2018 AMC 10A Problem 3

One day is $60 \cdot 60 \cdot 24 = 86400$ seconds. Rather than multiplying out $10!$, cancel by factoring:
$$86400 = 2^7 \cdot 3^3 \cdot 5^2, \qquad 10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.$$
Dividing, $\dfrac{10!}{86400} = 2 \cdot 3 \cdot 7 = 42$, so the blood lasts exactly $42$ days.

Counting from noon on January 1: $31$ days later is noon on February 1, and $11$ more days brings us to noon on February 12.

The answer is $\boxed{\textbf{(E)}\ \text{February 12}}$.

Converting a huge number of seconds to days is a division problem, and factorials divide cleanly when written in primes, so factor both numbers instead of computing $3{,}628{,}800$. The exact answer of $42$ days means the expiry time is also noon, so no "partial day" ambiguity arises. Calendar counting then works month by month.

Group the factorial into the day's factors directly: $10! = (10 \cdot 6)(9 \cdot 4 \cdot 5 \cdot 3)(8 \cdot ... )$ is awkward, but $10! = 60 \cdot 60 \cdot 24 \cdot 42$ can be checked by $60 \cdot 60 \cdot 24 = 86400$ and $86400 \cdot 42 = 3{,}628{,}800 = 10!$.

• Multiplying out 10! and dividing by 86400 by hand, or forgetting January has 31 days and landing on February 11.
• Counting January 1 itself as day one (an off-by-one), which shifts the answer to February 11, choice (D).