Show that ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ,9##

[Math100]

Homework Statement

For any integer $a$, show that $a^{2}-a+7$ ends in one of the digits $3, 7$, or $9$.

Relevant Equations

None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8$ or $9\pmod {10}$.
Note that $a^{2}\equiv 0, 1, 4, 9, 6, 5, 6, 9, 4$ or $1\pmod {10}$.
Thus $a^{2}-a+7\equiv 7, 7, 9, 9, 7, 7, 7, 9, 9$ or $7\pmod {10}$.
Therefore, $a^{2}-a+7$ ends in one of the digits $3, 7$, or $9$ for any integer $a$.

 

[fresh_42]

Homework Statement:: For any integer $a$, show that $a^{2}-a+7$ ends in one of the digits $3, 7$, or $9$.
Relevant Equations:: None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8$ or $9\pmod {10}$.
Note that $a^{2}\equiv 0, 1, 4, 9, 6, 5, 6, 9, 4$ or $1\pmod {10}$.
Thus $a^{2}-a+7\equiv 7, 7, 9, 9, 7, 7, 7, 9, 9$ or $7\pmod {10}$.

I get $7,7,9,3,9,7,7,9,3,9$ from $a^2-a+7=a(a-1)+7$ which I think is easier.

Therefore, $a^{2}-a+7$ ends in one of the digits $3, 7$, or $9$ for any integer $a$.

 

[Math100]

I get $7,7,9,3,9,7,7,9,3,9$ from $a^2-a+7=a(a-1)+7$ which I think is easier.

Indeed. I agree that, factoring can be easier.

 

[Math100]

But which is a more exact/accurate way to express this? Is it $a^{2}-a+7=a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3$ or $9\pmod {10}$, or $a^{2}-a+7\equiv a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3$, or $9\pmod {10}$?

 

[fresh_42]

But which is a more exact/accurate way to express this? Is it $a^{2}-a+7=a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3$ or $9\pmod {10}$, or $a^{2}-a+7\equiv a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3$, or $9\pmod {10}$?

I would go even further.

We only need to show that $a^2-a$ always ends on $\{0,2,6\}.$ Now $a(a-1)\equiv 0\pmod {10}$ if one factor is from $\{0,5\}$. That leaves us with $2\cdot 1=2\, , \,3\cdot 2=6\, , \,4\cdot 3=12\, , \,7\cdot 6=42\, , \,8\cdot 7=56$ and $9\cdot 8=72,$ all ending on $2$ or $6$.

This is a regular pattern: simplify if possible, and explain what you actually do.

 

[Math100]

I would go even further.

We only need to show that $a^2-a$ always ends on $\{0,2,6\}.$ Now $a(a-1)\equiv 0\pmod {10}$ if one factor is from $\{0,5\}$. That leaves us with $2\cdot 1=2\, , \,3\cdot 2=6\, , \,4\cdot 3=12\, , \,7\cdot 6=42\, , \,8\cdot 7=56$ and $9\cdot 8=72,$ all ending on $2$ or $6$.

This is a regular pattern: simplify if possible, and explain what you actually do.

And what about the sign? Equal sign is okay above or congruence sign?

 

[fresh_42]

And what about the sign? Equal sign is okay above or congruence sign?

I think equal is ok since it is modulo $10$ and everybody can see the last digit.

I made a mistake as I first wrote $a(a-1)=0$ for one factor in $\{0,5\}.$
This was wrong because what I meant was $a(a-1)\equiv 0\pmod{10}$ for one factor in $\{0,5\}.$
Here it is important since I only wrote the last digit, and not $\{0,20,30\}.$