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I believe the answer is E.

First I set up an equation:

n^2-8 = 3*11*k + R, where k is an integer and R is the remainder.

Next, plug in the roman numerals for R and inspect the resulting equation to see if its possible.

I; R = 0; yields the equation:
n^2 = 3*11*k + 8.

By testing values of k, I was not able to find an answer which yielded a perfect square, so I decided this one is not possible.

II; R = 15; yields the equation:
n^2 = 3*11*k +23

Similarly, by testing k = 1,2,3, I was not able to find a perfect square, so I decided this one is not possible.

III: R = 21, yields the equation:

n^2 = 3*11*k + 29.

I noticed that the remainder for the last two are both prime numbers, which led me to suspect that it was not possible to obtain a perfect square by adding a prime to a multiple of 33. Therefore, I guessed E.

However, can someone please explain the underlying principle of number theory which allows us to confirm that the answer is E? I spent too much time on this question.
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Answer is D. II and III only.
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 is B.

n2-8/33
n2/33 - 8/33

and since it is given that n is a positive integer and that the remainder cantr be negative, hence started substitution with n=3
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If n is a positive integer, which of the following could be the remainder when n^2 - 8 is divided by 33

n^ 2- 8 = r (mod 33)
n^2 = r + 8 (mod 33)

n^2 = r + 8 (mod 3)
n^2 = r + 8 (mod 11)

Case 1: n^2 = r + 8 (mod 3) = r + 2 (mod 3)
n = 0; n^2 = 0 = 0 (mod 3)
n = 1; n^ 2 = 1 = 1 (mod 3)
n = 2; n^2 = 4 = 1 (mod 3)
n = 3; n^2 = 9 = 0 (mod 3)
n = 4; n^2 = 16 = 1 (mod 3)

n^2 = r + 2 (mod 3) is either 0 or 1 but not 2.

r is never a multiple of 3.

Since 0, 15 & 21 are all multiple of 3.

IMO E
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We can break 33 as 3x11 and consider division only by 3 for now.
If we divide any number n by 3 the possible remainders are:

0 -> 0
1 -> 1
2 -> 2
3 -> 0

Now for n^2.
0 -> 0
1 -> 0
2 -> 4 ->1
SO possible remainders 0 and 1.
Now for n^2 - 8 if divided by 3 we get remainders as :

For n^2 leaving 0: n^2 - 8 will leave 1.
For n^2 leaving 1: n^2 - 8 will leave 2.
So looking at the options I can see that all the three numbers 0, 15 and 21 leaves remainders 0.
IMO the answer should be 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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imo E
since 33=3*11
so whenever we square any number and divide by 3 we get 0 or 1 as remainder not 2 1^2=1 %3 =1 4%3=1 etc
all answer choices are multiples of 3 so not possible that they will not be divisible by 3
n^2-8 when divided by 33 cannot leave those remainders since 3 divides all those numbers
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Say, \(n^2 - 8\) gives remainder \(R\) when divided by 33.
for simplicity let us take only \(n^2\) thus, \(n^2\) gives remainder \(8 + R\) when divided by 33.

However, 33 has factors 3 and 11, \(n^2\) must also satisfy congruency when divided by 3 and 11 also.

Now take 3 since it is smaller among the 2 factors of 33.

when \(n = 0\), \(n^2 = 0\) , remainder when divided by 3 = 0
when \(n = 1\), \(n^2 = 1\) , remainder when divided by 3 = 1
when \(n = 2\), \(n^2 = 4\) , remainder when divided by 3 = 1

Thus, we will find that \(n^2\) for any positive integer n , remainder when divided by 3 will be either 0 or 1

Now test the 3 cases in the answer choices.

I. For \(R = 0\), \(n^2\) gives remainder \(8 + R = 8\) when divided by 33.
Now let us check what remainder we will get when divided by 3, by further dividing 8 by 3 we get remainder 2. but we should get either 0 or 1. Hence \(R = 0\) is not possible.

