Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$

Question

While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)

$x^{(n+4)} \mod 10 = x^n \mod 10$

How can this be proven?

Answer 1

$x^4\equiv1\pmod{10}$ for $x$ and $10$ coprime, by Euler's theorem, since $\varphi (10)=4$.

(The proof for $x$ not coprime to $10$ can be found in @Bill Dubuque's answer.)

Answer 2

Consider how the difference, i.e., $$x^{n + 4} - x^{n} = x^{n} \left( x^4 - 1 \right) = x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)$$ behaves for all cases of $x \mod 10$ from $0$ to $9$, inclusive. For $x$ being $0$, the result is $0$. For $x$ being an even positive value, then $x^n$ is a multiple of 2, while either $x^2 + 1$ or $x^2 - 1$ is a multiple of 5, so together are a multiple of $10$. Finally, for $x$ being an odd value, consider the cases:

$1$: $x^2 - 1$ is $0$

$3$: $x^2 + 1$ is $10$, i.e., $0 \mod 10$

$5$: $x^n$ is a multiple of $5$ and $x^2 - 1$ is a multiple of $2$, so together shows congruent to $0$

$7$: $x^2 + 1$ is $50$

$9$: $x^2 - 1$ is $80$

This shows that the result is congruent to $0$ in all cases. The other answers here are generally shorter and simpler, so they're better if you're able to use them. However, this is a fairly general way to check basically any congruence operation where ever it can be relatively easily used (e.g., where the mod divisor is not a variable or too large).

Answer 3

This is true because of the periodicity of the last digits of numbers raised to powers. if you check cor $x=2,3,4\ldots n$ you will see that after every $4$th power the last digit is the same, $2^1 = 2$ and $2^5 = 32$ thus the last digit is the same. $x^n \mod 10$ is essentially asking for the last digit. Thus, $$x^{n+4} \mod 10 = x^n \mod 10$$ Some numbers have a periodicity of $2$ when raised to powers, for example, $9$ but the unit digit is still the same as the first power as the first.

Answer 4

We want to show that $x^{n+4}-x^n=x^n(x^4-1)$ is a multiple of $10$.

We first notice that if $x^n$ is odd then $x^4-1$ will be even and vice versa. Consequently, the product will always be even.

If $x$ is divisible by $5$ then the product is divisible by $10$ because we have an even product which is divisible by $5$.

If $x$ is not divisible by $5$ then $x=5m+k$ for some $m$ and $k$ where $k\in\{1,2,3,4\}$.

Now, $x^4-1=(5m)^4+4(5m)^3k+6(5m)^2k^2+4(5m)k^3+k^4-1$ and every term is divisible by $5$ with the possible exception of $k^4-1$.

But there are only four values of $k$ and if you raise each to the fourth power and subtract $1$ you get a multiple of $5$. Once again, this gives us an even number divisible by $5$ which is a multiple of $10$.

Answer 5

It is the special case $\,p,q,k = 2,5,4\,$ below.

Lemma $\,p\neq q\,$ primes $ $ & $\, \color{#c00}{p\!-\!1,\,q\!-\!1\mid k}\,\Rightarrow\, pq\mid\smash[t]{\overbrace{ x^n(x^k\!-\!1)}^{\large a}}\,$ for all $\,x\,$ & $\,n>0$

Proof $\ $ $\,p,q\,$ are coprime so $\,pq\mid a\iff p,q\mid a,\,$ by Euclid / unique prime factorization.

When $\, p\mid x\,$ then $\,p\mid x\mid x^n\mid a\,$ by $\,n>0,\,$ hence $\,p\mid a\,$ by transitivity of divisibility,

$ $ else: $\,\ p\nmid x\,$ so $\!\bmod p\!:\ x\not\equiv 0\,$ so $\,x^k \equiv (\color{#0a0}{x^{p-1}})^{\smash[t]{\Large\color{#c00}{\frac{k}{p-1}}}}\!\!\equiv\color{#0a0} 1\,$ by $\rm\color{#0a0}{Fermat},\,$ so $\,p\mid x^k\!-1\mid a$.

