Can you figure out this equation?

Notice that $P=1$ because it carries from $E+G$ . $F+H$ must equal 11 because two distinct numbers cannot sum to 1. $E+G$ must then equal 10. The sum of the numbers from 1 to 9 is 45. Therefore $A+B+C+D=45-10-11-1=23$

$\small\begin{array}{lr} &8\ 6 \\ -&5\ 4\\ \hline \\ &3\ 2 \\ + &7\ 9\\ \hline \\ &1\ 1\ 1 \end{array}$

Since two digit number can be denoted as $10x+y$ where $x$ and $y$ are its digits. Similary for above case where $a,b,c, d,e,f,g$ and $h$ are distinct we have as $\begin{aligned} & (10a+b) -(10c+d) = 10e + f \\& 10(a-c)+(b-d) = 10e +f\end{aligned}$ Note that obtained difference is one digit number and it's digits can be equated as follow : $\begin{aligned} a - c = e \qquad b-d = f \end{aligned}$ Also $\begin{aligned} 10(e+g) +(f+h) = 111p\end{aligned}$ Note that the maximum sum of two two digits can never exceed 200 so which leads us the conclusion that $p=1$ . Since for $p>1$ the sum is not possible. Hence, $\begin{aligned} & 10(a-c+g) +(b-d+h) = 111\\& 10(a-c+g) +(b-d+h) = 10\times 11 + 1 \phantom{ccc}\text{isn't true}\\& \text{since} \phantom{cc} a-c+g+{\color{#3D99F6}1}= 11 \qquad b-d +h = 11\\& \therefore 10(a-c+g) +(b-d+h) =10\times 10 +11 \end{aligned}$

Equating the respective digits we have $\begin{cases} & a-c +g =10 \implies e+g =10 \\& b-d+h = 11\implies f+h =11 \end{cases}$ since $p=1$ so neither $a,c$ or $b,d$ cannot be consecutive integers so there difference should lie between 1 to 9(inclusive). Note: $(a-c)/e$ is read as $(a-c)$ or $e$ $(a-c)/e = \begin{cases} 2\implies g = 8 \\ {\color{#D61F06}3 \implies g =7 }\\ 4 \implies g = 6 \\ 5\implies g = 5 \end{cases} \quad (b-d)/f =\begin{cases}{\color{#D61F06} 2\implies h = 9} \\ 3 \implies h = 8 \\ 4 \implies h = 7 \\ 5 \implies h = 6 \end{cases}$

Note that when $(b-d)/f =2, h=9$ is uniquely defined since there occurs no $9$ for $g$ (also reverse works ) and it's clear that $(a-c)/e \neq 2 , g\neq 8$ . since $g\neq 8$ So highest possible value for $g=7$ (among ,6,5,) and it only works that prevents the reappears of same number.

$\begin{array}{lr} &A\ B \\ -&C\ D \\ \hline \\ &E\ 2 \\ + &7\ 9\\ \hline \\ &1\ 1\ 1 \end{array}$
Then we can conclude that $e=4$ and only left digits are $4$ , $5$ , $6$ and $8$ since $e =3$ which is possibly yield by $a-c= 8-5$ and $b-d=6-4=2$ .

So, $a=8, b = 6, c=5, d = 4, e = 3 , f = 2 , g = 7$ and $h=9$ . Therefore, $a+b+c+d = \boxed{23}$ .

There are three possible solutions:

$\begin{array}{ccc} & 9 & 5 \\ - & 2 & 7 \\\hline & 6 & 8 \\+ & 4 & 3 \\\hline 1 & 1 & 1 \\ \end{array}~~~~~$ $\begin{array}{ccc} & 8 & 6 \\ - & 5 & 4 \\\hline & 3 & 2 \\+ & 7 & 9 \\\hline 1 & 1 & 1 \\ \end{array}~~~~~$ $\begin{array}{ccc} & 8 & 5 \\ - & 4 & 6 \\\hline & 3 & 9 \\+ & 7 & 2 \\\hline 1 & 1 & 1 \\ \end{array}$

in all of them $A+B+C+D = 23$ , found via code:

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P = perms(1:9);
R =[];          
for k = 1:factorial(9)
    if (10*P(k,1)+P(k,2)-10*P(k,3)-P(k,4)==10*P(k,5)+P(k,6))&&(10*P(k,5)+P(k,6)+10*P(k,7)+P(k,8)==111*P(k,9))
        R = [R; P(k,:),P(k,1)+P(k,2)+P(k,3)+P(k,4)];
    end
end

using $Mathematica$ :

$P$ must be $1$

Select[Permutations@Range[2,9],#[[2]]-#[[4]]==#[[6]]&&#[[1]]-#[[3]]==#[[5]]&&#[[6]]+#[[8]]==11&&#[[5]]+#[[7]]==10&]

returns {8, 6, 5, 4, 3, 2, 7, 9} which is the correct {a,b,c,d,e,f,g,h} tuple

Seeing John Ross’s elegant solution makes me feel I made this hard for myself, but here’s my approach:

Observations:

• P must be 1
• EF + GH = 111
• F + H = 11... possible combinations (5,6) (4,7) (3,8) (2,9)
• E + G = 10 ... possible combinations (4,6) (3,7) (2,8)
• A > 4

Since digits cannot be duplicated possible combinations for EG&FH and remaining digits for ABCD (not in order) are:

Seek combinations of ABCD that give a remainder that can be formed from EG&FH...

Within the first set we find one solution: 95 - 27 = 68, leaving GH to be 43.

Sum of digits ABCD is 23

$\\It is clear that P=1 since it carries from E+G . \\ Therefore F+H =11 and E+G=10. \\Taking F=2, H=9, E=3, G=7 we get A+B+C+D= 8+6+5+4 =23.$

首先，9个数字的和是45。

PPP首位的P是由E+G进位得来的，所以一定是1，即PPP是111。F+H=11，才能得到末位的1，十位的1则是由个位进上来的，所以E+G=10。

最后A+B+C+D=45-11-10-1=23。

至于算式，则是：

86-54=32,32+79=111

如果想得到这个式子，不妨从下面四个数入手，1不能使用，因为P已经是1了，9和8的位置得到限制，接着将数字和为10和11的成对考虑，就可以得到这个式子了。

The solution.

Step 1: As the three digit in PPP are the same. PPP can be 111, 222, 333...

        The largest possibility of EF+GH is 187. Therefore, PPP can only be 111.

Step 2: After knowing that P=1, we need to list all the possibility of EF+GH=111

         e.g. 24+87, 25+86, 26+85, 27+84...

Step 3: We need to know four things:

        (1) F+H=11     「2+9」、「3+8」、「4+7」、「5+6」、「6+5」、「7+4」、「8+3」、「9+2」

        (2) E+G=10      「2+8」、「3+7」、「4+6」、「6+4」、「7+3」、「8+2」

        (3) The biggest number for EF is 78. If AB=90 and CD=20, then EF=78, which is the greatest number

        (4) A-C must be positive. As this problem don't have negative number, A-C should be a positive number

Step 4: Keep trying.

Step 5: The answer is 86-54=32, 32+79=111

        A=8, B=6, C=5, D=4, E=3, F=2, G=7, H=9, P=1

Step 6: A+B+C+D

      =8+6+5+4

      =23

ANSWER: 23