II. For \(R = 15\), similarly \(8 + R = 23\), when divided by 3 gives remainder 2. Hence \(R = 15\) also not possible.

III. For \(R = 21\), again \(8 + R = 29\), when divided by 3 gives remainder 2. Hence \(R = 21\) also not possible.

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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Divisibility rules apply here.

If I reduce n^2 by 8, the remainder decreases by 8.
So if n^2 - 8 / 33 remainders are 0, 15, 21 then the n^2 remainders must be 8, 23, 29
Because 33 is divisible by 3, if n^2/33 leaves remainders 8,23,29 when divided by 33 then n^2 must also leave the same remainder as those numbers divided by 3.
The remainders are:
I. 8/3 -> remainder = 2
II. 23/3 -> remainder = 2
III. 29/3 -> remainder = 2
Which value for n^2 leaves a remainder of 2 when divided by three? None of them. Because every perfect square can only have a remainder of 1 when divided by 3. This is because every number that is divided by three can either be a multiple of 3 and leave a remainder of 0, or leave a remainder of 1 or 2 (if it left a remainder of 3, it would be the "next" multiple of 3). With perfect squares that are to be divided by 3, you are also squaring the distance from each number to the next multiple of 3. If the distance is 1, then the distance remains 1 because 1 squared is 1. If the distance is 2, then the distance become 4 but that in itself has a remainder of 1 when divided by three.
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I solved it using plugging in numbers. We have to check for remainders when n^2 - 8 is divided by 33.
I'll start with n = 3 (because 3^2 = 9). Rem will be 1. Similarly
n = 4, rem = 8
n = 5, rem = 17
n = 6, rem = 28
n = 7, rem = 8 again. Nowhere is 0, 15 or 21 in this cycle.
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Both will have remainders, it can't be zero
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Making use of modulus in this question starting from zero we observe that
None of the three numbers works. They all fail the mod 3 test because r+8= 2 (mod 3) Since 8 mod 3 is 2and the three numbers are divisible by 3
Hence 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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If n=0, n^2-8/33 = -8/33 =Remainder (25)
If n=1 , -7/33 =R(26)
If n=2, -4/33 =R(29)
If n=3, 1/33 =R(1)
If n=4, 8/33 =R(8)
If n=5, 17/33 =R(17)
If n=6, 28/33 =R(28)
If n=7, 41/33 =R(8)
If n=8, 56/33 =R(23)
If n=9, 73/33 =R(7)
If n=10, 92/33 =R(26)

0,15,21 is none in the list.

E) None of these
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I got E.

Essentially the question requires you to spot that n^2 -8 needs to be divisible by 3. As both 33 and the answer choices are divisible by 3. Then check if n^2 - 8 is ever divisible by 3. It isnt. n^2 only leaves remainders 0 or 1.

Alternative way is to try values and quickly realise that none of the values work, also note the question states COULD BE NOT MUST BE.
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i am going with e.
even if it is a long method, it is basic and foundational. we know dividend = divisor x quotient + remainder
to follow this simple rule with all the three options, n^2=33k+8, n^2=33k+23 and n^2=33k+29 there is no such possibility that fulfills any three condition making it none of the above.
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Ans E - None of these . N2- 8 = 33K + r . Substitute all reminders in options and check and see when on division by 3 does it give the same reminder as n2/3 . You will notice in all case it leaves a reminder of 2 which wont be equal to n2/3 in any case . Eg n2/3 = 33k/3 +(8+15)/3 = 33k/3+23 /3 = reminder (0+2) whereas n2/3 where n2 could be any perfect square lets say 25 when divided by 3 leaves 1 as reminder which does not match our RHS.
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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So, n^2 = r +8

n is not a +ve integer for either of 0,15,21 so 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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n^2 - 8 = 33k + (0, 15,21)
n^2 = 33k + (8,23,29) = 3(11k+(2,7,9)) + 2 = 3q+2(2 remainder)
which is not possible, because any number can be written in the below form,
n = 3p, 3p+-1
n^2 = 9p^2, 9p^2 +- 6p + 1
therefore, n^2 / 3 can only give remainders 0 or 1.
Therefore none of the options. (E)
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