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

Basically, the equation at hand looks as follows: x^2 - 8 = 33k + R Note that 33 = 11*3 , which means it's divisible by 3. The remainders we have - 0, 15 and 21 - are also divisible by 3. This means that for any of them to 'work', we need x^2 - 8 to be divisible by 3 as well. Let's see all possible remainders on the example of 9, 10 and 11: n = 9 (divisible by 3) n^2 - 8 = 81-8=73 -> leaves remainder 1 n = 10 (remainder 1) n^2 - 8 = 100-8 = 92 -> leaves remainder 2 n = 11 (remainder 2) n^2 - 8 = 121-8 = 113 -> leaves remainder 2 Therefore, for n^2-8 we can never have divisibility by 3, which means none of the remainders in question fit. The 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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Since 33 is a multiple of 3, the remainder after dividing by 33 would have to match what happens when we divide by 3.

Any square number n^2 can only leave a remainder of 0 or 1 when divided by 3.

If n^2 leaves 0, then n^2 −8 leaves 1.
If n^2 leaves 1, then n^2 −8 leaves 2.

So n^2 −8 can only leave remainders that are 1 or 2 when dividing by 3.

It can never be a multiple of 3. But 0, 15, and 21 are all multiples of 3, so none of them can be the remainder.

The answer is E. None of these.

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E

n^2 mod 3 -> always 0 or 1
(n^2 - 8) mod 3 -> always 1 or 2

all 3 options result in 0 -> none of the above

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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Correct Answer: E

The 30-Second Shortcut: Divisibility by 3
When dealing with a divisor like 33, check its prime factors. Since 33 is a multiple of 3, the logic of the equation must also hold up when divided by 3.

Let's set up the remainder equation (where Q is the quotient and R is the remainder):
n 2 −8=33Q+R
n 2=33Q+R+8

The Core Rule: A perfect square (n
2
) will always leave a remainder of either 0 or 1 when divided by 3. It will never leave a remainder of 2.

Since 33Q is completely divisible by 3, the remainder of the right side depends entirely on (R+8). Let's test the options for R:

I. If R = 0: The remainder part is 0+8=8. When divided by 3, 8 leaves a remainder of 2. (Impossible)

II. If R = 15: The remainder part is 15+8=23. When divided by 3, 23 leaves a remainder of 2. (Impossible)

III. If R = 21: The remainder part is 21+8=29. When divided by 3, 29 leaves a remainder of 2. (Impossible)

Since all three options force the perfect square to leave a remainder of 2 when divided by 3, none of them can be the correct remainder.

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For 0 remainder, n2-8 =33x
I tried with different values of x, as per pattern, we can’t observe any square number such that n2 = 33x +8
Hence 0 cannot be the case.

I tried similar methodology that got no case for 15 and 21 remainder too.

Hence (E) is the answer

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Let's think backwards to solve this.

If n^2 - 8 leaves remainder r when divided by 33, then n^2 must leave remainder r+8.

I. 0 -> n^2 must leave remainder 8.
But a perfect square can never leave a remainder of 8 when divided by 11.

II. 15 -> n^2 must leave remainder 23.
When 23 is divided by 3, the remainder is 2. But a perfect square can only leave a remainder of 0 or 1 when divided by 3.

III. 21 -> n^2 must leave remainder 29.
When 29 is divided by 11, the remainder is 7. But a perfect square can never leave a remainder of 7 when divided by 11.

Thus, none of the three are possible.

