#11
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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 . |
#12
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Re: Fun with Exponents
You beat me to the punch! I was just about to post the
binomial solution. |
#13
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Re: Fun with Exponents
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) |
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