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# How many consecutive digit 0s does have on the right end when it is written in base 12?

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How many consecutive digit 0s does $$1000_4!\cdot100_4!\cdot10_4!$$ have on the right end when it is written in base 12?

I've simplified that to 64!*16!*4!.

I know that 64! has 2^63 and 3^30. 16! has 2^15 and 3^6. 4! has 2^3 and 3^1.

I combine those to get (2^63)(2^15)(2^3)(3^30)(3^6)(3), which simplifies to (2^81)(3^39).

I know for every 12, there's a zero when written in base 12, so I make it (4^n)(3^m).

So I get (4^40)(3^39)(2), which becomes (12^39)(4)(2).

But it's incorrect. What am I doing wrong here?

Oct 1, 2020
edited by PocketThePenguin  Oct 1, 2020
edited by PocketThePenguin  Oct 1, 2020

#1
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What is it that you trying to find out? Do you want the number of "Trailing Zeros" in converting 1000! * 100! * 10! from base 4 to base 12?

Oct 1, 2020
#2
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1000!_4  *  100!_4  *  10!_4 =130 1302221110 0132333201 3122022132 1230301003 2233112303 0023022013 1201210330 1112231131 3211111013 1013222232 1332011100 3000000122 1021202112 0000000000 0000000000 0000000000 0000000000 [Which has 40 trailing zeros]. Now, convert this to base 12:

171886B 38B1954098 7094864863 34AA256B70 879A167705 4B2A283341 A280000000 0000000000 0000000000 0000000000 [Which has 37 trailing zeros]

Oct 1, 2020