Ans : E

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when we square any number and divide by 3 we get remainder 0 or 1. So when 8 is subtracted from n^2 the remainder will be 2 or 1. Since 3 is a factor of 33 anything that is divisible by 33 should be divisible by 3. If we look at the options all of them are multiples of 3 which would divide the number evenly so the answer is E

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Ans: 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

n^2 - 8 is divided by 33
means n^2-8 is divisible by both 11 and 3
In case of divisibility by 3, the remainders can be 1 and 2 only or there 11 multiples.
So None of these. ans

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Used Remainder logic for 3 for this question as we have 33 the divisor is 3*11 and hence anything that divisble by both of them will be dividend so we check this logic there no number whoese square or second power can generate 0 or 1 as remainder and for 11 they generate 0,1,4 and 9 but if we subtract the 8 out of it no answer choices mentioned pop up hence E

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The most logical way to solve this scenario

For ( n^2 - 8 ) has to be perfectly divisible by 33 it has to be divisible by 3 and 11

So lets check the whether ( n^2 - 8 ) is divisible by 3

When a number is divisible by 3 it can leave the remainder 0,1,2... No other possibility can exist.

So possible values of ( n^2 - 8 ) are 1,8,17,28,41,56. On calculating the remainders for the above values when divided 3 by are 1,2. In any case we can't get the remainder as Zero.

Since, We are 100 percent sure that ( n^2 - 8 ) is NOT divisible by 3.
Therefore, we can also conclude that ( n^2 - 8 ) is NOT divisible by 33.

So we can eliminate Answer choice 1

-------------------------

Now Lets Evaluate whether we can get remainder 15 or not.

Visualise this scenario properly

in order to get the remainder as 15
( n^2 - 8 ) Has to be addition of 15 and a multiple of 33.
In other words ( n^2 - 8 ) === 15 + 33(any integer multiple)

So we can take 3 as a common from the expression

( n^2 - 8 ) ===== 3 [5 +11(any integer multiple) ]

Therefore in order to get the remainder as 15 ( n^2 - 8 ) has to be a multiple of 3 which can't be the case as explained above.So we can eliminate answer choice 2.

Similarly, we can do for answer choice 3.

Please let me know if any of the steps is unclear.

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Basically, the equation at hand looks as follows:
$x^2 - 8 = 33k + R$

Note that $33 = 11*3$, which means it's divisible by 3.
The remainders we have - 0, 15 and 21 - are also divisible by 3. This means that for any of them to 'work', we need $x^2 - 8 $ to be divisible by 3 as well.

Let's see all possible remainders on the example of 9, 10 and 11:

n = 9 (divisible by 3)
$n^2 - 8 = 81-8=73$ -> leaves remainder 1

n = 10 (remainder 1)
$n^2 - 8 = 100-8 = 92$ -> leaves remainder 2

n = 11 (remainder 2)
$n^2 - 8 = 121-8 = 113$ -> leaves remainder 2

Therefore, for n^2-8 we can never have divisibility by 3, which means none of the remainders in question fit. The answer is E.

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None of the above is answer as per my understanding

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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Here if you open the algebera using the $a^2$-$b^2$ identity, we will get (n-$\sqrt{8}$)(n+$8\sqrt{}$),

since n = Integer, we wont get a integer result for this.

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N^2 - 8 is the dividend
33 is the divisor
let's take Q as the quotient and R as the remainder.

so i can create a equation: n^2 - 8 = 33Q + R
Which also equals to n^2 = 33Q + R + 8

That means the value of 33Q + R + 8 is a perfect square.

Now i thought let's see the factors of 33 - that's 3 & 11.

If we look at perfect squares:
1^2 = 1 when divided by 3 leaves a remainder 1
2^2 = 4 when divided by 3 leaves a remainder 1
3^2 = 9 when divided by 3 leaves a remainder 0
4^2 = 16 when divided by 3 leaves a remainder 1
5^2 = 25 when divided by 3 leaves a remainder 1
....and so forth.
So basically, any perfect square when divided by 3 should leave a remainder 1 or 0

In that case our RHS should also give us a remainder 1 or 0 when divided by 3, for it to satisfy the equation of N^2

So, trying all the cases given in the question:
Case I. R = 0
So, the equation becomes - n^2 = 33Q + 8 + 0
I know 33Q is divisible by 3, but 8 when divided by 3 leaves a remainder 2 which means, in this equation, it won't be a perfect square.

