I suppose the $a_n$ is actually $a(n)$ (i) consider $n=a_0+10a_1+10^2a_2+\cdots 10^ka_k$ Since $0\leq a_i<10$ , hence $n\geq 10^ka_k\geq a_1a_2\cdots a_k=a(n)$ (ii) From first part we can get $4\leq n\leq 14$ . If $n$ is single digit , from $4$ to $9$ , then by we can solve to get $n=4$ . If $n$ is double digit from $10$ to $14$ , by setting $n=10 + k$ then $a(n)=k$ . Plug back into the equation and we see no solution . Hence $n=4$ is the only one

(a) Let $k$ be the number of digits of $n$. Since $10^{k+1}-1 \geq 9^{k+1}$, we have: $n \geq \frac{10^{k+1}-1}{9} \geq 9^{k} \geq a(n)$. (b) Since $n \geq a(n)$ , we have: $n^{2} -17n +56 \leq n \Leftrightarrow n \in [4;14]$. If n is odd: $n^{2}-17n+56$ is even and $a(n)$ is odd (since $n \leq 14$). Contradiction. So $n \in (4;6;8;10;12;14)$. After checking the possibilities for $n$, we notice that $n=4$ is the unique solution.

$n=14$ is also a solution.

M4RI0 wrote: $n=14$ is also a solution. check again .... if $n=14$ , then $a(n)=4$ but not $14$ ....

The full answer is $n \in \lbrace 4,13 \rbrace$

giancarlo wrote: If n is odd: $n^{2}-17n+56$ is even and $a(n)$ is odd (since $n \leq 14$). Contradiction. Counter example: $n\in \lbrace 11, 13\rbrace$

shyong wrote: If $n$ is double digit from $10$ to $14$ , by setting $n=10 + k$ then $a(n)=k$ I think, that $a(n)=k+1$

(a) Let $n= a_0+10a_1+10^2a_2+\cdots+10^ka_k=\overline{a_ka_{k-1}\cdots a_1a_0}$ Now clearly$ a_n=a_1a_2\cdots a_k \le 10^ka_k $ Since each $a_i$< $10$. This finishes part a. (b) Note that $n^2-17n+56=(n^2-18n+56)-n=(n-4)(n-14) -n=a_n$ $\implies (n-4)(n-14)=a_n-n\le 0.....$ Hence,$n$ is bounded:$4\le n \le 14$ Now, It is easy to check that $n=4$ is the only possible solution. $\implies$ LHS =$ 16-68+56=4$ and RHS=$a_4 = 4$ This finishes part b and the problem. $QED$

We have a) $ n= a_0+10a_1+10^2a_2+\cdots+10^ka_k=\overline{a_ka_{k-1}\cdots a_1a_0} \ge 10^k*a_k>a_k*9^k \ge a_ka_{k-1}.....a_0 $. b)$n^2-17n+56=a_n \le n$ so $n^2-18n+56 \le0$ so $ 4\le n \le 14$ and now we have to calculate.n=4 the only solution

(a) Let $n_1$ be the first digit of $n$, and suppose $n$ has $k$ digits. The $n\ge n_1\cdot 10^{k-1}\ge n_1\cdot 9^{k-1}\ge a(n)$ with equality when $k=1$. (b) We have that $n\ge a(n)=n^2-17n+56$, so $(n-4)(n-14)\le 0$, implying $n \in [4,14]$. 1. On the range $[4,9]$, $n=a(n)$, so $n^2-18n+56=0\implies n=4$. 2. On the range $[10,14]$, $n=a(n)+10$, so $n^2-18n+66=0\to n=9+\sqrt{15}$, which is not an integer.

Part $(a)$:We will prove by induction,on the number of digits. Base case:true for all integers with one digit. assume true for all integers with $k$ digits. Now,it suffices to show, $a_{k+1}*10^{k+1}+....+a_0 \ge (a_k*10^k+\cdots+a_0)*a_{k+1} \ge a_{k+1}a_k \cdots a_0$ $\implies a_{k+1}10^{k+1} \ge (a_{k+1}-1)(a_k10^k+\cdots+a_0) $ which is obviously true. I think given part a ,part b becomes a lot easier.

四边形$ABCD$用满足$AD \nparallel BC$.$AD$和$BC$的中点分别是$M$和$N$.直线$MN$分别与对角线$AC$和$BD$相交于点$K$和$L$。证明三角形$AKM$的外接圆和$BNL$的外接圆在直线$AB$上有公共点.