# A different kind of math problem

A different kind of math problem

$$\text{If }(2+5=8),\ (8-3=1),\ \text{and}\ (4+3 = 9),$$

$$\text{Then, what is } (8-6+5) ?$$

Bonus:

$$\text{If }(8 - 2 + 7 + \text{P} = \text{A})$$,

$$\text{Then what is } (6 - 5 + 4)?$$

• I'm sure this is a duplicate, but it's very hard to search for ... Mar 16, 2020 at 11:44

These are 7-segment digital displays, where the result is an inclusive OR.
8-6 leaves only the top right vertical; adding that to 5 gives 9.

Bonus answer edited in after initial:

(8-2+7+P) = A
8-2 is top left and bottom right verticals.
+7 overlaps bottom right, so it's 7 + top left.
+P as expected adds the central horizontal and bottom left vertical to give a standard 7-segment display capital A.

So!

6-5 is bottom left vertical only.
Adding that to 4 makes a capital H.

• Interestingly, subtraction doesn't commute with addition - (8-6)+5 would give a different result to (8+5)-6. Mar 16, 2020 at 9:23

This is obviously not the answer you intended

In your three initial examples, if you double the smallest number on LHS
and subtract 1 from the largest number on LHS
all three examples work out.

Examples:

2+5=8 -> 2*2+5-1=8 -> 4+4=8
8-3=1 -> 8-1-3*2=1 -> 7-6=1
4+3=9 -> 4-1+3*2=9 -> 3+6=9

Therefore:

8-6+5=? -> 8-1-6+5*2 -> 7-6+10=11