induction

**edu_1313** · September 27th 2008, 01:27 PM

demostrate that the exponentiation of the number
12890625 is another number which its lasts numbers are 12890625.

exemple: 12890625^2 = 166168212890625

Why can i demostrate it by induction? i don't know what i have to write on the right side of the equation (12890625)^n = ?
to start the indutcion

thanks
PD: i'm sorry if my english is not too good, it isn't my usual lenguage.

**o_O** · September 27th 2008, 03:08 PM

$n = 12890625$

From your clue, notice that:
$\begin{array}{rcl}n^2 & = & 166168212890625 \\ n^2 & = & 1661682 \cdot 10^8 + 12890625 \\ {\color{red}n^2} & {\color{red}=} & {\color{red}10^8 m + n} \end{array}$

So assume the statement is true for $P_{k}$ , that is $n^k$ last 8 digits are $n$ i.e. $n^k = 10^8 s + n$ . So it remains to show that it is true for $n^{k+1}$
Multiply both sides by $n$ :
$\begin{array}{rcl} n^k & = & 10^8 s + n \\ n^{k+1} & = & 10^8 sn + n^2 \\ n^{k+1} & = & 10^8 sn + {\color{red}(10^8m + n)} \\ & \vdots & \end{array}$

Can you finish?

**edu_1313** · September 28th 2008, 04:42 AM

yes, now i've could finish. thanks

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