Multiplication of Large Numbers Calculator

Published on: February 11, 2024

Final Answer: Free • Full Steps: Free

This Multiplication of Large Numbers Calculator helps you multiply large numbers and shows the working clearly using column multiplication. Write the numbers in columns, then multiply the top number by one digit of the bottom number at a time, carrying past nine. Each pass makes a partial product row, shifted one place left. Adding those rows together gives the final answer. Following the same pattern for every digit is what makes long multiplication easy to picture, check, and repeat on your own with any large numbers.

Step-by-step method

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.
Take the rightmost digit of the bottom factor and multiply it by every digit of the top factor, moving right to left, carrying whenever a result goes past nine. That makes one partial-product row. Then move to the next digit of the bottom factor and do the same again, shifting each new row one more place to the left by adding a zero at its end. This repeats once for every digit of the bottom factor, moving right to left.
Do an addition of large numbers on the partial-product rows to get the final answer.

See Example 1 Hide Example 1

Problem

\(27 \times 93\)

Approach

Step-by-step method

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.
Take the rightmost digit of the bottom factor and multiply it by every digit of the top factor, moving right to left, carrying whenever a result goes past nine. That makes one partial-product row. Then move to the next digit of the bottom factor and do the same again, shifting each new row one more place to the left by adding a zero at its end. This repeats once for every digit of the bottom factor, moving right to left.
Do an addition of large numbers on the partial-product rows to get the final answer.

Step 1

Step 1 - 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.

	2	7
×	9	3

Step 2

Step 2 - Take the rightmost digit of the bottom factor and multiply it by every digit of the top factor, moving right to left, carrying whenever a result goes past nine. That makes one partial-product row. Then move to the next digit of the bottom factor and do the same again, shifting each new row one more place to the left by adding a zero at its end. This repeats once for every digit of the bottom factor, moving right to left.

In this problem: The bottom number \(93\) has \(2\) digits, so we will create \(2\) partial-product rows, moving right to left. The lettered steps below build them one digit at a time, with an extra step wherever a row has to shift first.

	2	7
×	9	3

Step 2a

Step 2a - Multiply this digit of the bottom factor by the digit of the top factor above it, adding any carry brought over. Write the ones digit of that result in the row, and carry the tens digit to the next column on the left.

In this problem: \(3 \times 7 = 21\) → write \(1\), carry \(\textcolor{red}{2}\)

	2
	2	7
×	9	3

		1

Step 2b

Step 2b - At the leftmost digit of the top factor there is no next column to carry into, so the whole result is written into the row.

In this problem: \(3 \times 2 + \textcolor{red}{2} = 8\) → write \(8\)

	2
	2	7
×	9	3

	8	1

Step 2c

Step 2c - Before multiplying, shift the new row left to match the place of the bottom-factor digit making it, by putting that many zeros at its end.

In this problem: The digit \(9\) is in the tens place, so we start this row with one 0 at its end.

	2	7
×	9	3

	8	1
		0

Step 2d

Step 2d - Multiply this digit of the bottom factor by the digit of the top factor above it, adding any carry brought over. Write the ones digit of that result in the row, and carry the tens digit to the next column on the left.

In this problem: \(9 \times 7 = 63\) → write \(3\), carry \(\textcolor{red}{6}\)

	6
	2	7
×	9	3

	8	1
	3	0

Step 2e

Step 2e - At the leftmost digit of the top factor there is no next column to carry into, so the whole result is written into the row.

In this problem: \(9 \times 2 + \textcolor{red}{6} = 24\) → write \(24\)

			2	7
×			9	3

			8	1
	2	4	3	0

Step 3

Step 3 - Do an addition of large numbers on the partial-product rows to get the final answer.

In this problem: Adding the \(2\) partial-product rows gives \(2511\).

		1
			8	1
+	2	4	3	0

	2	5	1	1

Problem

\(81 + 2430\)

Approach

Step-by-step method

Write the two addends in columns, aligned on the right (ones under ones). Put the + sign to the left of the bottom addend, and draw a line underneath.
Start at the rightmost column and add its two digits. Write the ones digit under the line, and if the total goes above nine, carry the tens digit to the next column on the left by writing it above that column. In every column after that, add the two digits plus any carry brought over. This repeats once for every column, moving right to left. If there is a carry after the leftmost column, write it as a new digit on the left of the current result to get the final answer.

Step 1

Step 1 - Start with the rightmost column (ones). Add the two digits, write the ones digit, and carry the tens digit (if any) to the next column.

In this problem: \(1 + 0 = 1\) → write \(1\) under the ones column.

			8	1
+	2	4	3	0

				1

Step 2

Step 2 - Move one column to the left and repeat the same process: add the two digits plus any carry, write the ones digit, and carry the tens digit (if any). Keep repeating until you finish the leftmost column.

In this problem: \(8 + 3 = 11\) → write \(1\) under the tens column, carry \(\textcolor{red}{1}\) to the hundreds column.

		1
			8	1
+	2	4	3	0

			1	1

Step 3

Step 3 - Move one column to the left and repeat the same process: add the two digits plus any carry, write the ones digit, and carry the tens digit (if any). Keep repeating until you finish the leftmost column.

