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I have been working through some past math competition problems and have had difficulty in solving some of the ones on number theory. Examples include:

1) If we need 27 cents can we make it using 5 cents and 8 cents?

2) Given $1. How many ways can you make it from 20 cents, 10 cents and 5 cent coins?

3) How may ways are there of making 4 (using addition) with numbers (non-negative) less than 4?

4) What is the highest non-attainable number when adding numbers from 6,9 and 20?

I was able to solve these problems with the use of trial and error however, this seemed highly inefficient due to the great number of such questions in the competitions. So, I was wondering whether there is a formula that would allow me to quickly solve them.

Help would be appreciated, Thank you :)

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Note: Here are two different answers of number 2). Both of them could be interesting when looking for a systematic approach. The first one provides a more general approach, the second one is presumably more efficient when fast calculation is needed.

In both cases it is helpful to observe that finding solutions in non-negative integers of \begin{align*} 5p+10q+20r=100 \end{align*} is equivalent to finding solutions of non-negative integers of \begin{align*} p+2q+4r=20 \end{align*}

So, in the following we are looking for the number of partitions of $20$ with elements from $\{1,2,4\}$.

First approach: Generating Functions

We develop a formula for the number of partitions of $n$ with elements from $\{1,2,4\}$. Question $2$ is answered by applying the formula to $n=20$. Zero or more $1$'s can be algebraically described as

\begin{align*} z^0+z^1+z^2+z^3+\cdots\tag{1} \end{align*} with the power of $z$ in (1) indicating the number of $1$'s. Zero or more $2$'s can be algebraically described as \begin{align*} z^0+z^2+z^4+z^6+\cdots\tag{2} \end{align*} with the power of $z$ in (2) indicating the number of used $2$'s in the partition.

The number of partitions of $n$ with elements from $\{1,2,4\}$ can thereby described as the coefficients of $z^n$ in the generating function $F(z)$ with \begin{align*} F(z)&=\left(1+z+z^2+z^3+\cdots\right)\left(1+z^2+z^4+z^6+\cdots\right)\left(1+z^4+z^8+z^{12}+\cdots\right)\\ &=\frac{1}{1-z}\cdot\frac{1}{1-z^2}\cdot\frac{1}{1-z^4}\\ &=\left(\frac{1}{1-z}\right)\left(\frac{1}{1-z}\cdot\frac{1}{1+z}\right)\left(\frac{1}{1-z}\cdot\frac{1}{1+z}\cdot\frac{1}{1-iz}\cdot\frac{1}{1+iz}\right)\\ &=\frac{1}{(1-z)^3}\cdot\frac{1}{(1+z)^2}\cdot\frac{1}{1-iz}\cdot\frac{1}{1+iz} \end{align*}

The coefficient of $z^n$ of the generating function $F(z)$ contains the information of the number of wanted partitions. The following calculation is more or less straightforward. We use partial fraction decomposition of $F(z)$ and develop each term as power series and finally extract the coefficient of $z^n$.

Using partial fraction decomposition we obtain \begin{align*} F(z)&=\frac{9}{32}\frac{1}{1-z}+\frac{1}{4}\frac{1}{(1-z)^2}+\frac{1}{8}\frac{1}{(1-z)^3}+\frac{5}{32}\frac{1}{1+z}\\ &\qquad+\frac{1}{16}\frac{1}{(1+z)^2}+\frac{1-i}{16}\frac{1}{1-iz}+\frac{1+i}{16}\frac{1}{1+iz}\\ &=\frac{9}{32}\sum_{k=0}^{\infty}z^k+\frac{1}{4}\sum_{k=0}^{\infty}\binom{-2}{k}(-z)^k +\frac{1}{8}\sum_{k=0}^{\infty}\binom{-3}{k}(-z)^k+\frac{5}{32}\sum_{k=0}^{\infty}(-z)^k\\ &\qquad+\frac{1}{16}\sum_{k=0}^{\infty}\binom{-2}{k}z^k+\frac{1-i}{16}\sum_{k=0}^{\infty}(iz)^k +\frac{1+i}{16}\sum_{k=0}^{\infty}(-iz)^k\tag{3}\\ \end{align*}

Since $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$ we observe \begin{align*} \binom{-2}{k}&=\binom{k+1}{k}(-1)^k=(-1)^k(k+1)\\ \binom{-3}{k}&=\binom{k+2}{k}(-1)^k=(-1)^k\frac{(k+2)(k+1)}{2} \end{align*}

In the following we use the coefficient of operator $[z^n]$ to denote the coefficient $a_n$ in $A(z)=\sum_{k=0}^{\infty}a_kz^k$. $$[z^n]A(z)=[z^n]\sum_{k=0}^{\infty}a_kz^k=a_n$$

