$n=1$ Then $\frac{10^1}{9} = \frac{10}{9}$ remainder = $1$.
For $n\geq2$, how does you do this?
I want to prove that last digit is always zero, of $10$ raised to power. How do I do that please by induction.
thanks you
$\endgroup$ 35 Answers
$\begingroup$If you know (and are allowed to use) modular arithmetic, it's as simple as:
$$10 \equiv 1 \pmod 9 \implies 10^n \equiv 1^n = 1 \pmod 9$$
EDIT: For a proof by induction, you already have the base case. The assumption is that $10^k = 9m + 1$.
So here's the inductive step: $10^{k+1} = 10^k\cdot 10 = (9m+1)\cdot 10 = (9)(10m) + 9 + 1 = 9(10m+1) + 1$
$\endgroup$ 1 $\begingroup$Hint:
$$10^n=999...9 +1 \,.$$ .
$\endgroup$ $\begingroup$In case you don't have modular arithmetic, there's always
$$ \ 10^n \ - \ 1^n \ \ = \ \ (10 \ - \ 1) \ (10^{n-1} \ + \ 10^{n-2} \cdot 1 \ + \ 10^{n-3} \cdot 1^2 \ + \ \ldots \ + \ 10 \cdot 1^{n-2} \ + \ 1^{n-1}) \ \ $$
or
$$ \ 10^n \ \ = \ \ 1 \ + \ 9 \ (10^{n-1} \ + \ 10^{n-2} \cdot 1 \ + \ 10^{n-3} \cdot 1^2 \ + \ \ldots \ + 10 \cdot 1^{n-2} \ + \ 1^{n-1}) \ \ . $$
$\endgroup$ $\begingroup$The answer by Deepak should be more than enough, but I sense that you are struggling to see how induction really fits in here. I am going to give a more drawn out answer, but I think it may be clearer for you.
Base case: We have that $10=9+1$ so this is fine.
Inductive step: We want to prove that $$ S(n) : 10^n = 9\ell+1, r\in\mathbb{Z} $$ holds for all $n\geq 1$. Fix some $k\geq 1$ and assume that $$ S(k) : 10^k = 9\ell+1, \ell\in\mathbb{Z} $$ holds. We want to show that $$ S(k+1) : 10^{k+1} = 9m+1, m\in\mathbb{Z} $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} 10^{k+1} &= 10\cdot 10^k\tag{manipulate}\\[0.5em] &= 10\cdot(9\ell+1)\tag{by $S(k)$}\\[0.5em] &= 9(10\ell+1)+1\tag{manipulate}\\[0.5em] &= 9m+1\tag{$m=10\ell+1$ and $m\in\mathbb{Z}$} \end{align} we see that the right side of $S(k+1)$ follows.
Thus, by mathematical induction, the statement $S(n)$ holds for all $n\geq 1$.
$\endgroup$ $\begingroup$You have already given the base case. Now assume it works for $n$: $10^n \equiv 1\pmod{9}$. Multiply both sides by $10$: $10^{n+1} \equiv 10\pmod{9}$. We also have $10 \equiv 1\pmod{9}$, so therefore $10^{n+1} \equiv 1\pmod{9}$. Induction concludes the proof.
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