Factorials in base 5

Before we actually head into the problem, let's create an outline of what steps we should take to solve this problem:

1. Determine what it means to have k trailing zeroes in base B for a number n. (Word problem ==> mathematical equation)

2. Determine k using the conclusions of (1) (Solving a mathematical equation)

3. Determine what it means to have d digits in a number n in base B. (Word problem ==> mathematical equation)

4. Determine d using the conclusions of (3) (Solving a mathematical equation)

5. Compute the answer using the results of (2) and (4) (Trivial matters; involves converting a word problem to a mathematical equation and solving the equation, but is of a much lower level than the previous steps.)

Now let us actually execute those steps.

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First, we must determine what it means to have k trailing zeroes in base B for a number n. Given that k, B and n are positive integers, the amount of trailing zeroes k is defined as: $k \colon B^{k} \mid n \wedge B^{k + 1} \nmid n$

Second, we must determine k. In terms of solving an equation, we must solve k for the above equation where $B = 5$ and $n = 53!$ . For this we simply use the rule that exponential terms distribute over multiplication. (More precisely: addition of reals of exponentiation of reals distributes over multiplication of reals. Now say that quick three times!) In Mathese, we say:

$B^{m + n} = B^{m} \cdot B^{n} \hspace{1.0cm} \text{<follows directly from definition of exponentiation,} \\ \hspace{4.5cm} \text{multiplication and addition of reals>}$

This rule means we can look for prime factors 5 of the factors of 53! and sum up the resulting terms. As anyone who has read an absolute beginner's guide to number theory will know (I haven't, but I do happen to know this), a number is divisible by 5 if its least significant digit is either 0 or 5. We can then divide by 5 and repeat this test to see how many prime factors of 5 there are in total. Obviously, we should stop as soon as the test returns negative. So we know we can simplify the problem: $k : B = 5, n = 53! = \\ k : B = 5, n = 50 \cdot 45 \cdot 40 \cdot ... \cdot 5$ We can do this computation easily ourselves: $k = 2 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 = 12$ Note that $5 \cdot 5 = 25$ , thus 25 and its multiples less than $5 \cdot 5 \cdot 5 = 125$ will contain two prime factors of 5.

Great! With both abstract and procedural thinking it is not hard (and conversely, lacking these skills will make it very hard) to determine the equation for the amount of trailing zeroes of the number n. Then, using some basic number theoretical properties and some definitions of basic arithmetic operations we were able to solve the equation.

Now, let's move on with step 3, in which we must come up with an equation that describes the amount of digits in n when written in base B.

The amount of digits d required to write n in base B is given by: $d = \lceil log _B (n) \rceil$

That's not only a general rule like earlier, it is an actual equation as I promised in the overview! Great, I get to keep my word half the time. Now we can simply compute $\lceil log _5 (53!) \rceil$ which gives us that $d = 100$ .

Great! Now, all we need to do is formulate our answer. The percentage of digits, call it x , of the number n that are trailing zeroes is given by: $x = \frac{k}{d} \cdot 100 \% = \frac{12}{100} \cdot 100 \% = \boxed{12 \%}$

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Note that I did not come up with a clean problem solving strategy (instead of simply a computation) for step 4, so if anyone has more prototypical solutions, be sure to upload them! Note, also, that I am a beginner problem solver, as in the phrase "started yesterday". Obviously I have some background in mathematics and problem solving, but not with enough rigor to be certain that my solutions are didactic instead of merely sufficient to solve the problem. Hence, I must warn you that you may learn more from looking at more adept problem solver's solutions when they will pose solutions to this problem.