2018 AMC 10A Problem 3

We need to turn $10!$ seconds into days. Multiplying out $10!$ and dividing is slow, so instead pull the pieces of one day out of $10!$.

1. Seconds in one day. $60$ seconds in a minute, $60$ minutes in an hour, $24$ hours in a day. So one day is $60 \cdot 60 \cdot 24$ seconds.
2. Find those factors inside $10!$. Write out
$$10! = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.$$
- $10 \cdot 6 = 60$.
- $5 \cdot 4 \cdot 3 = 60$.
- $8 \cdot 9 = 72 = 24 \cdot 3$. (We cannot use $8 \cdot 3$ for the $24$, because the $3$ is already taken.)
3. See what is left. After taking out $10, 6, 5, 4, 3, 8, 9$, the numbers left are $7, 2, 1$, and the step above left one extra factor $3$. So
$$10! = (60)(60)(24) \cdot 3 \cdot 7 \cdot 2 = (60 \cdot 60 \cdot 24) \cdot 42.$$
The blood lasts exactly $42$ days.
4. Count on the calendar. Start at noon on January 1. January has $31$ days, so $31$ days later is noon on February 1. That leaves $42 - 31 = 11$ more days, reaching noon on February 12.

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

A factorial is a product of small numbers, so it is easy to split into the factors you want ($60$, $60$, $24$) without ever computing the huge number $3{,}628{,}800$. Because the division comes out to exactly $42$ with nothing left over, the blood expires at noon, so there is no question about a partial day. When counting days across months, always stop at the first day of the next month and then count what is left.

Check the answer directly: $60 \cdot 60 \cdot 24 = 86{,}400$ seconds in a day, and $86{,}400 \cdot 42 = 3{,}628{,}800$, which is exactly $10!$. So $42$ days is right, and the calendar count is the same as above.