# The number of seconds in six weeks

Fun fact: 10! is the number of seconds in 6 weeks. We can prove this as follows: There are 7 days in a week, 24 hours per day, 60 minutes per hour, and 60 seconds per minute. Thus, there are

$$6 \cdot 7 \cdot 24 \cdot 60 \cdot 60$$ seconds in 6 weeks. 10! is $$10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$$We can figure out whether these two numbers are equal in at least two ways. The first, is to find the prime factorization of each, then compare. If the prime factorizations are equal, then the numbers are themselves equal. The second way is much nicer. We can simply cancel common factors until we reach an equation which is obviously true, or obviously false. For example, if we end up with 1 = 1, then the original statement is true. But if we ended up with 3 = 5, or any other equation which is obviously false, then the original statement must be false as well.

I notice right away that \(6 \cdot 7\) appears in both expressions, and so I cancel those. Now the first expression reads

$$24 \cdot 60 \cdot 60$$and the second expression reads

$$10 \cdot 9 \cdot 8 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$$Next, I notice that \(24,\) in the first expression, is equal to \(8 \cdot 3,\) in the second expression, and so I cancel those. Now the first expression reads

$$60 \cdot 60$$and the second expression reads

$$10 \cdot 9 \cdot 5 \cdot 4 \cdot 2 \cdot 1$$Next, I notice that each \(60\) is divisible by \(3,\) and so I break the \(9\) up, in the second expression, to reveal its factors of 3. Now the second expression reads

$$10 \cdot 5 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 1$$Finally, I notice that \(10 \cdot 3 \cdot 2 = 60,\) but also, \(5 \cdot 4 \cdot 3 = 60,\) and so the two expressions must be equal. And thus, truly, 10! is the number of seconds in 6 weeks.

Here's where I got the idea for this article. I think I improved upon the approach by making it appear more mechanical, or, in other words, less creative.