# Incorrect answer supplied to one of the Quiz problems

One of the quiz problems presented is to investigate whether 90!/(30!)^3 is a whole number. The answer is supposedly true: it is a whole number and a very weak argument is provided to show that. The argument given for proof is incorrect. it starts out with a proof, then doesn't complete the logic and simply restates the answer as a given. I have tested the premise that the division is a whole number and have found a contradiction to this premise. After factoring out the first 30! one is left with a quotient of (90 * 89 * ... * 32 * 31)/(30!)^2. I have continued the remaining factorization using all of the argument provided as "proof" and then continued until the numerator is not divisible by 3 any more. At that point there exists (among others) a residual factor in the denominator of 3^13. Since 3 is a prime number and all the factors of 3 in the numerator have been divided out, the residual quotient is not a whole number. My factorization algorithm is entered into an excel workbook and I'm happy to share this with any one who would like to see the proof that 90!/(30!)^3 is indeed NOT a whole number. Note by Detlev Tiszauer
1 year, 11 months ago

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Yes the given answer IS correct and I made an error in my factorization which led me to the wrong conclusion. For proof that the answer is indeed a whole number, I offer the following argument, which is easier for me to follow than any one I had seen before. The combinatorial (n | m) is represented by a number on the n'th row and m'th column of Pascal's triangle. It can be shown by induction that all entries on that triangle are whole numbers. The given problem can be represented as the product of two combinatorial expressions: (90 | 30) * (60 | 30) or equivalently ((60 + 30)! /(60! * 30!)) * ((30+30)! / (30! * 30!) = 90!/(30!)^3. Since each of the two combinatorial expressions is a whole number, their product is also a whole number and hence the given expression is a whole number as given by the posted answer.

- 1 year, 11 months ago

Hi Detlev, I believed you're referring to this problem. The answer is indeed correct. Please review the solution discussion to see how it's solved.

Staff - 1 year, 11 months ago

Thank you for your answer. Yes the answer is a whole number as I was able to prove to myself subsequent to the first attempt. For proof, I offer the following argument, which is easier for me to follow than that provided online. The combinatorial is represented by the numbers on a row of Pascal's triangle. From that triangle, it can be shown by induction that all entries are whole numbers. The given problem can be represented as the product of two combinatorial expressions: (90 | 30) * (60 | 30) or equivalently ((60 + 30)! /(60! * 30!)) * ((30+30)! / (30! * 30!) = 90!/(30!)^3. Since the each of the two combinatorial expressions is a whole number, their product is a whole number and hence the given expression is a whole number. Again thank you for your response and sorry for my error in the first post.

- 1 year, 11 months ago