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# Let . Prove that there exists a unique positive integer k such that is a perfect square.

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Let \(S = 1! 2! \dotsm 100!\). Prove that there exists a unique positive integer k such that \(S/k!\) is a perfect square.

Aug 2, 2020

#10
+140
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As a sort of lead in, consider the simpler situation S = 1!.2!.3!.4!.5!.6!.7!.8! = (1).(1.2).(1.2.3). ... .(1.2.3.4.5.6.7.8).

It's easier to see the product if it's written as,

1! = 1

2! = 1.2

3! = 1.2.3

4! = 1.2.3.4

5! = 1.2.3.4.5

6! = 1.2.3.4.5.6

7! = 1.2.3.4.5.6.7

8! = 1.2.3.4.5.6.7.8

S will equal the product of all of the numbers on the rhs of the equals signs.

Notice that all of the even numbers 2, 4, 6, 8 occur an odd  number of times, while the odd numbers 3, 5, 7 each occur an even number of times, (we can ignore the 1's).

\(\displaystyle S = 2^{7}.3^{6}.4^{5}.5^{4}.6^{3}.7^{2}.8^{1}, \\ = (2^{7}.4^{5}.6^{3}.8^{1}).(3^{6},5^{4}.7^{2}) ,\\ = (2^{6}.4^{4}.6^{2}).(2.4.6.8).(3^{6}.5^{4}.7^{2}), \\ =(2^{6}.4^{4}.6^{2}).2^{4}.(1.2.3.4).(3^{6}. 5^{4}.7^{2}), \\ = (2^{6}.4^{4}.6^{2}).2^{4}.4!.(3^{6}.5^{4}7^{2}).\)

Apart from the 4! in the middle everything else is a perfect square, so if we divide S by 4! the result will be a perfect square.

The same routine can be used for S = 1!.2!.3!. ... 100! .

The factorial in the middle will be 50!.

This only works if the final factorial is a multiple of 4. Try this for S = 1!.2!.3!.4!.5!.6! for example, and it doesn't work because the power of 2 that appears in the middle will be 3, so not a perfect square.

Aug 2, 2020

#1
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S=1!2!3! .... 100!

S=1*100 * 2*99 * 3*98 * ..... 100*1

S= 100!*100!

If you divide by 1!  then you have a perfect square.

Logic tells me that no other K will work but that is not a proof.

Anyway, one such number is  k=1

Aug 2, 2020
#2
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Hi Melody:  I wrote a very short computer code and it indicates that  S/50!  gives a perfect square, which is 3,438 digits long! It begins: 9 427 570 616 109 075 150 705 396 694 534 826 731 504.........etc. Of course 50! sits smack in the middle of [1! 2! 3!.....100!],  and it probably is no surprise! But how do you prove it mathematically? I don't know! Maybe Alan, heureka or yourself can do so.

Aug 2, 2020
#3
+111077
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mmm

If the maths I have presented above is correct I do not see how this could work because you would be left with some primes example 37 that do not have a pair.

Maybe there is something wrong with my logic that  S = 100!*100!

I am not doubting your output but it does not make sense to me.

Maybe another mathematician will comment.

Aug 2, 2020
#4
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melody i think your formula S=(100!)is incorrect (for example 2 appears more than twice in the product S) but i think if you divide S by 50! you get a perfect square

Guest Aug 2, 2020
#7
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Hi Melody: I wrote another short computer code, but this time only for numbers S= [1! 2! 3!.....20!], and again

when S is divided by 10!, it gives a pefect square. It appears that dividing the product S by the "median?" or the middle number gives a perfect square ! Why? I don't know.

Guest Aug 2, 2020
#5
+1130
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k can also equal 0

Aug 2, 2020
#6
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since 0!=1

jimkey17  Aug 2, 2020
#8
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Wolfram Alpha certainly gives an integer for

It returns the following integer:

So it looks like k = 50, though I don't know how to prove it's unique.

Aug 2, 2020
#12
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Hello Alan,

heureka  Aug 4, 2020
#13
+31085
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Thanks heureka, I've no idea why my images have sometimes been rejected and sometimes accepted during the last few days!

Alan  Aug 4, 2020
#9
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For some reason, which I don't quite understand, ALL the Prime Factors of the product[1! 2! 3!...100!] that have an ODD exponent become EVEN exponents when divided by 50! as can be seen in the following breakdown:

[(6941 digits) = 2^4731 × 3^2328 × 5^1124 × 7^734 × 11^414 × 13^343 × 17^250 × 19^220 × 23^174 × 29^129 × 31^117 × 37^91 × 41^79 × 43^73 × 47^61 × 53^48 × 59^42 × 61^40 × 67^34 × 71^30 × 73^28 × 79^22 × 83^18 × 89^12 × 97^4] /

[50!   = 2^47 * 3^22 * 5^12 * 7^8 * 11^4 * 13^3 * 17^2 * 19^2 * 23^2 * 29 * 31 * 37 * 41 * 43 * 47]

=  (6876 digits) =[ 2^4684 × 3^2306 × 5^1112 × 7^726 × 11^410 × 13^340 × 17^248 × 19^218 × 23^172 × 29^128 × 31^116 × 37^90 × 41^78 × 43^72 × 47^60 × 53^48 × 59^42 × 61^40 × 67^34 × 71^30 × 73^28 × 79^22 × 83^18 × 89^12 × 97^4].

Since ALL the Prime Factors, after division by 50!, have EVEN exponents, it therefore follows that the final result is a perfect square! Is it because all the Prime Factors of 50! are repeated in the remaining 51!....100! ?? I don't know.

Aug 2, 2020
#10
+140
+2

As a sort of lead in, consider the simpler situation S = 1!.2!.3!.4!.5!.6!.7!.8! = (1).(1.2).(1.2.3). ... .(1.2.3.4.5.6.7.8).

It's easier to see the product if it's written as,

1! = 1

2! = 1.2

3! = 1.2.3

4! = 1.2.3.4

5! = 1.2.3.4.5

6! = 1.2.3.4.5.6

7! = 1.2.3.4.5.6.7

8! = 1.2.3.4.5.6.7.8

S will equal the product of all of the numbers on the rhs of the equals signs.

Notice that all of the even numbers 2, 4, 6, 8 occur an odd  number of times, while the odd numbers 3, 5, 7 each occur an even number of times, (we can ignore the 1's).

\(\displaystyle S = 2^{7}.3^{6}.4^{5}.5^{4}.6^{3}.7^{2}.8^{1}, \\ = (2^{7}.4^{5}.6^{3}.8^{1}).(3^{6},5^{4}.7^{2}) ,\\ = (2^{6}.4^{4}.6^{2}).(2.4.6.8).(3^{6}.5^{4}.7^{2}), \\ =(2^{6}.4^{4}.6^{2}).2^{4}.(1.2.3.4).(3^{6}. 5^{4}.7^{2}), \\ = (2^{6}.4^{4}.6^{2}).2^{4}.4!.(3^{6}.5^{4}7^{2}).\)

Apart from the 4! in the middle everything else is a perfect square, so if we divide S by 4! the result will be a perfect square.

The same routine can be used for S = 1!.2!.3!. ... 100! .

The factorial in the middle will be 50!.

This only works if the final factorial is a multiple of 4. Try this for S = 1!.2!.3!.4!.5!.6! for example, and it doesn't work because the power of 2 that appears in the middle will be 3, so not a perfect square.

Tiggsy Aug 2, 2020
#11
+111077
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Thanks very much Tiggsy, makes perfect sense.

Aug 4, 2020