Science:Math Exam Resources/Courses/MATH312/December 2012/Question 03

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MATH312 December 2012

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Question 03
Find all positive integers n such that $\displaystyle n!$ ends with exactly 74 zeros in decimal notation.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To count the number of zeroes in this number, it suffices to count the number of factors of 10 the number has.

Hint 2
To accomplish hint 1, notice that the number of factors of 2 inside $\displaystyle n!$ must exceed the number of factors of 5 so it suffices to count the number of factors of 5 inside $\displaystyle n!$.

Hint 3
The number of 5s occurring in the expansion of $\displaystyle n!$ is given by $\displaystyle \left\lfloor {\frac {n}{5}}\right\rfloor +\left\lfloor {\frac {n}{5^{2}}}\right\rfloor +...+\left\lfloor {\frac {n}{5^{\lfloor \log _{5}(n)\rfloor }}}\right\rfloor$ This is true since you get a factor of 5 every 5 integers. You get another factor of 5 every 25 integers and so on.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

As displayed in the hints, it suffices to compute the values of n for which $\displaystyle \left\lfloor {\frac {n}{5}}\right\rfloor +\left\lfloor {\frac {n}{5^{2}}}\right\rfloor +...+\left\lfloor {\frac {n}{5^{\lfloor \log _{5}(n)\rfloor }}}\right\rfloor =74$. Clearly for $\displaystyle n=400$, we have that the first term in the above sum is 80 so it suffices to look smaller than this value. This reduces the above sum to $\displaystyle \left\lfloor {\frac {n}{5}}\right\rfloor +\left\lfloor {\frac {n}{5^{2}}}\right\rfloor +\left\lfloor {\frac {n}{5^{3}}}\right\rfloor =74$ We start with the value $\displaystyle n=200$. $\displaystyle \left\lfloor {\frac {200}{5}}\right\rfloor +\left\lfloor {\frac {200}{5^{2}}}\right\rfloor +\left\lfloor {\frac {200}{5^{3}}}\right\rfloor =40+8+1=49$. Trying $\displaystyle n=300$ yields $\displaystyle \left\lfloor {\frac {300}{5}}\right\rfloor +\left\lfloor {\frac {300}{5^{2}}}\right\rfloor +\left\lfloor {\frac {300}{5^{3}}}\right\rfloor =60+12+2=74$. That is an exact match and further we know that $\displaystyle n=299$ makes the above sum less than 74. Notice that increasing the above sum from 300 to 301,302,303 or 304 does not increment any of the terms but at 305, the first summand increases by 1. Thus the five possible values are $\displaystyle n\in \{300,301,302,303,304\}$ completing the question.

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