Bump!!!!

Dont't know enough for NT, but I though I should give it a try. As we know the representation of a number in numerical system with base $4$ is one and only. I believe it is enough to create the numbers $1, 2, 3$ under the given conditions, and add them to powers of $4$ to get any other number. $1=3^0, 2=3^0+4^0, 3=3^0+4^0+7^0$

parmenides51 wrote: Dont't know enough for NT, but I though I should give it a try. As we know the representation of a number in numerical system with base $4$ is one and only. I believe it is enough to create the numbers $1, 2, 3$ under the given conditions, and add them to powers of $4$ to get any other number. $1=3^0, 2=3^0+4^0, 3=3^0+4^0+7^0$ How would you express $8$ using this way?

You have a point, we need multiplication also, so my attempt was wrong. Thanks

Redacted....

$\displaystyle \mathbb{N} :$ the set of positive integers $\displaystyle D:$ the set of positive integers which have valid representation We can define the sequence $\displaystyle ( a_{n})$ as follows: $\displaystyle a_{1} =a_{2} =a_{3} =1,\ a_{n+3} =$ the $\displaystyle n$-th smallest element of the set $\displaystyle \left\{3^{x} |\ x\in \mathbb{N}\right\} \cup $$\displaystyle \left\{4^{x} |\ x\in \mathbb{N}\right\} \cup \left\{7^{x} |\ x\in \mathbb{N}\right\}$ for each $\displaystyle n\in \mathbb{N}$. For each $\displaystyle m\in \mathbb{N}$, we can define the set $\displaystyle D_{m} \subset D$ as follows: $\displaystyle e\in D_{m} \ \Leftrightarrow $ there exist a finite sequence $\displaystyle ( c_{n}) \in \{0,1\}^{m}$ such that $\displaystyle e\ =\sum ^{m}_{i=1} c_{i} a_{i}$ . Note that $D_{i} \subset D_{i+1} $ for all $i\in \mathbb{N}$. For each $\displaystyle x\in \mathbb{N} \ ( x\geq 2)$, let $\displaystyle f(x) =\sum ^{h}_{i=1} a_{i}$ where $\displaystyle h\in \mathbb{N}$ satisfy $\displaystyle a_{h} < x\leq a_{h+1}$. Lemma $\displaystyle f( x) \geq x$ Proof Let $\displaystyle b\geq 2$ an arbitrary integer. We can take $\displaystyle m\in \mathbb{N}$ such that $\displaystyle \ x\leq b^{m} < bx$. We know that $\displaystyle \sum\limits ^{m-1}_{i=0} b^{i} \ =\frac{b^{m} -1}{b-1} \geq \frac{x-1}{b-1}$ . So we get $\displaystyle f( x) \geq $ $\displaystyle \frac{x-1}{2} +\frac{x-1}{3} +\frac{x-1}{6} =x-1$. Since $\displaystyle x$ cannot be a power of $\displaystyle 3$ and a power of $\displaystyle 4$, we can say $\displaystyle f( x) \neq x-1.$ Hence we have $\displaystyle f( x) \geq x$ (Note that $\displaystyle f( x) \in \mathbb{N}$) Solution Let $\displaystyle S_{n} =\sum ^{n}_{i=1} a_{i} \ \ \ ( n\in \mathbb{N})$ Let $\displaystyle I_{n} =[ 1,S_{n}] \cap \mathbb{N} \ \ ( n\in \mathbb{N})$ We will prove that $\displaystyle I_{n} \subset D_{n}$ for each $\displaystyle n\in \mathbb{N}$ by Induction of $\displaystyle n$. We can easily verify that $\displaystyle I_{1} \subset D_{1}, I_{2} \subset D_{2}$, and $I_{3} \subset D_{3}$. Assume $\displaystyle I_{n} \subset D_{n}$ for some $\displaystyle n\in \mathbb{N} \ ( n\geq 3)$. Then we get that $\displaystyle \mathbb{N}\cap[ 1+a_{n+1} ,S_{n+1}] \subset D_{n+1}$. If we show $\displaystyle 1+a_{n+1} -S_{n} \leq 1$, then we have $\displaystyle I_{n+1} \subset D_{n+1}$ Since $\displaystyle S_{n} =f( a_{n+1}) \geq a_{n+1}$, we have $\displaystyle 1+a_{n+1} -S_{n} \leq 1+a_{n+1} -a_{n+1} = 1$. Hence $\displaystyle I_{n+1} \subset D_{n+1}$ and then $\displaystyle I_{n} \subset D_{n}$ for all $\displaystyle n\in \mathbb{N}$. Since $\displaystyle D_{n} \subset D$, $\displaystyle I_{n} \subset D_{n}$, and $\displaystyle \bigcup ^{\infty }_{n=1} \ I_{n} =\mathbb{N}$, we can conclude $\displaystyle \mathbb{N} =D$.

