Multiplication Structure Analysis

The problem involves multiplying a two-digit number, represented as $2a$, by another two-digit number, represented as $b2$. The digits $a, b, c, d$ are unknown.

The number $2a$ can be written algebraically as $20 + a$. The number $b2$ can be written as $10b + 2$. The final product is given as $8d6$, which algebraically is $800 + 10d + 6$. The multiplication process shown involves two partial products:

• The first partial product is $c6$, which results from $2a \times 2$.
• The second partial product is $84$, which is shifted one position to the left, representing $2a \times b0$.

Determine Digits 'a' and 'c'

The first partial product $c6$ is obtained by multiplying $2a$ by $2$. The unit digit of this product must be $6$. This implies that the unit digit of $2 \times a$ must be $6$. The possible digit values for $a$ are:

• If $a = 3$, then $2 \times 3 = 6$.
• If $a = 8$, then $2 \times 8 = 16$. The unit digit is $6$.

Now we find the corresponding digit $c$ for each case:

• If $a = 3$, then $2a = 23$. Multiplying by $2$: $23 \times 2 = 46$. Thus, $c = 4$.
• If $a = 8$, then $2a = 28$. Multiplying by $2$: $28 \times 2 = 56$. Thus, $c = 5$.

Find Digit 'b'

The second partial product is given as $84$, shifted left. This means the value $(2a \times b)$ is $84$. We test our possibilities for $a$:

• Case 1: If $a = 3$. We check if $(20+3) \times b = 84$. This means $23 \times b = 84$. Solving for $b$, we get $b = 84 / 23$, which is not an integer. Therefore, $a=3$ is incorrect.
• Case 2: If $a = 8$. We check if $(20+8) \times b = 84$. This means $28 \times b = 84$. Solving for $b$, we get $b = 84 / 28 = 3$. Since $3$ is a valid digit, $a=8$ and $b=3$ are the correct values.

From Case 2, we have $a=8$ and $b=3$. This also confirms $c=5$ from the previous step.

Calculate Digit 'd'

The final product is $8d6$. This is the sum of the two partial products. With $a=8$, $b=3$, and $c=5$, the partial products are $c6 = 56$ and $84$ (shifted left, meaning $840$).

Adding the partial products: $56 + 840 = 896$.

Comparing the calculated sum $896$ with the given final product $8d6$, we can identify the digit $d$. The hundreds digit is $8$, the units digit is $6$, and the tens digit $d$ corresponds to $9$. So, $d = 9$.

Compute Final Sum a + b

We have found the values for the digits $a = 8$ and $b = 3$. The question asks for the sum $a + b$.

Calculation: $a + b = 8 + 3 = 11$.