$107 \times 107 + 93 \times 93 = x$

### Concept & Formula This expression is in the form of $a^2 + b^2$. Notice that the numbers $107$ and $93$ are symmetrically distributed around a base of $100$. This allows us to use a powerful combined algebraic identity. $$(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$$ ### Step-by-Step Solution * Identify the symmetric base and variance: The base $x = 100$, and the variance $y = 7$. $$107 = 100 + 7$$ $$93 = 100 - 7$$ * Map this to our identity: $$(100 + 7)^2 + (100 - 7)^2 = 2 \times ((100)^2 + (7)^2)$$ * Calculate the much simpler inner terms: $$(100)^2 = 10000$$ $$(7)^2 = 49$$ * Perform the final multiplication: $$2 \times (10000 + 49) = 2 \times 10049 = 20098$$ ### Exam Strategy & Shortcut Use the unit digits of the squares. $107^2$ ends in $9$ (since $7^2=49$). $93^2$ ends in $9$ (since $3^2=9$). The sum $9 + 9 = 18$, meaning the final unit digit must be $8$. This doesn't narrow it down enough, so check the last two digits. Using base rules, the last two digits of $(100+7)^2$ are $49$, and $(100-7)^2$ are $49$. Sum them: $49 + 49 = 98$. The answer must end in $98$. Only options (c) $20098$ and (d) $21908$ remain. Since $100^2 + 100^2 = 20000$, the answer should be extremely close to $20000$, pointing directly to $20098$. ### Common Pitfall Failing to recognize the symmetry around 100 and stubbornly calculating $107 \times 107$ and $93 \times 93$ separately, burning precious exam minutes on arithmetic. ### Final Answer **Therefore, the correct answer is 20098.**