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

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

1 2 3 4 5 6

Arjita1507

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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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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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n^2 - 8/33

Remainder(n^2/33) - Remainder (8/33)

Remainder (8/33) will yield 8 as remainder

I. For the net remainder to be zero, remainder(n^2/33) should be 8. As 8 is not a perfect square, this this not possible.

II. For the net remainder to be 15, remainder (n^2/33) should be 15+8 = 23. As 23 is not a perfect square, this is not possible.

III. For the net remainder to be 21, the remainder (n^2/33) should be 21+8 = 29. As 29 is not a perfect square, this is not possible.

All the options are not possible.

Hence, the answer is Option E.

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There's some cyclicity game involved, and also odd even number rule. I am not sure, I got the answer manually.

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 know that,
D=dQ+R
Now I just applied all three remainder options to question
When R=0
n^2 - 8 = 33Q + 0
or n^2 = 33Q +8
No values of Q satisfied to give a perfect square,
Hence, Not I
Similarly I did for other two.

Hence, E

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D. Remainder is between 1 and n-1. n being the denominator.

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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Solving by cross checking the options and checking against condition - n is +ve integer.
nˆ2 = 33 (Option) + 8
for 0, nˆ2 = 8 Not possible, n cant be integer
for 15, nˆ2 = 295 , n cant be integer
for 21, nˆ2 = 699 , n cant be integer

Thus None of above, E.

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One way to solve this is by using modular arithmetic $n^2-8≡X(mod33)$, where X represents the reminder. To simplify the calculation, I used 3 and 11 since they are relative primes of 33 (3•11=33). The remainders of $ n^2(mod3)$ are 0 or 1, and then the remainders of $ n^2(mod11) $ are 0,1,3,4,5,9. When testing the options to see if they matched the previously mentioned reminders, I found that none of them matched even the first prime (mod 3)=0,1, because all three options resulted in a remainder of 2. Therefore, the 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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Any integer /3 has remainder 0,1,2
Square them 0,1,4
These will leave remainders of only 0,1. Hence all n^2 will leave 0,1 remainders

8 leaves remainder 2,
n^2-8 can leave 0-2,1-2==-2,-1==1,2(remainders)
n^2-8 can only leave 1 or 2

Therefore, none of the above.
IMO- 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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we can convert this into 33x + 8 + (options given) should be a perfect square.
None of the options give perfect square after adding to the equation.
So 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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Since n^2 -8 divided by 33 gives remainder r, I put it into the remainder formula to get n^2 - 8 =33x+r. Then I isolated n^2 by adding 8 to both sides.

n^2 = 33x + r + 8

Since 33 has factors of 3 and 11. we can use that 3 factor to filter the remainders of each term in the equation.

Squared numbers divided by 3 have a pattern of remainders of 0 and 1:
0^2 / 3 = 0/3, so remainder is 0
1^2 / 3 = 1/3, so remainder is 1
2^2 / 3 = 4/3, so remainder is 1
3^2 / 3 = 9/3, so remainder is 0
and so on.

33k will always be divisible by 3 and have a remainder of 0.

Lastly we will need the remainder for the term r+8 which we will find as we plug in the answer choices.

If we refer back to our equation n^2 = 33k + r + 8, we want the remainders of both sides to be equal. We can ignore 33k because the remainder is zero so we just need the remainder of r+8 to be either 0 or 1 to match n^2.

Now we plug in the remainder choices into r+8

0+8 = 8, 8/3 has a remainder of 2 so 0 is not a possible remainder.
15+8 = 23, 23/3 has a remainder of 2 so 15 is not a possible remainder
21+8 = 29, 29/3 has a remainder of 2 so 21 is not a possible remainder

The answer is E, none of these.

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E. none of these. Because all three candidate remainders leave remainder 0 when divided by 3, but n^2 - 8 can never leave remainder 0 when divided by 3 so none of them are possible.

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 answer choices, 0, 15, 21, are divided by 3, where n^2 - 8 cannot be divided by 3.
Why 3?
Divisor is 33, which can be divided by 3!
Let's test,
(4^2-8)/3 = 2, reminder 2
(5^2 - 8)/3 = 5, reminder 2
(6^2 - 8)/3 = 9, reminder 1
(9^2 - 8)/3 = 24, reminder 1
Therefore, it can be either 1 or 2, never 0.
Thus, neither of the choices is correct.
Answer (E)

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n^2 - 8 = 33k + r

if r=0 then n^2 = 33k + 8
if r=15 then n^2 = 33k + 23
if r=21 then n^2 = 33k + 29

To determine whether the square of a number can satisfy the equations we divide both terms by 3 and the remainder must be the same.

