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E - None of these as none of them leave the specified reminders
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As n^2 is going to be divided by 33 the value of n cannot be 6 as 36-8=28 which is less than 33 .

So, n is obviously more than 6 . if n is 7 , remainder is 8.

if n is 8, remainder is 23 , if n is more than 9 the remainder is obviously going to be more than all option ( 0,15,21) .

so correct ans is . E. NONE OF THESE
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Used substitution for n to check for remainder pattern. n upto 8, remainder is -ve. n>=8, the remainder will be 0, 1, 2 and so on. except for multiples of 11 or 3. Since 15 and 21 are both divisible by 3, they will never be remainder.
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Answer E. None of these.
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test with values of n = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
n^2= 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225

n^2-8 = -7,-4,1,8,17,28,41,56,73,92,113,136, 161, 188 , 217
factors of 33 = 33,66,99,132, 165, 198, 231
remainder possible = 1,4,7,8,17,28,23, 26, 14,29, 19

OPTION E none of these ; is correct

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 n2-8 / 33 can rem?

n2-8=r (33)
n2=r+8 (33)
r=0
n2=0+8
n2=8
n2=2
impossible


r=15
n2=15+8
n2=23
n2=2
impossible

r-21
n2=21+8
n2=29
n2=2
impossible

option E
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Let's check for
I. 0
For n^2 -8 to be perfectly divisible by 33 it has to be a multiple of 33 i.e 0,33,66, etc.
Considering n^2-8 = 33 here
n^2= 33+8
n^2= 41

which is not a perfect square; similarly, for other cases here in all the multiples of 33.
Hence, we can eliminate this option.

II. 15
n^2-8 must be 15 more than a multiple of 33.
n^2 should be (15+8) more than a multiple of 33, meaning 23 more than a multiple of 33.

now checking for perfect squares
33*0+23=23
33*1+23 = 56
33*2 +23= 89 and so on...
none of the results obtained are a perfect square hence, we can eliminate this option too.

III. 21
n^2 -8 must be 21 more than a multiple of 33.
n^2 should be (21+8) more than 33, meaning 29more than a multiple of 33.
checking for perfect quare
33*0+29= 29
33*1+29= 62 and so on
none of the results are a perfect suare hencem we can eliminate this option as well.


hence, the answer is E. none of the above.
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I got B only

Had to use the trial and error method to get this though
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n^2-8/33 should give either 0,15, 21.
for r\remainder =0
means, n^2= 33k +8 +0 , as 33= 11*3, let's check the divisibility by 3, it leave remainder 2. so 1 is not possible
n^2= 33k+8+15= 33k+23, divided by 3 leave remainder 2 again, so 15 is also not possible
n^2= 33k +8+21= 33k+29, divided by 3 leave remainder 2 again, so 21 also not possible
a perfect square can never leave remainder 2 when divided by 3, it always leave a remainder of 0 or 1
Hence answer is E
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I think the answer is E.

Working backwards with the options:

n^2 - 8/33 = n^2/33 - 8/33

There will always be -8 as remainder. So exploring options, we need to add +8 to the final remainder and check if it is possible for any square of number
1. 0+8 = 8 remainder cannot be possible
2. 15 + 8 = 23 is not possible remainder of n^2
3. 21 + 8 = 28 is not possible remainder of n^2
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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


 


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

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Let \(n^2= 33k + r + 8\)

we know any integer z can be expressed as follows:
z = 3m + p , p = {1,2,3}

when p = 0,
\(z^2 = 9m^2 \)
when divided by 3, reminder is 0

when p= 1,
\(z^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1 \)
when divided by 3 , reminder is 1

when p = 2,
\(z^2 = 9m^2+12m+4 = 3(3m^2 + 4m +1) +1 \)
when divided by 3, reminder is 1.

so for any integer z, \(z^2 \) divided by 3 always leaves reminder 0 or 1

lets check the options ,
I. r = 0,
\(n^2 = 33k + 8 = 3(11k + 2) + 2 \)
which is not possible

II. r = 15,
\(n^2 = 33k + 15 + 8 = 3(11k+7) + 2\)
which is not possible

III. r = 21,
\(n^2 = 33k + 21+ 8 = 3(11k+9) +2\)
which is not possible

so E. None of the above
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break it as n^2/33 and 8/33.
Now reminder when 8/33 = 8
and reminder of n^2/33 , for this try values for n=2,3,4,5,6
for n=2; value is non of the options.
for n=3; again 3^2 /33 = 9 rem and 9-8 =1 /33 and rem 1 not in the option. Similarly do for few more.
for n=4, 16-8 = 8 so not available
for n=5, 25-8 = 13 so not available
for n=6, 36-8 =28 again not available. so if we notice now , we got values more than 1,15 and 21 as well. Hence none of the options above.
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We want to know if the remainder of
n^2−8 when divided by 33 could be 0, 15, or 21.

Add 8 to each choice because the expression is
n^2−8
0 → 8
15 → 23
21 → 29

Since 33=3×11
33=3×11

Squares divided by 3 can only leave remainders:

0^2=0
1^2=1
2^2=1

So a square can only be 0 or 1 mod 3—never 2.

Check the new numbers:
8= 2
23=2
29=2

All three would require n^2=2 which is impossible


E. None of these
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so, here you can apply the remainder theorem formula, very basic which i used is quotient*divisor+remainder=dividend
here divisor= 33
qoutient=x
remainder(put the given values)
dividend=n**2-8
so it would look something like 33x+0=n**2-8
here, n say = 9
so, 81-8=73
multiple cases are possible for 15,21.
hence,none of these is the answer
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Can (n^2- 8)/33 leave a reminder of 0,15, 21? 33k + 0 / 33k + 15 / 33k + 21 are all divisible by 3. 33 is divisible by 3 and so is 0, 15, 21. Can n^2 - 8 ever be divisible by 3? We can try few numbers 1,2,3,4,5. It wont be. E is the answer.
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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
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N = integer positive
(n^2-8)/33 gives remainder of

n2-8 = 33 k+r
n2 = 33k+8 + R
For r=0, n2 = 33k+8,
We cannot get any integer value of K,so incorrect.
For r= 15, n2 =33k+ 8+ 15 → N2 = 33K +23, again not possible for k
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