A Factorial Product Problem

[@sternmg]

Magnificent! I was intimidated by the sheer number of factors, but your disciplined methodical consideration from the top down was the key and led directly to the solution. Thank you!

[@bash2357]

A nice variation is to ask what do you divide 1!x2!x....1000! by in order for the result to be a square number?

[@nathanisbored]

i think the phrasing "power of three" or "power of four" etc can be a bit confusing because when I think of "power of three" i usually think of 3^n not n^3. i guess it makes enough sense in context though

[@magnusPurblind]

Lol, I just watched the whole video without noticing. I think the eye takes precedence over the ear in cases like this

[@DrBarker]

Good point, it really ought to be "third power" or "fourth power".

[@GU-jt5fe]

Way to go Dr. Barker! BARK BARK BARK BARK BARK BARK

[@worldnotworld]

What a bizarrely out-of-the-blue but fun puzzle! How do you come up with these things? Do you discover or invent them? How are they invented? To invent them seems to me to be a form of puzzle-solving of its own, like designing crossword puzzles, or games in general. Might there even be a branch of mathematics, of some degree of rigor, concerning the _generation_ of puzzles? If so, it would seem to lie somewhere in the plane defined by rigors of formal mathematical proof crossed with the art of mathematical speculation and conjecture...

[@DrBarker]

This is something I find really fun - I'm glad it's taken your interest! For this problem, I'd seen similar problems like express 1! x ... x 100! as the product of a factorial and a square number (which can be solved using a similar argument to this problem), and wanted to create something similar, but a bit more complicated.

[@joeagorn]

Hey Dr. Barker. I really like your content and I've been watching for a while. I was wondering if you'd consider doing anything with Newton Polygons to factor polynomials. I've tried looking for material on this but it seems super rare and only for wacky abstract algebra stuff. I'm simply interested in reducing polynomials over the field of integers

[@ronbannon]

Nice job. By the way, you say term when you mean a factor.

[@quantumbuddha777]

I found a different solution where a=996

[@DrBarker]

Interesting! What values did you use for b and c?

[@quantumbuddha777]

@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.

[@quantumbuddha777]

@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.

[@quantumbuddha777]

@@DrBarker I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later.

[@quantumbuddha777]

@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996, b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497, c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125. We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.

[@Agent29416]

Couldn't you condince the last to 998x996!x995x994!x10x(3^222)