How to select 8 bits?

I am working on a simple counter that takes start, stop and increment values as decimal inputs. Here's the circuitry of the start value input of my counter: What the whole counter does is count from whatever start value you input by whatever increment you set and should stop if the count value is over the stop value. What I'm having trouble with is getting the inputs to get read as 8-bit values by the adder. My inputs are represented by switches, each switch represents the decimal numbers 1-9. The switches are organized in 2 columns, the first column represents the MSB and the second column represents the LSB.

On the above picture, the display is currently showing 11, switch one in column 1 is switched 'on' and switch one in column 2 is switched 'on' as well. The issue I have is, when I input the data into an 8-bit adder, the outputs it as 2. The adder reads them as two separate 4bit binary numbers. I know that using a 16-bit adder would work, but is there any other way to get the 8-bit adder to read the input values as 8-bit values or do I need a 16-bit adder?

• I had a hard time reading exactly what was going on. So let me take a shot at it. You have a total of 36 switches used to represent two numbers: a start value wasting 18 switches to make two BCD-coded values from 00 to 99 and an increment value using another 18 switches for two BCD-coded values also from 00 to 99. (This already sounds switch-crazy as those things are expensive.) So you need a way to cause two BCD-coded digits to be fused into a single binary 7-bit value before presenting the current value and the increment to an adder. Is that about it? – jonk May 25 '19 at 5:34
• eeeguide.com/decimal-adder-bcd-adder – Tony Stewart Sunnyskyguy EE75 May 25 '19 at 5:59