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 ⚠️ 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. M46-31

We can have 3 cases for the Number n^2-8, Checking for remainders 1. (n^2)-8 = 33a + 0 2. (n^2)-8 = 33b + 15 3. (n^2)-8= 33c + 21 therefore, a= (n^2) - 8/(3 * 11) - Eqn 1 b= (n^2) -8-15/(3*11) - Eqn 2 c= (n^2)-8-21/(3*11)- Eqn 3 Both a,b,c should be divisible by 3 and 11 Checking for divisibility by 3 Eqn 1: /3 = Eqn 2: /3 = Eqn 3: /3 = so we need to check if is divisible by 3 On putting Values for n we see that no value satisfies the condition. Thu we also don't need to check for 11 is they are not divisible by 3. Therefore 0,15,21 cannot be the remainders. Hence, Option E

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GMAT Club World Cup 2026 (Day 1): If n is a positive integer, which of

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Parry104

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06 Jul 2026, 10:01

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(n^2-8)/33.

take R as the remainder and Q as the quotient. (n^2-8)/33 = Q+R

N^2-8 = 33(Q+R). meaning N^2-8 needs to be divisible by 3 which wont happen for any integer. Hence E

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We can actually test the integers 1, 2, 3, 4, and 5 and with that we can come to a conclusion that none of these is the answer 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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I tried solving the question by trial and error and replaced n with the values of 1,2,3,4,.. 15 and did not get any such remainder as per the option and hence choose Option E

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I subtracted 8 from all the squares of numbers from 1-10 and divided them by 33, to get a remainder pattern. The answer choices did not show up in the pattern and all the odd remainders are prime.

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06 Jul 2026, 10:32

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I think none of these (Answer: E)

A number divisible by 33 should be divisible by 3 and 11. We will check if the number is divisible by 3. A perfect square divided by 3 can leave only 0 or 1 as the remainder.

2^2 /3 leaves remainder = 1
3^2 /3 leaves remainder = 0

So, n^2 - 8 leaves a remainder of 1 or 2 when divided by 3. Checking answers.

0 divided by 3 - leaves no remainder
15 divided by 3 - leaves no remainder
21 divided by 3 - leaves no remainder

n^2 - 8 can never be divisible by 3

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We need to check whether 0, 15,, or 21 could be the remainder when n^2 - 8 is divided by 33.
Now, 33 is a multiple of 3 and the possible remainders 0, 15, and 21 are also multiples of 3. So, if any one of them were the remainder when n^2- 8 were divided by 33, then n^2- 8 would also be divisible by 3.
For example, if n^2 - 8 = 33k + 21
=> n^2 - 8 = 3 (11k +7) which is divisible by 3.

That means, n^2 - 8 would have to be a multiple of 3. So, n^2 would be 8 more than a multiple of 3.
So, n^2 = 3p + 8 = 3p + 6 + 2.
So n^2 would leave a remainder of 2 when divided by 3.
This is impossible. since a perfect square when divided by 3 can only leave a remainder of 0 or 1.
The last point can be verified with examples.
0^2 leaves a remainder of 0 when divided by 3
1^2 leaves a remainder of 1 when divided by 3.
2^2 leaves a remainder of 1 when divided by 3
amd so on.

So, 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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If we split n^2-8/33 we get n^2/33 - 8/33

remainder of 8/33 = -25
therefore 33-25 = 8
so remainder for 8/33 = 8 ----------(1)

and now for n^2/33
we can take n = 1, 1/33 = rem is 1
and if we subtract 1-8 = -7

similarly for each n = 2,3,4...
we are not getting any of the options are the answers

therefore Ans: (E) None of these

Note: I know its not a great solution as I am new to it but the answer is right

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We can have 3 cases for the Number n^2-8, Checking for remainders

1. (n^2)-8 = 33a + 0
2. (n^2)-8 = 33b + 15
3. (n^2)-8= 33c + 21

therefore,

a= (n^2) - 8/(3 * 11) - Eqn 1
b= (n^2) -8-15/(3*11) - Eqn 2
c= (n^2)-8-21/(3*11)- Eqn 3

Both a,b,c should be divisible by 3 and 11
Checking for divisibility by 3

Eqn 1: [(n^2)-6-2]/3 = [{(n^2)-2}/3 -2]
Eqn 2: [(n^2) - 21-2]/3 = [{(n^2)-2}/3 - 7]
Eqn 3: [(n^2) - 27-2]/3 = [{(n^2)-2}/3 -9]

so we need to check if [(n^2)-2] is divisible by 3

On putting Values for n we see that no value satisfies the condition. Thu we also don't need to check for 11 is they are not divisible by 3.
Therefore 0,15,21 cannot be the remainders.

Hence, Option 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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The divisor is 33, which can be written as 3 x 11.
When any perfect square is divided by 3, the possible remainders are 0 and 1

Checking the options
a)0
If remainder is 0, n^2 leaves remainder of 8. Since 8 leaves remainder of 2 when divided by 3, this is not possible.

b)15
if the remainder is 15, then n^2 must leave a remainder of 23. Since 23 leaves remainder of 2 when divided by 3, this is not possible

c)21
if the remainder is 21, n^2 must leave a remainder of 29. Since 29 leaves remainder of 2 when divided by 3, this is also not possible.

