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9^0  01

9^1  09

9^2  81

   729

   6561

   59049

   531441

   4782969

  43046721

  387420489

  3486784401

  31381059609

Lo0ks like if even   the number ends in 1

  the tens pattern as 0  8  2  6  4       4  6  2   8  0      2004   would be the 5th in the sequence

 1 + 6 = 7          My guess

Here are the last 20 digits:  9^2004 mod 10^20 =2025682304 6793686561

9^2004   =   (10 - 1)^2004

So....the binomial expansion is

10^2004 - C(2004, 1) * 10^2003  + C(2004, 2)*10^2002 -  ..... - C(2004,2002)(10)^2  + C(2004, 2003) (10) -  1 

All of the terms except the last three end in  "000"...so they add nothing to the sum of the last two digits in the expansion

And the last three terms are

-200,700,600 + 20040 - 1  =

 abs   [ -600 + 40 -  1  ]

abs [ -601 + 40  ]    =

abs  [-561 ]  =

561

And the sum of the tens and the units digits in the expansion  =  7

What is the sum of the tens digit and the units digit in the decimal representation of \(9^{2004}\)?

\(\begin{array}{|rcll|} \hline && 9^{2004} \pmod{100} \\ &\equiv & \left(3^2\right)^{2004} \pmod{100} \\ &\equiv & 3^{2\cdot 2004} \pmod{100} \\ &\equiv & 3^{4008} \pmod{100} \quad &| \quad 3^{\phi(100)} \equiv 1 \pmod{100} \qquad \phi(100) = 40 \\ & & \quad &| \quad 3^{40} \equiv 1 \pmod{100} \\ &\equiv & 3^{40\cdot 100 + 8} \pmod{100} \\ &\equiv & \left(3^{40}\right)^{100} 3^8 \pmod{100} \\ &\equiv & 1^{100} 3^8 \pmod{100} \\ &\equiv & 3^8 \pmod{100} \\ &\equiv & 6561 \pmod{100} \\ &\equiv & \mathbf{61\pmod{100}} \\ \hline \end{array} \)

The sum of the tens digit and the units digit in the decimal representation of \(9^{2004}\) is \(6+1 =\mathbf{ 7}\)

Euler's Theorem gives us 9^2004phi100=9^4  mod 100=3^8 mod 100=6561 mod 100= 61 mod 100, and 6+1=7.