
So 15 is not a valid option.

Remainder of 33a+29 divided by 3 is 2
So 21 is not a valid option.

The correct answer is E
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Bunuel
If n is a positive integer, which of the following could be the remainder when n^2 - 8 is divided by 33?

I. 0
II. 15
III. 21

A. I only
B. II only
C. III only
D. II and III only
E. None of these


 


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Now the equation comes to n^2 - 8 = 33c + remainder , where c is a constant.

If a number is to be divisible b 33, it has to be divisible by 3 as well. and thus the possible remainders for division with 3 is 0, 1 or 2.

squaring them (as n is squared) gives 0, 1 and 4 ( which again gives remainder 1 on division by 3).

So the effective remainders are 0 & 1 only.

When the values are put in the equation and divided, all the 3 options give a remainder of 2 only.

And hence the answer should be E- none of the above
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testing from 1-12, there is no remainder for 0, 15, 21 --> E
Bunuel
If n is a positive integer, which of the following could be the remainder when n^2 - 8 is divided by 33?

I. 0
II. 15
III. 21

A. I only
B. II only
C. III only
D. II and III only
E. None of these


 


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for the GMAT World Cup Competition

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This is my solution.
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E. None of the above.
I tried with different values of n. There should be a smart way!
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Yes there is a smart way, Check my reply on this -
This question only checks basic observation - to check the divisibility of (n^2 - 8) by 33.
Step 1: Given = 'n' is a positive integer
Step 2: n^2 - 8 could be written as = [n - 8^(1/2)] [n + 8^(1/2)]
Step 3: As we know square root of 8 is not an integer and 'n' is a positive integer so the two brackets will form irrational numbers
Hence, no integral value of the remainder is possible and all three options contain integral numbers
puncuz
E. None of the above.
I tried with different values of n. There should be a smart way!
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Step 1. n^2-8>=33
n>6
Step 2. trial and error keeping n = 7,8,9,10,11,12
Step 3. Basis pattern noticed, none of the options satisfy

Bunuel
If n is a positive integer, which of the following could be the remainder when n^2 - 8 is divided by 33?

I. 0
II. 15
III. 21

A. I only
B. II only
C. III only
D. II and III only
E. None of these


 


This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 


⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.
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Can you explain the 2nd and 3rd case in detail? why 15+8?
ankushsambare
r be reminder when n^2 -8 by 33
then n^2 ~ r+ 8

check option which is giving perfect square when 8 is added.
I. r = 0
0+8 is not perfect square
II. r =15
15+8 not a prefect square
III. r =21
21+8 = 27 not a perfect square

So answer is E. I, II and III.
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Because r = 15 and from n^2 - 8, we get 8.

So 15+8 =23 not a perfect square.

Similarly, r = 21 and from n^2-8, we get 8,
So 21+8 =29 29 not a perfect square

ischiragkapoor
Can you explain the 2nd and 3rd case in detail? why 15+8?
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Hi ischiragkapoor,

You're looking at the step (from posts like ankushsambare's and Zeus_'s) that jumps from n2 − 8 to n2 = r + 8, and that "+8" feels like it comes from nowhere. Let me show exactly where it comes from.

Start with the basic remainder equation. If n2 − 8 leaves remainder r when divided by 33, then:

- n2 − 8 = 33k + r (dividend = divisor × quotient + remainder)

Now just add 8 to both sides to get n2 by itself:

- n2 = 33k + r + 8

That's the whole trick. Since 33k is a multiple of 33, this says n2 is (r + 8) more than a multiple of 33. So whatever remainder we're testing, we tack on the 8 that we had subtracted.

Case II (r = 15) and Case III (r = 21)

- r = 15: n2 = 33k + 15 + 8 = 33k + 23
- r = 21: n2 = 33k + 21 + 8 = 33k + 29

Now test each against 3 (since 33 = 3 × 11). A perfect square divided by 3 can only leave 0 or 1, never 2.

- 23 ÷ 3 leaves remainder 2 - impossible, so 15 is out.
- 29 ÷ 3 leaves remainder 2 - impossible, so 21 is out.

That's why both fail.

Quick check to lock in the "+8" move

Suppose x − 3 leaves remainder 4 when divided by 10. What remainder does x leave?

Write x − 3 = 10k + 4, then add 3: x = 10k + 7. You add back exactly what you subtracted. Same idea here - you subtracted 8, so you add 8 back.

Answer: E

ischiragkapoor
Can you explain the 2nd and 3rd case in detail? why 15+8?

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