Fun with Exponents

[jay_shark]

One way of doing this is to expand (10-1)^9 using the binomial theorem . This should be easy to do and you can determine very easily that the last two digits are 89 .

z1=9
z2=(10-1)^9 = 10^z1 - z1c1*10^(z1-1) +...z1c(z1-1)*10 -1
last two digits are 90-1=89 since z1=9 .


Now expand z3= [(10-1)^9]^9 again using the binomial theorem and using the fact that the last two digits of (10-1)^9 is of the form ...89 .The last 3 digits should be easy to find . You should arrive at -600+890 -1 =289

The last 4 digits of z4 = 4000-1600 +2890-1 =5289

The last 5 digits of z5 = -60000 +640000-11600+52890 -1 =45289 .

After this point it keeps on repeating and the answer becomes obvious .

[bigpooch]

You beat me to the punch! I was just about to post the
binomial solution.

[bigpooch]

If x is some "tower of nines" and with the notation for z[j]
as in your post, this is just to clarify (perhaps for any
readers out there):

Since x is odd, with C(n,r) denoting the combinations of n
choose r,

[10-1]^x = (-1) + C(x,1)*10 - C(x,2)*100 + C(x,3)*1000 - ...

so

z1 = -1
= 9 (mod 10)

z2 = -1 + C(9,1)*10 = -1 + 90
= 89 (mod 100)

z3 = -1 + C(89,1)*10 - C(89,2)*100 = -1 + 890 - 600
= 289 (mod 1000)

z4 = -1 + C(289,1)*10 - C(289,2)*100 + C(289,3)*1000
= -1 + 2890 - 1600 + 4000
= 5289 (mod 10000)

z5 = -1 + C(5289,1)*10 - C(5289,2)*100 + C(5289,3)*1000
-C(5289,4)*10000
= -1 + 52890 - 11600 + 64000 - 60000
= 45289 (mod 100000)