Thus in every case $\,p\mid a,\,$ and $\,q\mid a\,$ by $\,p,q\,$ symmetry (i.e. same proof works for $q).\,$

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Remark $ $ Above is a special case of this generalization of Euler-Fermat - which often proves handy.

Theorem $\ $ Suppose that $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{e_{1}}\cdots\:p_k^{e_k}\ $ and suppose that for all $\,i,\,$ $\ e\ge e_i\ $ and $\ \phi(p_i^{e_{i}})\mid f.\ $ Then $\ m\mid a^e(a^f-1)\ $ for all $\: a\in \mathbb Z.$

Proof $\ $ Notice that if $\ p_i\mid a\ $ then $\:p_i^{e_{i}}\mid a^e\ $ by $\ e_i \le e.\: $ Else $\:a\:$ is coprime to $\: p_i\:$ so by Euler's phi theorem, $\!\bmod q = p_i^{e_{i}}:\, \ a^{\phi(q)}\equiv 1 \Rightarrow\ a^f\equiv 1\, $ by $\: \phi(q)\mid f.\ $ Since all $\ p_i^{e_{i}}\ |\ a^e (a^f - 1)\ $ so too does their lcm = product = $m$.

Examples $\ $ You can find many illuminating examples in prior questions, e.g. below

$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$

$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$

$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$

$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$

Answer 6

If you can prove $x^4 \equiv 1 \pmod {10}$ that's enough as $x^{n+4}\equiv x^n\cdot x^4 \equiv x^n\cdot 1\equiv x^n\pmod {10}$

If you google Euler's theorem that will be true for all numbers relatively prime to $10$. Intuitively (but hand wavy) if $x$ and $10$ are relatively prime then $x^n$ will be relatively prime. There are only $4$ possible last digits that are relatively prime to $10$ so $x^k$ will cycle through the same four digits. (More or less.)

If $x$ is even or divisible by $5$. Well, if it's both then $x \equiv 0 \mod 10$ so $x^k \equiv 0 \pmod {10}$ and $x^{n+4} \equiv x^n \equiv 0\pmod {10}$. Likewise if it's odd but divisible by $5$ then $x^k \equiv 5 \pmod{10}$ and so $x^{n+4} \equiv x^n \equiv 5 \pmod {10}$.

Now if $x$ is even but not divisible by $5$ then. Well, $x^n$ is even and there are only $4$ even digits it could end with and it cycles through them.

More to the point $2*j \pmod 10 \equiv 2*(j \pmod 5)$ and there are only $4$ non-zero classes $\pmod 5$ so $x^k$ just cycles through those.

Answer 7

Because:

$x^{n+4} = x^{n-1}x^5$
$x^n = x^{n-1}x$
($ab\mod 10) = (a(b\mod 10) \mod 10)$

... you only have to prove that $(x^5 \mod 10)=(x\mod 10)$.

... which means: $((x \mod 10)^5\mod 10) = (x \mod 10)$.

Because there are only 10 possible values for $(x \mod 10)$ these 10 values can be calculated directly:

• $0^5\mod 10=0 \mod 10=0$
• $1^5\mod 10=1 \mod 10=1$
• $2^5\mod 10=32 \mod 10=2$
• $3^5\mod 10=243 \mod 10=3$
  ...
• $9^5\mod 10=59049 \mod 10=9$

Answer 8

Hint $$ x^{n} \left( x^4 - 1 \right) \equiv x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)\equiv \left( x - 1 \right) x \left(x+1\right) x^{n-1} \left( x^2 + 1 \right) \pmod{2}\\ x^{n} \left( x^4 - 1 \right) \equiv x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)\equiv x^{n} \left( x^2 - 1 \right) \left( x^2 -4 \right)\pmod{5}\\ \equiv x^{n-1} \left( x- 2 \right) \left( x- 1 \right) x\left( x+2 \right) \left( x+ 2 \right) \pmod 5 $$

Now use the fact that one of $x-1$ or $x$ is even, and that among 5 consecutive integers one is a multiple of 5.