Question

Show that the number of the form 7ⁿ , n∈N cannot have unit digit as zero.

Answer 1

We know the numbers leave their unit digit as the remainder on division by 10.

Since $\sf{7\equiv7\pmod{10},}$ 7 is one possible number which can be the unit digit of some powers of 7.

We see that,

$\longrightarrow\sf{7^2=49\equiv9\pmod{10}}$

$\longrightarrow\sf{7^3=7^2\times7\equiv9\times7=63\equiv3\pmod{10}}$

$\longrightarrow\sf{7^4=7^3\times7\equiv3\times7=21\equiv1\pmod{10}}$

Now we have the order of remainders with the powers: 7, 9, 3 and 1.

Since $\sf{7^4}$ is congruent to 1 modulo 10, the sequence of remainders is repeated for next powers of 7.

Thus, for $\sf{n\in\mathbb{W},}$

$\longrightarrow\sf{7^{4n+1}=(7^4)^n\times7\equiv1^n\times7=1\times7=7\pmod{10}}$

$\longrightarrow\sf{7^{4n+2}=(7^4)^n\times7^2\equiv1^n\times9=1\times9=9\pmod{10}}$

$\longrightarrow\sf{7^{4n+2}=(7^4)^n\times7^3\equiv1^n\times3=1\times3=3\pmod{10}}$

$\longrightarrow\sf{7^{4n}=(7^4)^n\equiv1^n=1\pmod{10}}$

These imply the unit digit of any number in the form $\sf{7^n}$ for $\sf{n\in\mathbb{N}}$ cannot be zero.