Let n~ be defined for all positive integers n as the remainder when (n

Ok so I was initially not able to do this problem because I didn't understand how to divide 31!/32...

Now that you guys have made the problem more clear I have a question-

5!/15 must be an integer? Because 5x4x3x2x1 contains the factors of 15 (5 x 3)

When I do 17!/30 in a calculator the result is not an integer?

First of all, 17!/30 = 11856247603200 = integer, but what 17!/30 has to do with this problem?

I trying to demonstrate the concept in this problem, which is new to me, in a different example- I think the calculator's notation just makes it appear as 1.something^e even though that doesn't necessarily mean it's not an integer. It's clear now.

For big numbers do not user calculator (unless it's not advanced) or Excell, use Wolframalpha: https://www.wolframalpha.com/

Hope it helps.

Simple- all this question is asking is if you have 31!/32 then what is the remainder? You don't necessarily have to expand 31!- 32 fits in because you have 16 and 2 so there is no remainder

Thus
"A"

\(= \frac{(32-1)!}{32}\)

\(= \frac{31!}{32}\)

Numbers \(8 * 4 = 32\), will cancel out the denomintor and hence the reminder will be ZERO.

Hence, Answer is A

Bunuel VeritasPrepKarishma

32! has various multiples of 32 built in say 8 * 4, 16* 2. Will not various no of multiples affect my remainder?
For eg, even if 8*4 will cancel out 32 in denominator leaving remainder = 0, I still have one multiple of 16*2, correct?

32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5

Since we can safely say that there are at least five 2s in 31! (for example, 31! has the factors 16 = 2^4 and 8=2^3), the remainder is zero.

Alternate Solution:

32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5

We want to know the remainder when 31! is divided by 2^5. If we can establish that there are at least 5 factors of 2 in 31!, then we will know that 2^5 evenly divides into 31!, which means that the remainder would be 0. Let’s determine if we can find at least 5 twos in 31!:

31! = 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x … x 1

31! = 31 x (2 x 15) x 29 x (2 X 2 x 14) x 27 x (2 x 13) x 25 x (2 x 2 x 2 x 3) x … x 1

Notice that we have found seven 2s, which is two more than what we needed. Thus, we know that 2^5 will evenly divide into 31!, leaving a remainder of 0.

Answer: A

Hi,

So basically as long as there are enough prime factor powers in the numerator to cancel out all the prime factor powers in the denominator, the remainder is 0 correct?

Yes. If the numerator contains the same (or more) primes as denominator and the (positive integer) powers of primes in the numerator are at least as big as the powers of the same primes in the denominator, then the result of the division will be an integer, so the remainder will be 0.

Hi All,

This prompt is an example of a 'Symbolism' question - the prompt 'makes up' a math symbol, tells you what it means and asks you to perform a calculation using it. Based on the information in the prompt, we're told that....

32~ is the remainder when (32-1)! is divided by 32.

Now, there's no way that the GMAT would expect you to calculate the value of 31!, so we have to think in terms of what 31! actually is.

Here's an example that's a bit easier to deal with:

4! = (4)(3)(2)(1) = 24

What numbers divide EVENLY into 24?

1, 2, 3, 4, 6, 8, 12 and 24

You can clearly see why 1, 2, 3 and 4 divide in - they're in the 'chain' of numbers that are multiplied together.
6 divides in because (2)(3) = 6 - and you can see the (2) and the (3) in the 'chain'
Similarly, 8, 12 and 24 are also 'combinations' of the numbers in the 'chain', so they divide evenly in too.

31! has LOTS of numbers in it, so it's evenly divisibly by LOTS of different integers. If you were to write out 31!, you would see a (2) and a (16). This means that (2)(16) = 32 divides evenly into 31!, so there will be a remainder of 0.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel, long time no hear no see

Here is my solution.

to define if \(31!\) is divisible by \(32\) make prime factorization of \(32\) so we get this --- > \(2^4\)

\(31!\) certainly will have four \(2\) as prime factors, hence \(\frac{31!}{32}\) we get integer.

if this solution correct. What if similar question would have remainder - how could i find a remainder ?

i would appreciate your explanation

thank you

32~ = 31!/31

Now if 31 from denominator is consumed completely this means that remainder is 0.

A

I understand the solution to this problem and why the remainder = 0. My confusion arises from how the question is worded. It says, n~ = remainder, but then goes onto saying what the value for the remainder is (32~). Isn't that then asking for the value of the multiple when (n-1)!/n = R32? I understand that it doesn't make sense since 31! divides evenly into 32 and the remainder is 0. Can someone help me understand the WORDING of the question, as stated, I understand the solution itself.

From the question:

"n~ (is) the remainder when (n - 1)! is divided by n"

We want to find "32~", so replace "n" everywhere above with "32":

"32~ (is) the remainder when (32 - 1)! is divided by 32"

So we want to find the remainder when 31! is divided by 32. Since 31! is divisible by 32, the remainder is 0.

Official Explanation:
32~ = the remainder that emerges from 31!/32, according to the definition given.
However, 31! will in fact be a multiple of 32, because along the multiplicative way to 31!, we multiply together a 2 and a 16 (or alternatively, because we multiply together a 4 and an 8, or because we multiply at least five even numbers together).
When we divide a multiple of 32 by 32, our remainder will be 0.

32~ = \(\frac{(32-1)! }{ n}\)

= \(\frac{31!}{32}\)

= \(\frac{31!}{2^5}\)

Since there are 5 even integers within 31!, the remainder will be zero. Answer is A.