In the left term, n can be:
+ multiple of 3 (3a): n^2 = 9a^2 and dividing by 3 the remainder is 0
+ (3a+1): n^2 = (3a+1)^2 = 9a^2 + 6a + 1 and dividing by 3 the remainder is 1
+ (3a-1): n^2 = (3a-1)^2 = 9a^2 - 6a + 1 and dividing by 3 the remainder is 1

In the right term:
33k + 8 = 33k + 6 + 2: dividing by 3 the remainder is 2
33k + 23 = 33k + 21 + 2: dividing by 3 the remainder is 2
33k + 29 = 33k + 27 + 2: dividing by 3 the remainder is 2

left term remainder can be 0 or 1
right term remainder can be 2

So it is impossible for any of the three options to be possible for any n.

IMO E

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E - None of these

Lets say n^2 - 8 = 33k + r where r is the remainder. Now lets substitute the value of options with r and check whether n^2 could be expressed in the form of 33k + r + 8

Option A: n^2 = 33K + 8 => Not Possible
Apply the similar logic for 15 and 21 and we find that none of these is possible

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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Possible values of n2 - 8 when divided by 33

Since, 33 = 3*11,

Possible values of reaminder, when n^2 divided by 3
0^2 = 0
1^2 = 1
2^2 = 1

So, n^2 = 0 or 1 (reaminder when divided by 3)

Thus,

remainder when n^2 - 8 divided by 3

0-8 = (remainder 1)
1-8 = (remainder 2)

So, the remainder can never be divisible by 3.

Checking options:
0 is divisible by 3 - Not our answer
15 is divisible by 3 - Not our answer
21 is divisible by 3 - Not our answer.

Answer : 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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n^2-8 divided by 33, remainder?

n^2-8=33*quotient+remainder

I. n^2-8=33*quotient+0 -> n^2=33*quotient+8
II. n^2-8=33*quotient+15 -> n^2=33*quotient+15+8=33*quotient+23
III. n^2-8=33*quotient+21 -> n^2=33*quotient+21+8=33*quotient+29

Cheking divisibility of n^2 by 3 (as 33=3*11):
if n=3,6,9,12... n^2 is divisible by 3 and remainder is 0
if n=1,4,7,10... n^2 divided by 3 the remainder is 1
if n=2,5,8,11... n^2 divided by 3 the remainder is 1

but:
33*quotient+8 divided by 3, the remainder is 2
33*quotient+23 divided by 3, the remainder is 2
33*quotient+29 divided by 3, the remainder is 2

In one case 0 or 1 and in the other case 2, no coincidence.

Answer E

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n^2-8=answer options
solving, no answe gives me an interger
so answe is none of the above

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Hi Everyone

Correction answer for this question - option E

Explanation - To check divisibility of n^2 -8 by 33 we need to check its divisiblity by 3 and 11 (as 33=3 x 11)

Check divilsiblity by 3 - Every +ve number is in a form of 3K, 3K+1, 3K+2 and if you sqaure these numbers and divide by 3 ,the outcome is only either 0 or 1

for example - 4%3=1 , 9%3=0, 49%3=1

Now if a number leaves remainder 0 is reduced by 8 and divided by 3 the remainder will be always 1
and if a number that leaves remainder 1 is reduced by 8 and divided by 3 the remainder is always 2

Therefore if number not divisible by 3 then it wont be divisible by 33

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E.
because the remainder of the sum of two numbers is sum of each number's remainder. The remainder of minus eight divided by thirty-three is minus eight. So, any additional remainder minus eight will have to be checked against the given remainders. n^2 can never be eight, so I is out. 15+8 = 23 it can also never be 23, so II is out and finally 21+8=29, It can also never be this number, so II is out.

Only E. None of these remains.

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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Given

N^2 = 33k + r + 8

We will test each value for R and see which comes out to be the perfect square

1st Test:

R = 0

Possible values of n^2: 8, 41, 74, 107, 140, 173 ....
None of them are perfect square

2nd Test:

R=15

Possible values of n^2: 23, 56, 89, 122, 155, 188 ....

None of them are perfect square

3rd Test:
r= 21

possible values of n^2: 29, 62, 95, 128, 161, 194 ....

From this we can conclude that 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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