Therefore E is the answer.

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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cant be zero as 41 is not square of an integer. other than that at n=9 we get 15 and at n=11 we get 22. so only correct remainder is 15 (B).

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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06 Jul 2026, 11:21

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E. None of these

So N^2 -8 / 33

1. 0 - not possible , because we have a positive integer
2. 15 - Not possible because
Any number divided by 33 must be divided by 3
So we are following mod(3)

n= 0,1 or 2 (mod)

so n^2 -8 = reminder
n^2 = reminder +8
=15 +8
=23

same logic using for 3rd option

n^2 = 21+8
=29

Both are not giving the correct ans

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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06 Jul 2026, 11:24

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(n^2-8)/33 can be written as n^2/33-8/33.

8/33 leaves remainder of 8.
1) to leave remainder of 0, n^2/33 will have to leave remainder of 8 so that both remainders cancel out and overall remainder is 0.
we check the possible values of n^2 if this is the case: {8,41,74,107,...}. none of these integers are perfect squares hence 0 as overall remainder is not possible.

2) similarly using same reasoning, possible values of n^2 for overall remainder of 15 is: {23,56,89,122,...}. none of these are perfect squares so 15 is not possible.

3) possible values of n^2 for overall remainder of 21 are {29,62,95,...}. none of them are perfect square so 21 is not possible.

Answer E

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I just took anything above 7^2 square up to 12^2 and saw the trend that nothing satisfied. Hence none of these.

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06 Jul 2026, 11:36

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33=11*3 (We will check remainders with mod 3 first)

I. 0
n^2 -8 =0
n^2 =8
8(mod 3) = 2

II. 15
n^2 -8 =15
23(mod 3) = 2

III. 21
29(mod3) = 2
A perfect square can never be 2(mod3). So 8,23,29 can't be squares mod33. Therefore none of given remainders is possible.

E

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My answer E- None of these-

By writing the below 3 equations-

I) n^2=33p+8
II) n^2= 33q+23
III) n^2= 33r+29.

Since in all three options, we cannot obtain a square by substituting 1,2,3,4 for p,q and r. Further, we can see 8,23 and 29 have nothing in common with 33.

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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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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we can write n^2 = 33(k) + 8 + (remainder). here k is a constant. we take 3 cases each for remainders 0, 15 and 21.
a. for rem = 0 we get n^2 = 33k +8
b. for rem = 15 we get n^2 = 33k + 23
c. for rem = 21 we get n^2 = 33k + 29
even if we put in sample values or take any common terms, neither of the cases will result in a perfect square for n. Hence answer is E (none of these)

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06 Jul 2026, 11:57

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We need to find the possible remainders when $n^2 - 8$ is divided by 33 ( n being a positive integer)

Now,
$33=3x11$
first let's find the possible remainders when $n^2$ is divided by 3, and the possible remainders when $n^2$ is divided by 11
And then using the Chinese Remainder Theorem, we will combine both the cases and then subtract 8 to get the desired answers.

As $1^2=1, 2^2=2, 3^2=9, 4^2=16, 5^2=25,$ and so on...
Possible Remainders when $n^2$ is divided by 3 are: {0,1}

Also, Possible Remainders when $n^2$ is divided by 11 are : {0,1,3,4,5,9}
Now, using the Chinese Remainder Theorem and testing the number from 1 to 33 using the above cases, we get
the possible remainders when $n^2$ is divided by 33 are: {0,1,3,4,9,12,15,16,22,25,27,31}

Now subtracting 8 from each of the above values to find the remainders when $n^2 -8$ is divided by 33,
we get the possible remainder = {1,4,7,8,14,17,19,23,25,26,28,29}

Hence the correct answer is E. None of these

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06 Jul 2026, 12:21

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When tried calculating the remainder using n= 1 to 10, the remainder was 7 and then directly 26, hence, in none of the cases did the given numbers 0, 15 and 21 pop up, hence, the answer is "None of these".

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For n^2-8, let remainder be r when divided by 3. r can be 0,1,2. Since for n^2 can give 0,1,2 remainder. n^2-8 can leave -8, -7, -4 as remainder which is 1,2,2. Therefore n^2-8 can be remainder if it leaves modulo 1,2 when divided by 3. Since all the options are divisible by 3, there fore answer is E.(None of these)

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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06 Jul 2026, 12:30

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One principle which can help solve this question faster instead of testing values:

*** A perfect square is either divisible by 3 or leaves a remainder of 1 when divided by 3

Eg Take perfect squares of 4 , 9 , 16 , 225 ---> if you divide them by 3 you ll get remainders or 1 , 0 , 1 , 1 respectively

I: n^2 - 8 = 33k or n = (33k-8)^(1/2) If you divide this expression you will get a remainder of 2 i.e 33 is anyways divisible by 3. if you divde 8 by 3 the remainder is 2. So this number cannot be a perfect square and as per the question n is positive integer hence n^2 has to be an perfect square.

II Similarly, This gives you n^2 = (33k+15+8)^(1/2) i.e (33k+23)^(1/2)... similar case as above 23/3 gives a remainder of 2 Hence not a perfect square

III n^2 = (33k+29)^(1/2) Again 29/3 = 2 Hence n wont be a perfect square

Answer is 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

This question was provided by GMAT Club
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