Let's consider the quantity of 7 rubles. Obviously, the fourth coin cannot be used to make 7 rubles, since it is worth more than 8. It cannot be made with only 1's, because at most 6 of each denomination can be used. Since each power is irrational, both the second and third coin must be used. These are at more than 2 and 4 rubles, respectively. Thus $7 = \alpha + \alpha^2$, so $\alpha = \frac{\sqrt{29}-1}{2}$. However, the quantity of 12 rubles cannot be formed, either using the fourth coins or just combinations of the second and third. Thus not all natural numbers of rubles can be formed.

Ksun48, actually $12=5+a+a^2$.

Here's a shorter solution. Again, take $\alpha = \frac{\sqrt{29} - 1}{7}$ and use induction. Assume that we had a representation $c_0(\alpha)^0 + c_1(\alpha)^1 \cdots + c_n (\alpha)^n$. We do the following after adding one; change every instance of a number with an index larger than $6$ from $c(\alpha)^n$ to $(c-7) (\alpha)^n + \alpha^{n+1} + \alpha^{n+2}$. Observe that this reduces the sum of the coefficients after every operation, so it must eventually terminate, since the coefficients stay positive. Hence, the algorithm terminates and we may induct.

Here is a non-inductive solution that I feel really shows what's going on in this problem. We will use $\alpha=\frac{\sqrt{29}-1}{2}$, which has minimal polynomial $\alpha^2+\alpha-7$. Though it's not necessary, let's begin by showing that this is the only value of $\alpha$ that could work. Claim: If $\alpha$ works, then we must have $\alpha=\frac{\sqrt{29}-1}{2}$. Proof: Suppose we are able to make $7$. Then, we can't use $\alpha^k$ for $k\ge 3$ as they are greater than $8$. Furthermore, we can't use two coins of $\alpha^2$ since the total would be again greater than $8$. However, we must use at least one $\alpha^2$, otherwise $\alpha$ would be rational. Thus, we have \[\alpha^2+k\alpha+r=7\]for some $k\in\{1,2,\ldots,6\}$ (since we want $\alpha^2$ to be irrational) and $r\in\{0,1,2,\ldots,6\}$. It's easy to see that the only quadratic for which there is a solution $\alpha>2$ is $\alpha^2+\alpha=7$, so we're done. $\blacksquare$ We have the following (technically unnecessary) claim which allows us to solve the problem. Claim: We are able to make the positive integer $n$ if and only if we have some $a_0,\ldots,a_m\in\mathbb{R}$ such that the coefficients of \[P(X)=(X^2+X-7)(a_m X^m+a_{m-1}X^{m-1}+\cdots+a_1X+a_0)+n\]are all in $\{0,1,\ldots,6\}$. Proof: The if direction is trivial, so we show the other direction, which supposes that \[n=P(\alpha)\]for some polynomial $\alpha$ with all coefficients in $\{0,1,\ldots,6\}$. Thus, $P(X)-n$ has a root $\alpha$, so it must be divisible by $X^2+X-7$, which implies the claim. $\blacksquare$ Now, the coefficients of the above $P(X)$ are \[(n-7a_0,a_0-7a_1,a_0+a_1-7a_2,\ldots,a_{m-2}+a_{m-1}-7a_m).\]Thus, we win by selecting $a_0=\lfloor n/7\rfloor$, $a_1=\lfloor a_0/7\rfloor$, and \[a_t=\left\lfloor\frac{a_{t-1}+a_{t-2}}{7}\right\rfloor.\]Note that this works since the above recursion clearly terminates due to size reasons. Remark: Note the final choice of $a_i$s is essentially forced, making this problem extremely rigid, since basically everything is forced from the start.

Nice problem! I claim that the positive root of $\alpha^2+\alpha=7$ works. To construct a natural number $n$, first begin with $n$ coins of denomination $1$, and then if there are at least $7$ coins of denomination $\alpha^k$ (including $\alpha^0=1$) at any point, exchange $7$ of them with singular coins of denomination $\alpha^{k+1},\alpha^{k+2}$. This process strictly decreases the number of coins at each step, and hence must terminate, at which points we will have at most $6$ coins of each denomination.

