A number has 2014 digits, all of which are the digit 3. If the number is divided by 101, what is the remainder?

Question

A number has 2014 digits, all of which are the digit 3. If the number is divided by 101, what is the remainder?

Answer 1

Note that $3333$ is divisible by $101$. So the number consisting of $2012$ $3$'s followed by $00$ is divible by $101$. So ...

Answer 2

Note that $$ 3333=101\times 33$$On the other hand $$2014 = 4\times 503 +2$$

Thus the remainder is $33$ which is the last two digits of the number.

Answer 3

The number is $3\cdot \frac {10^{2014}-1}{10-1}=\frac {10^{2014}-1}{3}=\frac {100^{1007}-1}{3}.$

Modulo $101$ we have $100^{1007}-1\equiv (-1)^{1007}-1\equiv -2\equiv 99.$

Since $\gcd(3,101)=1,$ if $\frac {A}{3}$ and $\frac {B}{3}$ are integers and $A\equiv B\mod {101}$ then $\frac {A}{3}\equiv \frac {B}{3}\mod {101}.$(See footnote).

Therefore $\frac {10^{2014}-1}{3}\equiv \frac {99}{3}\equiv 33 \mod {101}.$

Footnote. Let $\frac {A}{3}=x$ and $\frac {B}{3}=y.$ Then $3x\equiv A\equiv B\equiv y \mod {101},$ so $3x\equiv 3y \mod {101},$ so $101$ divides $3(x-y),$ so $101$ divides $(x-y),$ so $x\equiv y \mod {101}.$