Any one of the integers ## 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ## can occur?

[Math100]

Homework Statement

Prove the following statement:
Any one of the integers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ can occur as the units digit of $a^{3}$.

Relevant Equations

None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8$, or $9\pmod {10}$.
Thus $a^{3}\equiv 0, 1, 8, 27, 64, 125, 216, 343, 512$, or $729\pmod {10}$.
Therefore, anyone of the integers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ can occur as the units digit of $a^{3}$.

 

[fresh_42]

That's true.

 

[fresh_42]

I think you should still prove the following:

Given two positive integers $a,n.$

$A.$ $\operatorname{gcd}(a,n)=1$
$B.$ There are integers $s,t$ such that $1=s\cdot a+t\cdot n.$
$C.$ There is an integer $b\in \mathbb{Z}$ such that $a \cdot b \equiv 1 \pmod n.$

Prove that all three statements are equivalent.

Hint: It is sufficient to show $A \Longrightarrow B \Longrightarrow C \Longrightarrow A$ or $A \Longrightarrow C \Longrightarrow B \Longrightarrow A$

 

[Mark44]

Homework Statement:: Prove the following statement:
Any one of the integers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ can occur as the units digit of $a^{3}$.
Relevant Equations:: None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8$, or $9\pmod {10}$.
Thus $a^{3}\equiv 0, 1, 8, 27, 64, 125, 216, 343, 512$, or $729\pmod {10}$.

Except for 0, 1, and 8, the rest are not numbers modulo 10.

I would write the above as:
Let a be the units digit of some integer. Then a is 0, 1 2, 3, 4, 5, 6, 7, 8, or 9.
$a^3$ is one of 0, 1, 8, 27, 64, 125, 216, 343, 512, or 729, respectively.
Then the units digit of $a^3$ is 0, 1, 8, 7, 4, 5, 6, 3, 2, or 9, and the statement is proven.

Therefore, anyone of the integers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ can occur as the units digit of $a^{3}$.