If n is a prime number and n 3, which of the following

Question

If n is a prime number and n ≠ 3, which of the following could be the remainder when 100! + n is divided by 3?

I. 0
II. 1
III. 2

A. II only
B. III only
C. I and II only
D. II and III only
E. I, II and III

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Bunuel - I checked out reminders concepts in the math book https://gmatclub.com/forum/compilation- ... 86714.html
but the below idea wasn't mentioned

You can add and subtract remainders directly, as long as you correct excess or negative remainders. "

if x leaves a remainder of 4 after division by 7, and y leaves a remainder of 2 after division by 7, then x +y leaves a remainder of 4 + 2 = 6 after division by 7.

Similarly 100! leaves a remainder 0 on division by 3 so we are only interested in the remainder when N divided by 3,which will be actual remainders of 100! + n is divided by 3

if n =1 remainder 1 so overall remainder is 1
if n = 2 remainder 2 so overall remainder is 2
if n = 6 remainder = 0 so overall remainder is 0

All 3 are possible right then why Princeton says D?

Cheers.

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Official Answer

D

PareshGmat · Accepted Answer

\frac{100! + n}{3} = \frac{100!}{3} + \frac{n}{3}

\frac{100!}{3}  is a perfect division

\frac{n}{3}  may give remainder either 1 or 2 as n is prime

Answer = D

Bunuel · Answer

Notice that we are told that n is a prime number and n ≠ 3. Thus, n cannot be 1.

n also cannot be 6 or any other multiple of 3, thus the remainder cannot be 0.

It can be 1 for n=7 and 2 for n=5.

Answer: D.

Hope it's clear.

Jp27 · Answer

ohh my god. Ok thanks.

actleader · Answer

My answer is D.

if a number n is prime,
and together with that does not equal 3
we can divide it on 3 whithout a remainder.

0 can't be the answer.

we can also check it with, for example 5! and list of primes such as 2!,5!,7!...

in your example
 if n = 6  - 6 is not prime

ChiranjeevSingh · Answer

Hi,

A minor correction in your post:
for n=2, the remainder will be 2, not 1.
for n=7, remainder will be 1.

With respect,
CJ

nityamskhurana · Answer

Another way to look at it is:

100!+n where n ≠ 3, since 100! will be a factor or 3, so we just have to care about n.
Hence, if n=2 then remainder of 2/3 is 2.
for any value of n>3, and n being prime it can be written as (6k+1) or (6k-1).
Hence, factor (6k+1)/3 will give remainder as 1, and (6k-1) would leave remainder as 2.

Please correct me if I am wrong.

Regards,
Nityam

doe007 · Answer

100! is divisible by 3. So we need to find out the remainder when the prime number n is divided by 3.

For n = 2 or 5, remainder is 2.
For n = 7, remainder is 1.

n cannot be 3 as specified and cannot be any other multiple of 3 as n is prime. So the remainder cannot be 0.

Answer is D.

jemiljacob09 · Answer

100! is sure to have last digit as 0. (100 x 99 x 98 x......)
Therefore the question narrows down to n/3, where n is prime. 
Since n is not equal to 3, n/3 will leave a remainder for all cases.

Hence the answer is Option D

Please let me know if this approach is correct.

ScottTargetTestPrep · Answer

Since 100! is divisible by 3, then the remainder when 100! + n is divided by 3 is same as when n is divided by 3. If n is 5, then the remainder is 2. If n is 7, then the remainder is 1. However, the remainder can’t be 0 since n is neither 3 nor it is a multiple of 3 (since n is a prime ≠ 3).

Answer: D

narenbs · Answer

Since n is a prime number greater than 3, n can be expressed as 6p + 1 or 6p - 1. Therefore 100! + n can be expressed as either 100! + 6p + 1 or 100! + 6p - 1. When divided by 3 these expressions leave remainders of either 1 or -1 (i.e. 2). Therefore remainders can be 1 or 2. Therefore answer is D.

bumpbot · Answer

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If n is a prime number and n 3, which of the following

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