Multiplication of Large Numbers Calculator

This calculator multiplies two numbers of any size and shows the work using the classic column multiplication method (the one you do by hand).

Step-by-step method

1. Step 1 (Set up): Write the two factors in columns, aligned on the right (ones under ones). Put the × sign to the left of the bottom factor, and draw a line underneath.
2. Step 2 (Multiply to make rows): Take the rightmost digit of the bottom factor. Multiply it by each digit of the top factor, moving from right to left. When you multiply, you may get either a one-digit or a two-digit result. If the result is one digit, write that digit in the current column under the line. If the result is two digits, write the right digit (the ones digit) in the current column under the line, and carry the left digit (the tens digit) by writing it above the next column to the left. In the next multiplication to the left, add the carry (if any) to the result before writing the next digit in the row, and carry the tens digit (if any) to the next column. If there is a carry left after the last multiplication in the row, write that carry at the far left of the row, under the line, to complete the row. Then move to the next digit of the bottom factor (one place to the left) and do the same thing again to make the next row — but shift the row one place left by adding one 0 to the bottom-right of that row. If you move another digit left, shift two places by adding two 0s to the bottom-right of that row, and so on until there are no more digits in the bottom factor.
3. Step 3 (Add the rows): Add all the rows you wrote under the line (the partial products) to get the final answer. Add in columns from right to left, carrying when needed. If Step 2 produced only one partial-product row, then Step 3 has nothing to add — that single row is the final answer.

Example 1: 27 × 93

Step 1 (Set up): Write the two factors in columns, aligned on the right (ones under ones). Put the × sign to the left of the bottom factor, and draw a line underneath.

In this problem: We write 27 on top and 93 below it, aligned on the right, then draw the line.

Step 2 (Multiply to make rows): Take the rightmost digit of the bottom factor. Multiply it by each digit of the top factor, moving from right to left. When you multiply, you may get either a one-digit or a two-digit result. If the result is one digit, write that digit in the current column under the line. If the result is two digits, write the right digit (the ones digit) in the current column under the line, and carry the left digit (the tens digit) by writing it above the next column to the left. In the next multiplication to the left, add the carry (if any) to the result before writing the next digit in the row, and carry the tens digit (if any) to the next column. If there is a carry left after the last multiplication in the row, write that carry at the far left of the row, under the line, to complete the row. Then move to the next digit of the bottom factor (one place to the left) and do the same thing again to make the next row — but shift the row one place left by adding one 0 to the bottom-right of that row. If you move another digit left, shift two places by adding two 0s to the bottom-right of that row, and so on until there are no more digits in the bottom factor.

In this problem: The bottom number 93 has 2 digits, so we will create 2 partial-product rows: Step 2a, Step 2b, and so on.

Step 2a (Start with the rightmost digit): Start with the rightmost digit of the bottom factor and create the first partial-product row (multiply across the top factor from right to left, carrying when needed).

In this problem: For Step 2a, we use the digit 3 from the bottom number.
3 × 7 = 21 → write 1, carry 2
3 × 2 + 2 = 8 → write 8, carry 0

This row (before shifting) is 81.

Step 2b (Move left and repeat): Move one digit to the left in the bottom factor and repeat to create the next row. Shift the row left by adding zeros at the end (one 0 for the tens digit, two 0s for the hundreds digit, and so on). Repeat this for any remaining digits (Step 2c, 2d, ...).

In this problem: For Step 2b, we use the digit 9 from the bottom number.
9 × 7 = 63 → write 3, carry 6
9 × 2 + 6 = 24 → write 4, carry 2
Leftover carry 2 becomes a new digit on the left.

This row (before shifting) is 243.

Shift: This digit is in the tens place, so we shift the row left by one place (add one 0 at the end).

Step 3 (Add the rows): Add all the rows you wrote under the line (the partial products) to get the final answer. Add in columns from right to left, carrying when needed. If Step 2 produced only one partial-product row, then Step 3 has nothing to add — that single row is the final answer.

In this problem: Now we add the partial-product rows to get the final product, carrying when needed.

Final answer: 27 × 93 = 2511

Example 2: 123 × 45

Step 1 (Set up): Write the two factors in columns, aligned on the right (ones under ones). Put the × sign to the left of the bottom factor, and draw a line underneath.

In this problem: We write 123 on top and 45 below it, aligned on the right, then draw the line.

Step 2 (Multiply to make rows): Take the rightmost digit of the bottom factor. Multiply it by each digit of the top factor, moving from right to left. When you multiply, you may get either a one-digit or a two-digit result. If the result is one digit, write that digit in the current column under the line. If the result is two digits, write the right digit (the ones digit) in the current column under the line, and carry the left digit (the tens digit) by writing it above the next column to the left. In the next multiplication to the left, add the carry (if any) to the result before writing the next digit in the row, and carry the tens digit (if any) to the next column. If there is a carry left after the last multiplication in the row, write that carry at the far left of the row, under the line, to complete the row. Then move to the next digit of the bottom factor (one place to the left) and do the same thing again to make the next row — but shift the row one place left by adding one 0 to the bottom-right of that row. If you move another digit left, shift two places by adding two 0s to the bottom-right of that row, and so on until there are no more digits in the bottom factor.

In this problem: The bottom number 45 has 2 digits, so we will create 2 partial-product rows: Step 2a, Step 2b, and so on.

Step 2a (Start with the rightmost digit): Start with the rightmost digit of the bottom factor and create the first partial-product row (multiply across the top factor from right to left, carrying when needed).

In this problem: For Step 2a, we use the digit 5 from the bottom number.
5 × 3 = 15 → write 5, carry 1
5 × 2 + 1 = 11 → write 1, carry 1
5 × 1 + 1 = 6 → write 6, carry 0

This row (before shifting) is 615.

Step 2b (Move left and repeat): Move one digit to the left in the bottom factor and repeat to create the next row. Shift the row left by adding zeros at the end (one 0 for the tens digit, two 0s for the hundreds digit, and so on). Repeat this for any remaining digits (Step 2c, 2d, ...).

In this problem: For Step 2b, we use the digit 4 from the bottom number.
4 × 3 = 12 → write 2, carry 1
4 × 2 + 1 = 9 → write 9, carry 0
4 × 1 = 4 → write 4, carry 0

This row (before shifting) is 492.

Shift: This digit is in the tens place, so we shift the row left by one place (add one 0 at the end).

Step 3 (Add the rows): Add all the rows you wrote under the line (the partial products) to get the final answer. Add in columns from right to left, carrying when needed. If Step 2 produced only one partial-product row, then Step 3 has nothing to add — that single row is the final answer.

In this problem: Now we add the partial-product rows to get the final product, carrying when needed.

Final answer: 123 × 45 = 5535