Question

in the addition problem below, some digits are missing, as indicated by the blanks. if the problem is done correctly, what is the sum of the missing digits?
_ 4 5
8 _ 8
+ 1 5 _
_ 3 2 9

the sum of the missing digits is
(simplify your answer )

Explanation:

Step1: Solve units place digit

Let the missing units digit be $x$. We know $5 + 8 + x$ ends in 9.
$5+8+x=19$ (since the sum must be 19 to have a units digit 9 and carryover 1)
$13+x=19$
$x=19-13=6$

Step2: Solve tens place digit

Let the missing tens digit be $y$. Account for carryover 1 from units place: $4 + y + 5 + 1$ ends in 2.
$10+y=12$ (sum is 12, carryover 1 to hundreds place)
$y=12-10=2$

Step3: Solve hundreds place digits

Let the top hundreds digit be $a$, bottom sum hundreds digit be $b$. Account for carryover 1 from tens place: $a + 8 + 1 + 1 = 10 + 3$ (sum is 13, carryover 1 to thousands place)
$a+10=13$
$a=13-10=3$
The thousands digit of the total sum $b$ is the carryover 1, so $b=1$.

Step4: Sum all missing digits

Missing digits are $a=3$, $y=2$, $x=6$, $b=1$.
Sum: $3+2+6+1=12$

Answer:

12