+0

# Help

+1
391
7
+550

What is the sum of the tens digit and the units digit in the decimal representation of 9^2004?

Jul 8, 2019

#1
+24762
0

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

Jul 8, 2019
edited by ElectricPavlov  Jul 8, 2019
#2
+2

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

Jul 8, 2019
#3
+2

At the top of the pyramid, I believe 90 should equal 1.

.

Guest Jul 8, 2019
#4
+24762
0

You are correct....I originally had 01 in there, but it got deleted while I was formatting my answer before posting.....fixed it...THANX !   ~EP

ElectricPavlov  Jul 8, 2019
#5
+111396
+2

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

Jul 8, 2019
edited by CPhill  Jul 8, 2019
#6
+25325
+2

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 $$6+1 =\mathbf{ 7}$$

Jul 9, 2019
#7
+4586
0

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.

Jul 10, 2019