Case II. R = 15
the equation becomes - n^2 = 33Q + 8 + 15 = 33Q + 23
23 when divided by 3 leaves a remainder 2. So again not a perfect square. This remainder is also not possible.

Case III. R = 21
the equation becomes - n^2 = 33Q + 8 + 21 = 33Q + 29
29 when divided by 3 leaves a remainder 2. So not a perfect square either. This remainder is not possible either.

Thus, the answer is E. None of these.

Just for better understanding of what we did to solve this, we can also take another example. So for instance n = 9 and n^2 = 81
We can break 81 = 33 x 2 + 8 + 7, this is the same as saying n^2 = 33Q + 8 + R, with Q = 2 and R = 7
In this example, 33 x 2, when divided by 3, leaves the remainder 0
and 8+7 = 15, when divided by 3, also leaves a remainder 0
So we can conclude that our RHS is a perfect square.

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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IMO : E
since this question has directly given us the remainder options we can go for substituting the value of the remainder in the equtaion formed like Rem(n^2-8/33) =?
n^2-8 = 33k +Rem
here trying with all three : rem =0. doesnt satidfy
rem = 15, doesnt satisfy
rem = 21 also doesnt satisy
hence option E.

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Check the quotient with the divisors factors with the remainder options given as follows:-
For 1st option
(n^2 -8) / 33 = remainder which is 0 in this case.
so, n^2 which on putting 0 in above equation give as 8 should be divisible by 33 (3 * 11).
Hence, starting with smallest factor 8 is not divisible by 3 so this option gets eliminated.

For 2nd option
(n^2 - 8)/33 = 15 which gives n^2 as 23. Again same process on dividing it by 3 smallest factor of 33 it is not divisible.
Hence, this option also gets eliminated.

For 3rd option
Same process gives n^2 as 29. On dividing it with 3 we get 2 means it is not divisible.

All three option gets eliminated.

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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33= 3 × 11

n^2 - 8 shall be divisible by 3 or 11

n^2 gives remainder of 0 or 1 when devided by 3. n^2-8 is n^2-2, non devisible by 3.

So 0, 15 or 21 which all are factors of 3 cannot be the remainders

Ans E

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Resolving it to simplest form,
n^2 - 8 =~ r % 33 -> will give remainder
n^2 =~ r + 8 % 3*11

0 ->
0+8 = 8
for n^2 we need to check the square mod -> consider x as a number, x% 3 will give 0,1,2. x^2 % 3 will give 0,1.
So, for 8 % 3 -> 2, which is not a valid square mod of 3, thus this number isn't a valid remainder otherwise it must have a residue of 0 or 1 for % 3.

similarly for 15 and 21,
23, 29 %3 -> 2 which isn't a valid residue for square mod 3.

Thus all options are invalid.
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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Ans - E None of these. I have just utilized the Answer choice substitution method.
0-8/33 is not correct, similarly 225-8/33 not correct not divisible by 3 (if we sum up the digit 217) and similarly for 21^2 441-8/33 not divisible so none of these is the correct answer in my opinion. Thanks!

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 got option choice E ......
HERE to get a reminder it should not get divisible by the factors of 33 ,,, 33 = 3 * 11 ,,
-- so ,, here the numbers should not be divisible by 3 so ,,
--- Now ,, 0 is divisible by 3
15 is divisible by 3
21 is divisible by 3 ,, so none can be a reminder

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Lets take n^2-8 = 33k + r

We will assume values of k as (1,2,3....) and r as 0,15,21 and check if any of the answer is possible

Lets take an example of r = 0
n^2 - 8 = 33k + 0 => n^2 = 33k + 8

Now assuming the values of k as 1,2,3... will give us n^2 = 41, 74, 107...
Since, none of these numbers are perfect squares, we can eliminate 0

Repeat the same step for r = 15 and 21 and we will get to the same solution. Hence -> 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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