Choose $\alpha$ to be a root of the equation $\alpha^2+\alpha=7$. If some number has a representation using some $\alpha^k$ at least $7$ times, choose a group of $7$ of these coins and interchange them with $\alpha^{k+2}+\alpha^{k+1}$. Applying this operation until each coins appears at most $6$ times is sufficient.

Only post #3 shows that $\alpha^k\notin \mathbb{Q}\forall k\in\mathbb{N}$. Yeah. It's forced that $\alpha^2+\alpha=7$ but the positive root to this happens to work. You can first show via induction that $\alpha^k$ is irrational for all $k$, this is because it a linear combination of $1$ and $\alpha$. In fact, we can define the sequences $(a_k)$ and $(b_k)$ of integers such that $$\alpha^k=a_k+b_k\alpha$$Then $$\alpha^{k+1}=(a_k+b_k\alpha)\alpha=a_k\alpha+b_k(7-\alpha)=7b_k+(a_k-b_k)\alpha$$Therefore we have the recurrence relations $$a_{k+1}=7b_k$$$$b_{k+1}=a_k-b_k=7b_{k-1}-b_k$$$$a_0=1, b_0=0$$To show that $b_k\ne 0$, suppose otherwise. Then $7b_{k-2}=b_{k-1}$. Then $$7b_{k-2}=7b_{k-3}-b_{k-2}\implies b_{k-2}=\frac{7}{8}b_{k-3}$$$$\frac{15}{8}b_{k-3}=7b_{k-4}\implies b_{k-3}=\frac{56}{15}b_{k-4}$$$$\cdots$$$$b_1=cb_0$$For some $c$, which is false as $b_0=0, b_1=1$. The way you show that any $n$ has a representation is algorithmically constructing one using $7\alpha^k=\alpha^{k+1}+\alpha^{k+2}$. Repeatedly applying this identity starting with $n=n\alpha^0$ will eventually yield $$n=a_0\alpha^0+a_1\alpha^1+a_2\alpha^2+\cdots$$With all $a_i\le 6$, as desired.

Consider trying to make a total of 7. Since $\alpha$ is irrational, you can't just have a combination of 1s and $\alpha$s, and 6 1s is too small. Therefore, since $\alpha^3>7$, the $\alpha^2$ coin must be used. Furthermore, since it has value greater than 4, only one $\alpha^2$ coin can be used. Then, since $\alpha^2$ is also irrational, we can't just put 1s to finish, so we must have another $\alpha$. We are already at $>6$ with a $\alpha^2$ and an $\alpha$, so we cannot add any more coins since it would put us over 7. Therefore, the only way that 7 is reachable is if $\alpha^2+\alpha=7$, or $$\alpha=\frac{\sqrt{29}-1}{2}.$$ We claim that this value of $\alpha$ works. Consider obtaining a value $n$. Express $n$ in base 7, and take the sum of the corresponding powers of $\alpha^2+\alpha$. Of course, this may contain coefficents larger than 6. However, given any $a^k$ whose coefficent is at least 7, we can decrease its coefficent by 7 and increase the coefficents of $\alpha^{k+2}$ and $\alpha^{k+1}$ by 1 each. This operation decreases the sum of coefficents by 5, so it must eventually terminate, leaving a state with no coefficents at least 7, hence done.

Let $\alpha = \frac{\sqrt{29}-1}2$. I claim this works. Notice that $\alpha+\alpha^2 = 7$. Consider the following algorithm $n$: start with $n$ coins of denomination $1$ ruble, and every turn, if there exist $7$ coins of the same denomination $\alpha^k$, where $k$ is nonnegative, replace them with two coins of denominations $\alpha^{k+1}$ and $\alpha^{k+2}$, respectively. This process terminates finitely as the number of coins strictly decreases. When it does terminate, it leaves a representation of $n$ rubles satisfying the given condition.

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