Now we are ready to extract the coefficient of $[z^n]$ from the generating function $F(z)$ in (3) \begin{align*} [z^n]F(z)&=\frac{9}{32}+\frac{1}{4}\binom{-2}{n}(-1)^n+\frac{1}{8}\binom{-3}{n}(-1)^n+\frac{5}{32}(-1)^n\\ &\qquad+\frac{1}{16}\binom{-2}{n}+\frac{1-i}{16}i^n+\frac{1+i}{16}(-i)^n\\ &=\frac{9}{32}+\frac{1}{4}(n+1)+\frac{1}{16}(n+2)(n+1)+\frac{5}{32}(-1)^n\\ &\qquad+\frac{1}{16}(-1)^n(n+1)+\frac{1-i}{16}i^n+\frac{1+i}{16}(-i)^n\\ &=\ldots\\ &= \begin{cases} \frac{1}{16}\left(n^2+6n+7-2(-1)^{\frac{n+1}{2}}\right)\qquad \qquad& n \text{ odd}\\ \frac{1}{16}\left(n^2+8n+14+2(-1)^{\frac{n}{2}}\right)\qquad \qquad& n \text{ even} \end{cases} \end{align*}

Finally we obtain for $n=20$ $$[z^{20}]F(z)=\frac{1}{16}\left(20^2+8\cdot 20+14+2(-1)^{\frac{20}{2}}\right)=\frac{576}{16}=36$$

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Second approach: Iteratively from $\{1,2\}$ to $\{1,2,4\}$

This is an elementary approach by first obtaining a formula $a_n$ which contains the number of partitions with elements from $\{1,2\}$. Once we have found $a_n$ we use it as basis to find a formula for the number of partitions with elements of the extended set $\{1,2,4\}$.

In order to find $a_n$ we use the Ansatz: $$n=2p+r\qquad\qquad r=0,1$$ We see that $n-r=2p$ is even and the number of possible summands $2$ of a partition of $n=2p+r$ is $0,1,2,\ldots,p$. Therefore we obtain \begin{align*} a_n=p+1=\frac{n-r}{2}+1=\left\lfloor\frac{n}{2}\right\rfloor+1 \end{align*} We observe \begin{align*} r= \begin{cases} 1&\qquad n\text{ odd}\\ 0&\qquad n\text{ even} \end{cases} \end{align*} So, $r=\frac{1-(-1)^n}{2}$ and we obtain this way the number $a_n$ of partitions of $n$ containing elements from $\{1,2\}$ \begin{align*} a_n&=\frac{n-r}{2}+1=\frac{2n+3+(-1)^n}{4}\tag{4} \end{align*} In order to determine the number $c_n$ equal the number of partitions of the extended set $\{1,2,4\}$ we use a similar approach as before. We start with $$n=4p+r\qquad\qquad r=0,1,2,3$$ We get \begin{align*} c_n=a_n+a_{n-4}+a_{n-8}+\ldots+a_{n-4p}\qquad\qquad n\geq 1 \end{align*} Setting $n=20$ we obtain using formula $a_n$ of (4) \begin{align*} c_{20}&=a_{20}+a_{16}+a_{12}+a_8+a_4+a_0\\ &=\frac{44}{4}+\frac{36}{4}+\frac{28}{4}+\frac{20}{4}+\frac{12}{4}+\frac{4}{4}\\ &=11+9+7+5+3+1\\ &=36 \end{align*}

Note: We calculated $c_n$ without deriving a closed formula for it. But, we could also derive one from $$c_n=\sum_{j=0}^{p}a_{n-4j}$$ and the formula for $a_n$ of (4) by considering the different values of $r=0,1,2,3$.

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Begin with an intuitive approach: 27 Plus 8 equals 35 35 is a multiple of 5. 27 times 8 ends in five, which is also a multiple of 5. The purpose of turning over numbers is to help the mind to see alternatives. Then you can understand what lies beneath equations.

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For the first question, see the artilce on the coin problem. You will need the notion of a gcd and other tools from elementary number theory.
The third question is answered by the partition function. We have $p(4)=5$. If you want the summands less than $4$, then we have \begin{align*} 4 & = 1+1+1+1,\\ & =1+1+2,\\ & =1+3,\\ & = 2+2. \end{align*} There is a general formula for $p(n)$ and an asymptotic formula, namely $$ p(n) \sim \frac {1} {4n\sqrt3} \exp\left({\pi \sqrt {\frac{2n}{3}}}\right) \mbox { as } n\rightarrow \infty. $$

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Some of the following answers are non-formulaic. Luckily, they are one-liners.

  1. If we need $27¢$, can we make it using $5¢$ and $8¢$ coins?

The $8¢$ pieces must number from $0$ to $3$, so the second digit must be $\{0,8,6,4,5,3,1,9\}$ which is not $7$, so no.

  1. How many ways can you make $\$1$ from $20¢$, $10¢$ and $5¢$ coins?

Because of the regular intervals of the coins, this is a path problem where O = $20¢$, X = $10¢$ and V = $5¢$:

$0                     $1.00
| O   O   O   O   O    |
| X X X X X X X X X X  |
| VVVVVVVVVVVVVVVVVVVV |

Order does not matter, so moves are either right or down. Because one cannot always exit the top row, assume that there are only $n = 5$ junctions and multiply by 2 for the extra options on the bottom row. There is 1 path terminating on the top row, $\binom{n}{1} = n$ paths terminating on the middle row, and $2 \cdot \binom{n+1}{2} = n(n+1)$ terminating on the bottom row. $f(n) = 1 + 2 n + n^2 = (n+1)^2$, so $f(5) = 36$

From the bottom row up, there are $n = 10$ junctions, so rather than an undercount which may be multiplied into compliance, the overcount, $\binom{n/2}{n/2-2} = \dfrac{n}{2} \left(\dfrac{n}{2}-1\right)$, must be subtracted yielding $f(n) =1 + n + \left(\dfrac{n}{2}\right)^2$ for even $n$.

