Thrifty numbers Part 3

One example is sufficient to support the fact of maximum of 9 distinct integers:

$\boxed{1} + \boxed{7} = \boxed{8}$

$\boxed{9} - \boxed{5} = \boxed{4}$

$\boxed{2} \times \boxed{3} = \boxed{6}$

$\boxed{8} \div \boxed{4} = \boxed{2}$

Full set applied. The last row is flexible!

Answer: $\boxed{9}$

$\large{ \boxed{8} \, + \, \boxed{1} \, = \, \boxed{9} \\ \boxed{7} \, - \, \boxed{5} \, = \, \boxed{2} \\ \boxed{4} \, \times \, \boxed{2} \, = \, \boxed{8} \\ \boxed{6} \, \div \, \boxed{3} \, = \, \boxed{2} \\ }$ It is clearly seen, that I have used all the given digits.

By example

3+4=7

9-4=5

2x3=6

8/1=8

Yet another solution with all $9$ digits.

$\large{ \boxed{2}\,+\,\boxed{4}\,=\,\boxed{6}\\ \boxed{8}\,-\,\boxed{3}\,=\,\boxed{5}\\ \boxed{3}\,\times\,\boxed{3}\,=\,\boxed{9}\\ \boxed{7}\,\div\,\boxed{1}\,=\,\boxed{7}\\ }$

Well as the puzzle had stated, "What is the maximum possible distinct number of integers used?" It would be quite logical that the answer would be 9 without any work needed as you were given numbers from 1 to 9. 0 does not count so you only have 9 numbers that are available. You were given details and assumptions so there is a chance that some of the statements given were false.

I found it easiest to start with the division problem and work up. My solution:

$\boxed{5} + \boxed{4} = \boxed{9}$ $\boxed{6} - \boxed{3} = \boxed{3}$ $\boxed{7} \times \boxed{1} = \boxed{7}$ $\boxed{8} ÷ \boxed{4} = \boxed{2}$