Factorial Product?

Question

Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers. For example 6! can factorial can be expressed as the product of 6-3 = 3 positive integers, namely 8, 9, and 10 (6! = 8*9*10).

Source: 1990 AIME #11

XR: This is an old contest problem so it should have an answer as well as a method of attaining that result without the use computers.

By the way, the correct answer is 23.

XR: The question says n! has to be the product of "n - 3" consecutive integers. In the case where n = 23, 23! can be expressed as the product of 23 - 3 = 20 consecutive integers.

Answer 1

We require
n! = k*(k-1)*(k-2)*...*(k-{n-4}) [NOTE k = largest integer]
= k! / (k - {n-3})!
So (k + 3 - n)! = k! / n!, ...(1)
obviously k > n

So we need to find the largest pair of (k,n) that satisfy the above equation.

Let's do a little manipulation and let k = n + m
From eqn 1,
(m+3)! = (m+n)! / n! = (n+m)*(n+m-1)*...*(n+1)
= product of m consecutive integers, n+m being the largest.

So interestingly if the property holds for (n, k), it also holds for (m+3, n+m) or (k-n+3, k).

The lowest value for which this works is n = 4, k=24, m=20
So it also works for n=23, k=24, m=1.
Applying the recursive property again takes us back to n = 4.

It also works for n = 6, k = 10, m = 4
The recursive property takes us to n=7, k = 10, m = 3.

So for every n value for which it works, there exists another corresponding n value which shares the same k value.

As n gets larger, the difference, m, between n and k gets smaller because the number of terms in the product approaches n. ie as n --> ∞,
n-3 --> ∞ also,
hence k - n --> 0+, or m --> 0+

With this logic, we can conclude that n=23 is the maximum, because it already has the lowest permissible value of m(=1). Any higher value of n would require m < 1 and hence would require non integral values of k.

NOTE there are other ways of showing that n>23 cannot work. We know that 23! = 24*23*...*5 = 24!/4!.
Conside n > 23, say n = 23 + p, with m = 1.
For it to work, the following equality must be true:
(23+p)! =? (24+p)! / (4+p)!
But we know that (23+p)! = (24+p)! / (24+p)
So we must find p such that (4+p)! = 24+p.
Clearly the only such value is p=0.
For p ≥ 1, (4+p)! > 24+p
hence (23+p)! > (24+p)! / (4+p)!

In order for n>23 to work, m must be greater than 1, hence k>25. We know that this pair must match with another pair (m+3, k). We also know by trial that the only n values from 1-23 that work are 4, 6, 7, 23. There is no n value in this range that has a k value >24.

Conclusion: n = 23 is the highest positive integer that answers the question.

Answer 2

Been staring at this all day: I didn't want to give an answer unless i could add something useful, but here's my thinking...

It is a truly interesting question, and I totally admire the insight that Bracha brought in his answer. I can't do better at this point!

However, I cannot see why Bracha's solution of n=23 is the *largest* solution, as required by the question. Why did analysis stop at solutions of the form (n+1)!/24, when other possibilities may exist from considering (n+2)!/120, (n+3)!/720 and all the rest? After all, the solution for n=6 is up in the realm of (n+4)(n+3)(n+2).

So, while I admire both the question and the response given, I feel that we haven't gone far enough. Please explain further - I'd love to know the answer!!

**Thanks Dr D : Read yours through carefully, and it looks watertight to me! So to be pedantic, the answer is n=23, since 23! = 25852016738884976640000 = 24*23*22*...*7*6*5 ( a string of 20 consecutive integers).

Answer 3

I am doing a spreadsheet to solve this, but only spend 15 min on it since I need to get other stuff done. Will come back and report the results..
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Edit: Thanks to Dr. D that I realize this is a simple quadratic equation!!! (but I have no exact idea of what Dr. D is trying to say) Though, the discrete function has helped on obtaining values. It does not help in finding the solution any easier, nor am I sure of the final data. Anyone's got access to a Super Computer around here? Anyhow, my discrete function is for n!=k [k-1] [k-2]. I had no clue what Dr. D is trying to write, but his explaination gave me the idea about discrete functions (electrical engineering know-how, though I am Mechanical engineer by trade).

With that said, what it means we can also rewrite the equation in n! = k³ - 3k² + 2k, while keeping in mind that k must be an integer. This equation helped tremendously with finding the values and also narrow down the range of search very quickly. With this basic equations in mind, I started the search at the cubic root for every n!. The spreadsheet did not find 23 to be one of the answers. Instead, 38, 39, 41, 47, 95, 103, 110, 142 and 154 were found instead.

Unfortunately, my answers could not be verified by a scientific calculator (I never own a graphing calculator. Did not start to use calculator until I start learning trig. For prior math calculations, I alwasy rely on brain power. Though, I don't think a graphing calculator would help) Here are my problems with the answers I found:

First of all, values in PC are not stored discretely, which is ironic since digital informations are always stored discretely, one bit at a time. However, to store a very large number say 9.9 x 10ª with a being 100. You would require 500 bits of data, or 62.5 Bytes (1,024 Bytes per KB; 1,024 KB per MB; 1,024 MB per GB and so on). Usually, one number is stored within 1 Byte of data, which is plenty for most uses.

To store large numbers for this problem, instead of storing the precise value, the computer has taken the task of store the discrete value, which is the primary goal. PCs we used were instruct to save the values in scientific expression format instead. This is where the problem starts. First of all, scientific expression format has no intention of displaying the precise value of the number. Secondly, digital data storage can never store decimals well, so there are two layers of data problems already.

At lower numbers, the numbers come out okay, but the spreadsheet stars havingg problems. After it found a possible match, after singling out a largest number for a larger number, say 39!. The spreadsheet continued to list what more largest numbers, which suggest that the value stored for cubic root of 39! and (³√39!)+1 are the same. Also, the spreadsheet quits storing values larger than 9.99 x 10^307 which in this case, 171!.

Unless we can get access to a super computer, I don't think any of us can find the answer.
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Edit 2:

Finally fully grasp the problem, I think people are getting very confused on what the question is asking for and I don't think whoever dreamed up this problem obviously is not thinking too clearly either. First of all, the very same equation: n!=k [k-1] [k-2] can be transformed into

n! (k-3)! = k!, to simplify furthur:
2 n! (k-3)(k-4)....(k-(n+1)) = k!
In the case of 6!, we would have
10! = 2 x 6! x 7 x 6 x 5 x 4 x 3

3! is a special case in which k and n are identical. Also, 23 is a completely wrong answer, I don't know why people are obessing over it. We are looking for THREE consecutive numbers, not TWENTY!! To double check my answers, I tried to identify the numbers precisely by taking the cubic root of the factorials again, and line up the numbers. Here are the ones that the spreadsheet pulled off before the digital data storage problem got into the way again,

38! = 805700234291877 x 805700234291878 x 805700234291879
39! = 2732299854169810 x 2732299854169811 x 2732299854169812
*Remember: to verify the answers, make sure your calculators or computers can handle the number precisely down to every last digit.

Number 41, 64, 76 & 103 has the same bizzare outputs, with which I can't verity the results.

The original problem designers and most of people who answer this question have serious problems understanding the factorial calculations and how discrete calculation works, which is not mathethatics at its purest at alll.


XR

Answer 4

The smallest product of n-3 consecutive positive integers is (n-3)! That is always too small. Similarly (n-2)!, (n-1)!/2 and n!/6 are too small. The next smallest is (n+1)!/24 = n! (n+1)/24. For n+1 > 24 that is too big and hence there are no such products for n > 23. For n = 23 it works.

This is a great problem!