  1. How may ways are there of making 4 (using addition) with numbers (non-negative) less than 4?

As both positive rationals and irrationals have infinite ways to add to 4, I'm assuming natural numbers: Simplest approach: $1+1+1+1 = 2+1+1 = 2+2 = 3+1 = 4$; 4 ways.

  1. What is the highest non-attainable number when adding numbers from 6,9 and 20?

Given some natural number $n$, $2 \cdot 6 = 12$ and $9+6 = 15$, so every $6 + 3n$ is attainable. Similarly, $26 + 3n$ and $46 + 3n$ are attainable. The highest unattainable number is $43$.

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The fundamental question at work here is "what is the (typed) value of unity". Each of the solution results in $Z$ units of some $\Omega$ "coin" which is valued at $\pi$, the same as the original coins. I use $\alpha_n$ to denote the value of the original coins and $x_n$ to denote the number of coins.

$$ \begin{eqnarray} \Omega &=& \left(\sum_{n=1}^{k-1} \frac{\alpha_nx_n}{Z}\right) + \frac{\alpha_k\left(Z-\sum_{n=1}^{k-1}x_n\right)}{Z} && \\ &=& \alpha_k - \frac{\sum_{n=1}^{k-1} (\alpha_k-\alpha_n)x_n}{Z}\\ \frac{1}{Z} &=& \frac{\alpha_k - \Omega}{\sum_{n=1}^{k-1} (\alpha_k-\alpha_n)x_n}\\ Z &=& \sum_{n=1}^{k-1} x_n + \frac{\pi-\sum_{n=1}^{k-1} \alpha_nx_n}{\alpha_k} \\ \Omega Z &= \pi \end{eqnarray} $$

A good starting point is to note that $Z * \frac{1}{Z} = 1 $

Exercise 1

$$ \begin{eqnarray} \Omega &=& 5a/Z + 8(Z-a)/Z && \\ &=& 8 - \frac{3a}{Z}\\ \frac{1}{Z} &=& \frac{8 - \Omega}{3a}\\ Z &=& a + (27-5a)/8 \\ \Omega Z &=& 27\\ 1 <= &\Omega& <= 27 &&\\ 0<=&a&<= 5 &&\\ 0<=&Z-a&<= 3 &&\\ 1<=&Z&<= 8 && \end{eqnarray} $$

Here the key is to use modulo arithmetic to solve for $Z = a + b$. I start with $Z * \frac{1}{Z} = 1 $ and simplify to get:

$$ \begin{eqnarray} \frac{8 - \Omega}{3a}\frac{8a + 27-5a}{8} &=& 1 \\ (8 - \Omega)(3a + 27) &=& 24a \\ (24 - 3\Omega)a-24a &=& 27(8 -\Omega) \\ (27-3a)\Omega &=& 216 \end{eqnarray} $$

Exercise 2

The trick here is to partition the problem:

$$ \Omega_k = \frac{Z_1\Omega_1+\alpha_k\left(Z_k-Z_1\right)}{Z_k} $$

Exercise 3

$$ \begin{eqnarray} \Omega &=& a/Z + 2b/Z + 3(Z-(a+b))/Z && \\ &=& 3 - \frac{2a + b}{Z}\\ \frac{1}{Z} &=& \frac{3 - \Omega}{2a + b}\\ Z &=& a + b + (4-a-2b)/3 \\ \Omega Z &=& 4\\ 1 <= &\Omega& < 4 &&\\ 0<=&a&<= 4 &&\\ 0<=&b&<= 2 &&\\ a+b+c-Z&=&0 \\ 1<=&Z&<= 4 && \end{eqnarray} $$

Here the key is that $\Omega Z = 4$, with $Z \in \{1,2,3,4\}$ so $\Omega \in \{4.0, 2.0, 1.\overline{3},1.0 \}$, taking into account the constraints each choice of $Z$ places on $a, b, \text{ and } c$.

Exercise 4

I'm not sure that this question belongs in the same category. The question requires two numbers $a$ and $b$, one odd and one even. Assuming $b>a$, if the gcd of the numbers is 1, then a sequence of length $b+1$ exists made from expressions starting at $a*(b)+b*(0)$ and ending at $a*(0)+b*(q)$, where $0\le q \le a$ is chosen to keep the sum between $a*(b-1)+1$ and $a*b$. Each value must be unique, so this makes the extreme upper limit $a*(b-1)-1$. If $a*(b-1)-1 \equiv 0 \text{ mod } b$, then the value is $a*(b-2)-1$.

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