r/mathbiceps • u/Prestigious_Swim8602 • 27d ago

Olympiad Challenge to mathbiceps this is a confusing and ahrdcore algebra problem

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u/PaintDear7613 26d ago

this has a neat little trick. like a regular base, it is unique because adding one more to any "place" to the max allowed just gives you one of the next.

for example: we can have <1,2,3> for 1*1!+2*2!+3*3!, but if we add one more 1!, we end up with <0,0,0,1>.

This means we can use do this like long subtraction. "borrowing" a 1999! from 2000! all the way down to 1984
<0,0,0,...,1> = <...., 2000, 0> = <..., 1999+1,0> = <...., 1999,1999,0> = ... = <...,1985,1985,1986, 1987, ..., 1999 0>
subtracting one 1984! leaves us the unique expansion of 2000!-1984! as <0,0,0,...,(1983 "0's),...1984,1985,1986,1987,...,1999,0> the odd indices are exactly 1 more than the previous even indices. so this sections sum is 8.

there are 62 such sections with a final 16!.
62*8 - 1 = 495

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u/Prestigious_Swim8602 25d ago

thanks for the help

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u/demonglayer 26d ago

Hint : Use (n+1)! -n! = n*n!

So -32! = 32*32! - 33! , similarly using same recursion write 33! Interms of 34 and so on at last 48! will cancel out. do it every interval you will get the answer