Animated Solution for Mathematics - Binomial Theorem: The remainder, when $7^{103}$ is divided by 17, is ————————

The challenge is to find the remainder when $7^{103}$ is divided by $17$. Attempting to calculate this value directly is impossible, so we utilize the elegance of modular arithmetic to simplify the expression.

Our first step is to make the exponent manageable. Since $103$ is odd, we extract one factor of $7$ to make the remaining exponent even:

We focus on the base $7^2=49$. This transformation allows us to work with a number that has a convenient relationship with our divisor, $17$.

We observe that $17\cdot 3=51$. Since $49$ is two units less than $51$, we can express this in modular arithmetic as:

Substituting this into our expression, we get $7\cdot (-2{)}^{51}(\mathrm{mod}17)$. Because $51$ is an odd power, the negative sign is preserved, resulting in:

We now simplify $2^{51}$ by finding a power of $2$ close to a multiple of $17$. We know that $2^4=16$, which is congruent to $-1(\mathrm{mod}17)$. We rewrite $2^{51}$ as:

Applying the modular equivalence, the expression becomes:

Since $12$ is an even number, $(-1{)}^{12}=1$. The expression simplifies to $-7\cdot 8=-56$.

We have arrived at $-56(\mathrm{mod}17)$. To find the positive remainder, we identify the nearest multiple of $17$ greater than $56$, which is $17\cdot 4=68$.

Alternatively, we can add multiples of $17$ to $-56$ until we reach a positive value:

The remainder is $12$. By identifying proximity to multiples and utilizing negative remainders, we have tamed the giant exponent.