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Re: Fun with Exponents
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 . |
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