Calculator Killer 5

Number Theory Level 3

How long is the string of consecutive 0's at the end of this number?

( 9 ! ) ( 10 ! ) \huge (9!)^{(10!)}

1 0 10 10^{10} 100 100 10 10 100 ! 100! 10 ! 10!

Zandra Vinegar by Zandra Vinegar

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1 solution

Ankit Nigam
Dec 24, 2015

Since 9! has one trailing 0, we can write 9! as 10y (where y is an integer). Therefore ( 10 y ) 10 ! (10y)^{10!} = y 10 ! y^{10!} . 1 0 10 ! 10^{10!} . Also if we observe y here (which is equal to 36288), last digit of powers of y will be either 2,4,6 and 8. Thus we can say that this number has 10! trailing zeroes.

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