exam: HMMT

problem: (a) Decompose 1 into unit fractions.

solution:
$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
(b) Decompose $\frac{1}{4}$ into unit fractions.

solution_match: Answer:


solution:
$\frac{1}{8}+\frac{1}{12}+\frac{1}{24}$
(c) Decompose $\frac{2}{5}$ into unit fractions.


solution: $\frac{1}{5}+\frac{1}{10}+\frac{1}{15}+\frac{1}{30}$


problem: Explain how any unit fraction $\frac{1}{n}$ can be decomposed into other unit fractions.

solution: $\frac{1}{2 n}+\frac{1}{3 n}+\frac{1}{6 n}$


problem: (a) Write 1 as a sum of 4 distinct unit fractions.

solution:
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}$
(b) Write 1 as a sum of 5 distinct unit fractions.


solution:
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{43 \cdot 42}$
(c) Show that, for any integer $k>3,1$ can be decomposed into $k$ unit fractions.


solution: If we can do it for $k$ fractions, simply replace the last one (say $\frac{1}{n}$ ) with $\frac{1}{n+1}+\frac{1}{n(n+1)}$. Then we can do it for $k+1$ fractions. So, since we can do it for $k=3$, we can do it for any $k>3$.

solution_match: Solution:


problem:
Say that $\frac{a}{b}$ is a positive rational number in simplest form, with $a \neq 1$. Further, say that $n$ is an integer such that:

$$
\frac{1}{n}>\frac{a}{b}>\frac{1}{n+1}
$$

Show that when $\frac{a}{b}-\frac{1}{n+1}$ is written in simplest form, its numerator is smaller than $a$.

solution: $\quad \frac{a}{b}-\frac{1}{n+1}=\frac{a(n+1)-b}{b(n+1)}$. Therefore, when we write it in simplest form, its numerator will be at most $a(n+1)-b$. We claim that $a(n+1)-b<a$. Indeed, this is the same as $a n-b<0 \Longleftrightarrow a n<b \Longleftrightarrow \frac{b}{a}>n$, which is given.


problem:
An aside: the sum of all the unit fractions

It is possible to show that, given any real M , there exists a positive integer $k$ large enough that:

$$
\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M
$$

Note that this statement means that the infinite harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, grows without bound, or diverges. For the specific example $\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.

solution:
Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2 n}+\ldots+\frac{1}{2 n}=\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get

$$
\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4
$$

so, adding in $\frac{1}{1}$, we get

$$
\sum_{n=1}^{256} \frac{1}{n}>5
$$

so $k=256$ will suffice.


problem:
Now, using information from problems 4 and 5 , prove that the following method to decompose any positive rational number will always terminate:
Step 1. Start with the fraction $\frac{a}{b}$. Let $t_{1}$ be the largest unit fraction $\frac{1}{n}$ which is less than or equal to $\frac{a}{b}$.
Step 2. If we have already chosen $t_{1}$ through $t_{k}$, and if $t_{1}+t_{2}+\ldots+t_{k}$ is still less than $\frac{a}{b}$, then let $t_{k+1}$ be the largest unit fraction less than both $t_{k}$ and $\frac{a}{b}$.
Step 3. If $t_{1}+\ldots+t_{k+1}$ equals $\frac{a}{b}$, the decomposition is found. Otherwise, repeat step 2 .
Why does this method never result in an infinite sequence of $t_{i}$ ?

solution:
Let $\frac{a_{k}}{b_{k}}=\frac{a}{b}-t_{1}-\ldots-t_{k}$, where $\frac{a_{k}}{b_{k}}$ is a fraction in simplest terms. Initially, this algorithm will have $t_{1}=1, t_{2}=\frac{1}{2}, t_{3}=\frac{1}{3}$, etc. until $\frac{a_{k}}{b_{k}}<\frac{1}{k+1}$. This will eventually happen by problem 5 , since there exists a $k$ such that $\frac{1}{1}+\ldots+\frac{1}{k+1}>\frac{a_{k}}{b_{k}}$. At that point, there is some $n$ with $\frac{1}{n}<t_{k}$ such that $\frac{1}{n}>\frac{a_{k}}{b_{k}}>\frac{1}{n+1}$. In this case, $t_{k+1}=\frac{1}{n+1}$.
Suppose that there exists $n_{k}$ such that $\frac{1}{n_{k}}>\frac{a_{k}}{b_{k}}>\frac{1}{n_{k}+1}$ for some $k$. Then we have $t_{k+1}=\frac{1}{n_{k}+1}$ and $\frac{a_{k+1}}{b_{k+1}}<\frac{1}{n_{k}\left(n_{k}+1\right)}$. This shows that once we have found $n_{k}$ such that $\frac{1}{n_{k}}>\frac{a_{k}}{b_{k}}>\frac{1}{n_{k}+1}$ and $\frac{1}{n_{k}} \leq t_{k}$, we no longer have to worry about $t_{k+1}$ being less than $t_{k}$, since $t_{k+1}=\frac{1}{n_{k}+1}<\frac{1}{n_{k}}<$ $t_{k}$, and also $n_{k+1} \geq n_{k}\left(n_{k}+1\right)$ while $\frac{1}{n_{k}\left(n_{k}+1\right)} \leq \frac{1}{n_{k}+1}=t_{k+1}$.
On the other hand, once we have found such an $n_{k}$, the sequence $\left\{a_{k}\right\}$ must be decreasing by problem 4. Since the $a_{k}$ are all integers, we eventually have to get to 0 (as there is no infinite decreasing sequence of positive integers). Therefore, after some finite number of steps the algorithm terminates with $a_{k+1}=0$, so $0=\frac{a_{k}}{b_{k}}=\frac{a}{b}-t_{1}-\ldots-t_{k}$, so $\frac{a}{b}=t_{1}+\ldots+t_{k}$, which is what we wanted.

