Standard Algorithm: Regrouping

Core Concepts

Master the standard algorithm by learning how to 'carry over' or regroup when products are 10 or more.

Standard Algorithm: Regrouping

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When do you need to regroup (or carry) in multiplication?

You need to regroup when the product of the digits in any place value column is $10$ or greater.

In $127 \times 4$, the first step is $7 \times 4 = 28$. How do you write this and regroup?

A vertical multiplication problem of 127 times 4 displaying the first step, with an 8 written in the ones place of the answer and a regrouped 2 placed above the tens column.

Write the $8$ in the ones place of the answer. Regroup (carry) the $2$ by writing it above the tens place digit (the other $2$).

After regrouping the $2$ in $127 \times 4$, what is the next step?

The vertical multiplication problem 127 times 4 shows multiplying the tens digit 2 by 4 and adding the carried-over 2 to equal 10, demonstrating the tens-place step in the standard algorithm.

Multiply the tens digit: $2 \times 4 = 8$. Then, add the regrouped number: $8 + 2 = 10$. This result is for the tens place.

For $127 \times 4$, the tens calculation resulted in $10$. How do you write this and regroup again?

Vertical multiplication setup of 127 times 4 showing 08 below the line, with the previous regrouped 2 crossed out above the tens place and a regrouped 1 carried above the hundreds digit.

You have $10$ tens. Write the $0$ in the tens place of the answer. Regroup (carry) the $1$ to the hundreds place.

What is the final step and answer for $127 \times 4$?

Vertical multiplication of 127 by 4 using the standard algorithm, showing the final step where 1 times 4 plus the regrouped 1 equals 5, completing the product of 508.

Multiply the hundreds digit: $1 \times 4 = 4$. Add the regrouped $1$: $4 + 1 = 5$. Write $5$ in the hundreds place. The final answer is $508$.

Solve $38 \times 6$ using the standard algorithm.

1. Ones: $8 \times 6 = 48$. Write $8$, regroup $4$. 2. Tens: $3 \times 6 = 18$. Add regrouped $4$: $18 + 4 = 22$. Write $22$. Answer: $228$

Solve $475 \times 3$ using the standard algorithm.

1. Ones: $5 \times 3 = 15$. Write $5$, regroup $1$. 2. Tens: $7 \times 3 = 21$. Add regrouped $1$: $21 + 1 = 22$. Write $2$, regroup $2$. 3. Hundreds: $4 \times 3 = 12$. Add regrouped $2$: $12 + 2 = 14$. Write $14$. Answer: $1,425$

What is a common mistake when regrouping?

A common mistake is adding the regrouped number *before* multiplying. Always multiply first, then add the number you carried over.

Show the correct order for the tens place in $86 \times 4$.

Correct: Multiply $8 \times 4 = 32$, then add the regrouped $2$ (from $6 \times 4 = 24$) to get $34$. Incorrect: Adding $8 + 2 = 10$, then multiplying $10 \times 4 = 40$.

Solve $2,649 \times 5$ using the standard algorithm.

1. Ones: $9 \times 5 = 45$. Write $5$, regroup $4$. 2. Tens: $4 \times 5 = 20$. $20+4=24$. Write $4$, regroup $2$. 3. Hundreds: $6 \times 5 = 30$. $30+2=32$. Write $2$, regroup $3$. 4. Thousands: $2 \times 5 = 10$. $10+3=13$. Write $13$. Answer: $13,245$

Solve using the standard algorithm: $$ 809 \times 7 $$

1. Ones: $9 \times 7 = 63$. Write $3$, regroup $6$. 2. Tens: $0 \times 7 = 0$. $0+6=6$. Write $6$. 3. Hundreds: $8 \times 7 = 56$. Write $56$. Answer: $5,663$

What happens if your last multiplication step requires regrouping?

You simply write down the entire result. For example, in $621 \times 5$, the last step is $6 \times 5 = 30$. Since there are no more places to carry to, you write down $30$ to get the answer $3,105$.

Solve using the standard algorithm: $$ 3,182 \times 8 $$

1. Ones: $2 \times 8 = 16$. Write $6$, regroup $1$. 2. Tens: $8 \times 8 = 64$. $64+1=65$. Write $5$, regroup $6$. 3. Hundreds: $1 \times 8 = 8$. $8+6=14$. Write $4$, regroup $1$. 4. Thousands: $3 \times 8 = 24$. $24+1=25$. Write $25$. Answer: $25,456$

Standard Algorithm: Regrouping

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