I am sorry for reviving this one, but doesn't anyone have another solution?

Exactly the same reasoning as ELMO Shortlist 2013 N3

anybody who has another solution?

Another way:Assume that N is a integer we want of this form. The base case 1 obviously works. Let ${x_1}>{x_2}>{x_3}$ be the largest powers of 3,4,7 less than $N-3$ in some order. If one of the inequalities of the form 3≤$N-({x_1}+....+{x_k})<{x_k}+3$ (1≤k≤3) Is true then we will be done,since we can subtract ${x_1},{x_2},....,{x_k}$ from n to get $N'$ and then apply the inductive hypothesis,the construction for $N'$ cannot use any of the (${x_1},{x_2},....,{x_k}$) since $N'-3<{x_k}$ Also $N-3>{x_1}$ And this is true since ${x_1}+{x_2}+2{x_3}$$\ge(N-3)*(\frac{1}{3}+$$\frac{1}{4}+$$\frac{1}{7}+$$\frac{1}{7}$)$>N'-3$ So ${a_0}=N-3$ and ${a_i}={x_i}$ for 1≤i≤3 Hence proved.

Let's prove the following lemma: Lemma: The following sequence $1={a_1}\leq{a_2}\leq {a_3} \leq...$ satisfies: $a_{n+1} \leq a_{1} + a_{2} + ... + a_{n} +1$ If and only if we can form every integer positive as sums of different terms of this sequence. Prove: Suppose that we can form every integer positive, if $a_{n+1} > a_{1} + a_{2} + ... + a_{n} +1$ for some n, so take the number $a_{1} + a_{2} + ... + a_{n} +1$ and it cannot be expressed as sums of different terms of this sequence. For the second part, suppose that $a_{n+1} \leq a_{1} + a_{2} + ... + a_{n} +1 \forall n \in \mathbb{N}$, we are gonna do an induction. The base is $a_{1} = 1$ and suppose that for a number k we can expressed all number between 1 and $a_{1} + a_{2} + ... + a_{k}$ just using $a_{1}, a_{2}, ..., a_{k}$. How $a_{k+1} \leq a_{1} + a_{2} + ... + a_{k} +1$, so we can take a number between $a_{1} + a_{2} + ... + a_{k} +1$ and $a_{1} + a_{2} + ... + a_{k} + a_{k+1}$, subtract $a_{k+1}$ and make the construction using our hypothesis, so we take $a_{k+1}$ back and we finish the induction. For our problem, take the sequence $(a_{n})$ to be the nth smallest term of the set $\displaystyle \left\{3^{x} |\ x\in \mathbb{Z_{+}}\right\} \cup \left\{4^{x} |\ x\in \mathbb{Z_{+}}\right\} \cup \left\{7^{x} |\ x\in \mathbb{Z_{+}}\right\}$. We just need to prove that $a_{n+1} \leq a_{1} + a_{2} + ... + a_{n} +1 \forall n \in \mathbb{N}$, and we will prove with induction. The base it's just see that $a_{1} = a_{2} = a_{3} = 1 = 3^0 = 4^0 = 7^0$, now suppose that $a_{i+1} \leq a_{1} + a_{2} + ... + a_{i} +1 \forall i \leq k-1$, so we will prove that $a_{k+1} \leq a_{1} + a_{2} + ... + a_{k} +1$, for that take $3^x, 4^y, 7^z$ the highest powers of 3, 4, and 7 that appeared in the sequel until $a_{k}$, so $a_{1} + a_{2} + ... + a_{k} +1 = 1 + 3^1 + ... + 3^x + 1 + 4^1 + ... + 4^y + 1 + 7^1 + ... + 7^z + 1 =$ $\frac{3^{x+1}-1}{2} + \frac{4^{y+1}-1}{3} + \frac{7^{z+1}-1}{6} + 1$. Let be m = min{$3^{x+1}, 4^{y+1}, 7^{z+1}$}, so $a_{k+1} = m$ and how $a_{1} + a_{2} + ... + a_{k} +1 = \frac{3^{x+1}-1}{2} + \frac{4^{y+1}-1}{3} + \frac{7^{z+1}-1}{6} + 1$ $ \geq \frac{m-1}{2} + \frac{m-1}{3} + \frac{m-1}{6} - 1 = m-1+1 = m = a_{k+1}$, so the results follow for our lemma. Cono sur in my city