## Juicy Numbers [100]

A juicy number is an integer $j>1$ for which there is a sequence $a_{1}<a_{2}<\ldots<a_{k}$ of positive integers such that $a_{k}=j$ and such that the sum of the reciprocals of all the $a_{i}$ is 1 . For example, 6 is a juicy number because $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, but 2 is not juicy.

In this part, you will investigate some of the properties of juicy numbers. Remember that if you do not solve a question, you can still use its result on later questions.


problem: Explain why 4 is not a juicy number.

solution: If 4 were juicy, then we would have $1=\ldots+\frac{1}{4}$. The $\ldots$ can only possible contain $\frac{1}{2}$ and $\frac{1}{3}$, but it is clear that $\frac{1}{2}+\frac{1}{4}, \frac{1}{3}+\frac{1}{4}$, and $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$ are all not equal to 1 .


problem: It turns out that 6 is the smallest juicy integer. Find the next two smallest juicy numbers, and show a decomposition of 1 into unit fractions for each of these numbers. You do not need to prove that no smaller numbers are juicy.

solution: 12 and $151=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}, 1=\frac{1}{2}+\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$.


problem:
Let $p$ be a prime. Given a sequence of positive integers $b_{1}$ through $b_{n}$, exactly one of which is divisible by $p$, show that when

$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}
$$

is written as a fraction in lowest terms, then its denominator is divisible by $p$. Use this fact to explain why no prime $p$ is ever juicy.

solution:
We can assume that $b_{n}$ is the term divisible by $p$ (i.e. $b_{n}=k p$ ) since the order of addition doesn't matter. We can then write

$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n-1}}=\frac{a}{b}
$$

where $b$ is not divisible by $p$ (since none of the $b_{i}$ are). But then $\frac{a}{b}+\frac{1}{k p}=\frac{k p a+b}{k p b}$. Since $b$ is not divisible by $p, k p a+b$ is not divisible by $p$, so we cannot remove the factor of $p$ from the denominator. In particular, $p$ cannot be juicy as 1 can be written as $\frac{1}{1}$, which has a denominator not divisible by $p$, whereas being juicy means we have a sum $\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}=1$, where $b_{1}<b_{2}<\ldots<b_{n}=p$, and so in particular none of the $b_{i}$ with $i<n$ are divisible by p.


problem: Show that if $j$ is a juicy integer, then $2 j$ is juicy as well.

solution: Just replace $\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}$ with $\frac{1}{2}+\frac{1}{2 b_{1}}+\frac{1}{2 b_{2}}+\ldots+\frac{1}{2 b_{n}}$. Since $n>1,2 b_{1}>2$.


problem: Prove that the product of two juicy numbers (not necessarily distinct) is always a juicy number. Hint: if $j_{1}$ and $j_{2}$ are the two numbers, how can you change the decompositions of 1 ending in $\frac{1}{j_{1}}$ or $\frac{1}{j_{2}}$ to make them end in $\frac{1}{j_{1} j_{2}}$ ?

solution:
Let $1=\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}=\frac{1}{c_{1}}+\ldots+\frac{1}{c_{m}}$, where $b_{n}=j_{1}$ and $c_{m}=j_{2}$. Then

$$
1=\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n-1}}+\left(\frac{1}{b_{n} c_{1}}+\frac{1}{b_{n} c_{2}}+\ldots+\frac{1}{b_{n} c_{m}}\right)
$$

and so $j_{1} j_{2}$ is juicy.