Problems & Puzzles: Puzzles

Puzzle 637. Factorials as product of palindromes As a follow up of Puzzle 636 now we ask for these factorials that are expressible a a product of two or more palindromes (two or more digits, please). Giovanni Resta sent one example: 15! = 88 * 525 * 585 * 48384 Q. Can you find more examples (please just send your smallest & your largest)?

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Contributions came from Giovanni Resta, Jan van Delden & Hakan Summakoğlu 

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Giovanni wrote:

100! = 48384^3*53535*57375^2*405504*2087802*2572752*5703075*
29344392*84277248*86766768*460979064*512272215*521818125*
526010625*650797056*8828118288*57348900984375*8212971179212

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Jan wrote:

40!=171*414*525*585*595*5775*44544*57375*405504*2572752*4095995904

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Hakan wrote:

Largest: 50! = 55 * 252 * 575 * 585 * 595 * 5005 * 6336 * 18081 * 27072 * 44544 * 48384 * 52325 * 55955 * 57375 * 405504 * 4811184

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On Dec. 30, 2022 Adam Stinchcombe wrote:

While paging through the past puzzles, I noticed the palindromic products had factors with a different number of digits.  With a random aka non-exhuastive aka maybe not smallest I found:

 

39!  is   equal to  63536*76167*44544*11011*21312*52325*57375*48384*57375*48384
 

and the factors are all palindromes with the same number of digits (some amount of symmetry).  Next up, maybe the same number of digits and no repeat factors.

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On Jan 10, 2023, Emmanuel Vantieghem wrote:

...There is an immediate follow up of Giovanni Resta's solution :

101! = 101*100!

 = 101*48384^3*53535*57375^2*405504*2087802*2572752*5703075*
29344392*84277248*86766768*460979064*512272215*521818125*
526010625*650797056*8828118288*57348900984375*8212971179212

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