The Answer Is Quite Large

Note that for $0 \le a,b \le 8$ , $9 \times a + b$ represents all integers between 0 and 80 inclusive. So if $a \neq b$ , we can use this representation to give the number.

When $a = b$ , the integer is $10a = a \times 5 \times 2$ , thus if $a \neq 2, 5$ , we can use this representation to give the number.

When $a = 2$ , the number is $20 = 5 \times 4 + 0$ . When $a = 5$ , the number is $50 = 8 \times 7 - 6$ . Thus in fact, all positive integers not greater than 80 can be represented in this way.

For 81, suppose not all circles are multiplication signs. We claim its value can't be greater than 80. One way to do this is with case by case analysis (there are only 15 cases; most of them involve " $a - b \le a$ " and " $a \div b \le a$ ", and the rest uses $a \times b \le 72$ only reachable when $\{a, b\} = \{8, 9\}$ , and $a \times b \le 63$ otherwise). Thus we have $a \times b \times c = 81$ . Thus $a, b, c$ must be perfect powers of 3, and thus must be $1, 3, 9$ in some order. But then $1 \times 3 \times 9 = 27 \neq 81$ , contradiction. So 81 is the smallest positive integer that cannot be represented in such way.

consider the forms

$9n-m$ , $9n+m$ where m,n represent distinct digits

The interval $9n+m$ and $9(n+1)-p$ overlap

if $n=m=x$ and $(n+1)=p=y$

for equality we have $9x+x=9y-y$

$\Rightarrow 10x=8y$ or

$(x,y)=(4,5)$

we can see that all numbers from $1$ to $79$ except $40$ can be represented using one of these forms

but , $40=8 \times5+0$

Also note that $79$ is the largest number that can be written in the form $a\times b +c$

thus any larger number has to be of the form $a\times b\times c$

checking for $80$ , we get $80=2\times5\times8$

now $81=3^4$ thus we get ,

$3^a\times3^b\times3^c=3^4$

$\Rightarrow 3^{a+b+c}=3^4$ or

$a+b+c=4$

now a,b,c needs to be distinct and $0\leq a,b,c\leq2$

thus we find that $a,b,c$ must be a permutation of $(0,1,2)$

but,

$0+1+2=3\neq4$

thus the number 81 is impossible to represent .