Solve the following by using the lattice method.

Imagine a grid with 3 columns (for the digits of 678) and 2 rows (for the digits of 58). Each cell in the grid is divided by a diagonal line from the top-right to the bottom-left.

Multiply each digit from the top number by each digit from the side number. Write the tens digit of the product in the upper triangle of the cell and the units digit in the lower triangle.

• For the first row (multiplied by 5):

  • $6 \times 5 = 30 \rightarrow$ Cell (row 1, col 1) gets 3 (top), 0 (bottom).
  • $7 \times 5 = 35 \rightarrow$ Cell (row 1, col 2) gets 3 (top), 5 (bottom).
  • $8 \times 5 = 40 \rightarrow$ Cell (row 1, col 3) gets 4 (top), 0 (bottom).

• For the second row (multiplied by 8):

  • $6 \times 8 = 48 \rightarrow$ Cell (row 2, col 1) gets 4 (top), 8 (bottom).
  • $7 \times 8 = 56 \rightarrow$ Cell (row 2, col 2) gets 5 (top), 6 (bottom).
  • $8 \times 8 = 64 \rightarrow$ Cell (row 2, col 3) gets 6 (top), 4 (bottom).

The filled grid values (top/bottom) are:
Row 1: $3/0 \quad 3/5 \quad 4/0$
Row 2: $4/8 \quad 5/6 \quad 6/4$

Starting from the bottom-right, sum the numbers in each diagonal band. If a sum is 10 or more, write down the units digit and carry over the tens digit to the next diagonal sum.

• Diagonal 1 (rightmost): 4
• Diagonal 2: $0 + 6 + 6 = 12$. Write 2, carry 1.
• Diagonal 3: $5 + 4 + 5 + 8 + 1 (\text{carry}) = 23$. Write 3, carry 2.
• Diagonal 4: $0 + 3 + 4 + 2 (\text{carry}) = 9$.
• Diagonal 5 (leftmost): 3

Read the digits from the top-left diagonal sum to the bottom-right diagonal sum.
The digits are $3, 9, 3, 2, 4$.

The final answer is: