Animated Solution for Mathematics - Binomial Theorem: The remainder, when $7^{103}$ is divided by 23, is equal to:

The problem asks us to evaluate $7^{103}(\mathrm{mod}23)$. Calculating this directly is impossible, so we utilize Modular Arithmetic to find the remainder.

Our objective is to simplify the base $7$ by finding a power that is congruent to a small integer modulo $23$.

We examine the powers of $7$: $7^1=7$ $7^2=49\equiv 3(\mathrm{mod}23)$ $7^3=343$

We observe that $23\times 15=345$. Since $343=345-2$, we establish the congruence:

We rewrite the original expression $7^{103}$ to utilize this identity:

Since the exponent $34$ is even, the negative sign in $(-2{)}^{34}$ becomes positive. The expression simplifies to:

To reduce $2^{34}$, we use the property $2^5=32$. Since $32\equiv 9(\mathrm{mod}23)$, we can write:

Substituting this back into our main expression, we obtain:

First, we reduce $112$ modulo $23$. Since $23\times 4=92$ and $23\times 5=115$, we have $112\equiv -3(\mathrm{mod}23)$. Now we evaluate $9^6(\mathrm{mod}23)$ using the property $(9^3{)}^2$:

Dividing $729$ by $23$: $23\times 31=713$. Thus, $729\equiv 729-713=16(\mathrm{mod}23)$.

Our expression now becomes:

Reducing $256$ modulo $23$: $256=23\times 11+3$, so $256\equiv 3(\mathrm{mod}23)$. The final calculation is:

To express this as a positive remainder, we add $23$:

The final remainder is 14.