The smallest perfect square that is divisible by $7!$ is

### Concept & Formula A perfect square must have even powers for all of its prime factors. To find the smallest perfect square divisible by a number, break that number into its prime factorization and multiply by the exact missing prime factors needed to make all exponents even. $$ N = p_1^{e_1} \times p_2^{e_2} \dots $$ ### Step-by-Step Solution First, expand $7!$ (7 factorial): $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ Break it down into its fundamental prime factors: $7! = 7 \times (2 \times 3) \times 5 \times (2^2) \times 3 \times 2$ Combine the prime factors by adding their exponents: $2^{(1 + 2 + 1)} \times 3^{(1 + 1)} \times 5^1 \times 7^1$ $= 2^4 \times 3^2 \times 5^1 \times 7^1$ Check the exponents for parity (even/odd). The powers of $2$ ($4$) and $3$ ($2$) are already even. The powers of $5$ ($1$) and $7$ ($1$) are odd. To make the entire expression a perfect square, we must multiply by $5^1$ and $7^1$ so their powers become $2$ (an even number). Multiplier needed $= 5 \times 7 = 35$. Calculate the final perfect square: We know $7! = 5040$. Smallest Perfect Square $= 5040 \times 35$. $5040 \times 35 = 176400$. ### Exam Strategy & Shortcut $7!$ contains the prime numbers $2, 3, 5, 7$. Count the highest powers conceptually: $2$ appears in $2, 4, 6$ (total power 4, even). $3$ appears in $3, 6$ (total power 2, even). $5$ appears once (odd). $7$ appears once (odd). You immediately know you must multiply $7!$ by $35$. Multiplying $5040 \times 35$ quickly gives $176400$. ### Common Pitfall Forgetting to break composite numbers like $4$ and $6$ down into their prime components when finding the factorization. This leads to incorrect power counts and a wrong final multiplier. ### Final Answer Therefore, the correct answer is **176400**.