In this problem: \(0 + 4 + \textcolor{red}{1} = 5\) → write \(5\) under the hundreds column.

		1
			8	1
+	2	4	3	0

		5	1	1

Step 4

Step 4 - Move one column to the left and repeat the same process: add the two digits plus any carry, write the ones digit, and carry the tens digit (if any). Keep repeating until you finish the leftmost column.

In this problem: \(0 + 2 = 2\) → write \(2\) under the thousands column.

			8	1
+	2	4	3	0

	2	5	1	1

Final Answer

\(2511\)

Final Answer

\(27 \times 93 = 2511\)

See Example 2 Hide Example 2

Problem

\(123 \times 45\)

Approach

Step-by-step method

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.
Take the rightmost digit of the bottom factor and multiply it by every digit of the top factor, moving right to left, carrying whenever a result goes past nine. That makes one partial-product row. Then move to the next digit of the bottom factor and do the same again, shifting each new row one more place to the left by adding a zero at its end. This repeats once for every digit of the bottom factor, moving right to left.
Do an addition of large numbers on the partial-product rows to get the final answer.

Step 1

Step 1 - 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.

	1	2	3
×		4	5

Step 2

In this problem: The bottom number \(45\) has \(2\) digits, so we will create \(2\) partial-product rows, moving right to left. The lettered steps below build them one digit at a time, with an extra step wherever a row has to shift first.

	1	2	3
×		4	5

Step 2a

In this problem: \(5 \times 3 = 15\) → write \(5\), carry \(\textcolor{red}{1}\)

		1
	1	2	3
×		4	5

			5

Step 2b

Step 2b - Multiply this digit of the bottom factor by the digit of the top factor above it, adding any carry brought over. Write the ones digit of that result in the row, and carry the tens digit to the next column on the left.

In this problem: \(5 \times 2 + \textcolor{red}{1} = 11\) → write \(1\), carry \(\textcolor{red}{1}\)

	1	1
	1	2	3
×		4	5

		1	5

Step 2c

Step 2c - At the leftmost digit of the top factor there is no next column to carry into, so the whole result is written into the row.

In this problem: \(5 \times 1 + \textcolor{red}{1} = 6\) → write \(6\)

	1	1
	1	2	3
×		4	5

	6	1	5

Step 2d

Step 2d - Before multiplying, shift the new row left to match the place of the bottom-factor digit making it, by putting that many zeros at its end.

In this problem: The digit \(4\) is in the tens place, so we start this row with one 0 at its end.

	1	2	3
×		4	5

	6	1	5
			0

Step 2e

Step 2e - Multiply this digit of the bottom factor by the digit of the top factor above it, adding any carry brought over. Write the ones digit of that result in the row, and carry the tens digit to the next column on the left.

In this problem: \(4 \times 3 = 12\) → write \(2\), carry \(\textcolor{red}{1}\)

		1
	1	2	3
×		4	5

	6	1	5
		2	0

Step 2f

Step 2f - Multiply this digit of the bottom factor by the digit of the top factor above it, adding any carry brought over. Write the ones digit of that result in the row, and carry the tens digit to the next column on the left.

In this problem: \(4 \times 2 + \textcolor{red}{1} = 9\) → write \(9\)

		1
	1	2	3
×		4	5

	6	1	5
	9	2	0

Step 2g

Step 2g - At the leftmost digit of the top factor there is no next column to carry into, so the whole result is written into the row.

In this problem: \(4 \times 1 = 4\) → write \(4\)

			1
		1	2	3
×			4	5

		6	1	5
	4	9	2	0

Step 3

Step 3 - Do an addition of large numbers on the partial-product rows to get the final answer.

In this problem: Adding the \(2\) partial-product rows gives \(5535\).

	1
		6	1	5
+	4	9	2	0

	5	5	3	5

Problem

\(615 + 4920\)

Approach

Step-by-step method

Write the two addends in columns, aligned on the right (ones under ones). Put the + sign to the left of the bottom addend, and draw a line underneath.
Start at the rightmost column and add its two digits. Write the ones digit under the line, and if the total goes above nine, carry the tens digit to the next column on the left by writing it above that column. In every column after that, add the two digits plus any carry brought over. This repeats once for every column, moving right to left. If there is a carry after the leftmost column, write it as a new digit on the left of the current result to get the final answer.

Step 1

Step 1 - Start with the rightmost column (ones). Add the two digits, write the ones digit, and carry the tens digit (if any) to the next column.

In this problem: \(5 + 0 = 5\) → write \(5\) under the ones column.

		6	1	5
+	4	9	2	0

				5

Step 2

In this problem: \(1 + 2 = 3\) → write \(3\) under the tens column.

		6	1	5
+	4	9	2	0

			3	5

Step 3

In this problem: \(6 + 9 = 15\) → write \(5\) under the hundreds column, carry \(\textcolor{red}{1}\) to the thousands column.

	1
		6	1	5
+	4	9	2	0

		5	3	5

Step 4

In this problem: \(0 + 4 + \textcolor{red}{1} = 5\) → write \(5\) under the thousands column.

	1
		6	1	5
+	4	9	2	0

	5	5	3	5

Final Answer

\(5535\)

Final Answer

\(123 \times 45 = 5535\)