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5th Grade · 5.NBT.B.5

The multiplication standard algorithm in 5th grade

The multiplication standard algorithm is the written method most parents learned at school. You stack the numbers, multiply by one digit at a time and add the rows. In 5th grade, your child learns to use it quickly and correctly for multi-digit multiplication, such as 254×36254 \times 36.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm.

What it means, in plain words

An algorithm is a fixed set of steps that always works. In the standard algorithm, your child writes the larger number on top and lines up the ones digits. Each digit of the bottom number makes one row, called a partial product. The rows are added at the end to give the product, the answer to the multiplication.

Two details cause most of the trouble. The first is carrying, also called regrouping. When a step gives a two-digit number, such as 7×6=427 \times 6 = 42, the child writes the 2 and carries the 4 to the next column. The carried digit is added after the next multiplication, not before. The second detail is the zero at the start of the second row. The tens digit of the bottom number stands for tens, so the child writes a 0 in the ones place before multiplying by it.

In 4th grade, children multiplied with area models and wrote out every partial product. The multiplication standard algorithm is a shorter way to write the same steps. A quick estimate, made by rounding both numbers, tells your child whether the answer is about the right size.

Your child should be able to:

  • Multiply a 3-digit or 4-digit number by a 2-digit number with the standard algorithm
  • Carry digits correctly and add them after each multiplication
  • Write the 0 in the ones place before multiplying by the tens digit
  • Estimate the product first to check that the answer makes sense

Example

A 3-digit by 2-digit multiplication: what is 254×36254 \times 36?

  1. Multiply by the 6 ones first: 254×6=1,524254 \times 6 = 1{,}524. This is the first row.
  2. The 3 in 36 means 3 tens. Write a 0 in the ones place, then multiply by 3. The second row is 7,620, which is 254×30254 \times 30.
  3. Add the rows: 1,524+7,620=9,1441{,}524 + 7{,}620 = 9{,}144.
  4. Check with an estimate: 250×40=10,000250 \times 40 = 10{,}000, which is close.
  5. So 254 times 36 is 9,144.

Where children usually get stuck

  • For 254×36254 \times 36, your child writes the second row as 762, lined up under the first row, and gets 2,286.

    The 3 in 36 is treated as 3, not 30. Without the 0 in the ones place, the second row is worth ten times too little.

    What helps. Ask what the 3 in 36 is worth. Saying "30" out loud before the second row reminds your child to write the 0 first.

  • For 47×647 \times 6, your child writes 482 instead of 282.

    The child adds the carried 4 to the tens digit first and then multiplies: (4+4)×6=48(4 + 4) \times 6 = 48. The carry belongs after the multiplication: 4×6=244 \times 6 = 24, plus 4 is 28.

    What helps. Ask your child to say each step aloud: "multiply, then add the carry". Writing the carried digit small and crossing it out once it is used also helps.

  • For 38×2438 \times 24, the first row is right (152), but the second row reads 790 instead of 760.

    The 3 carried in the first row is still written above the tens column. The child adds it again while working on the second row.

    What helps. Cross out the carried digits as soon as the first row is done. Some children write the carries for each row in a different color.

Try four practice questions

Real 5th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

  1. 1Warm-up
    This multiplication is partly completed using the standard algorithm. One partial product row is missing.
    23×1492000‾\displaystyle \begin{array}{r}23 \\ \times 14 \\ \hline 92 \\ \underline{\phantom{000}}\end{array}
    Select the whole-number value of the missing partial product.
    A
    230\displaystyle \begin{array}{r}230\end{array}
    B
    023\displaystyle \begin{array}{r}\phantom{0}23\end{array}
    C
    220\displaystyle \begin{array}{r}220\end{array}
    D
    092\displaystyle \begin{array}{r}\phantom{0}92\end{array}
    Show the answer and how to solve it

    The missing row is 230 because the blank row represents 23 multiplied by 10.

    1. 1The row 92 is the first partial product, so it comes from the ones digit 4.
    2. 2Multiplying 23 by 4 matches the row that is already shown.
    3. 3The missing row must use the tens digit 1, so this row is 23 times 10.
    4. 4The choice showing 230 completes the missing line.

    Common mistake. A common mistake is choosing 23. That multiplies by 1 instead of 10, so the tens-place zero is missing.

    The idea underneath. Each partial-product row matches the place value of the multiplier digit, so the tens digit creates a tens row, not just another ones row.

  2. 2On level
    Liam used the standard algorithm to solve the multiplication shown below.
    312×23
    Select the statement that best describes the multiplication shown.
    A
    The first partial product is the first incorrect part.
    B
    The second partial product is the first incorrect part.
    C
    The final sum is the first incorrect part.
    D
    The work is correct.
    Show the answer and how to solve it

    The correct choice is "The first partial product is the first incorrect part" because 312×3=936312 \times 3 = 936, not 636636.

    1. 1Check the multiplication from top to bottom. The first row shows the product of 312 and 3 ones, so start there.
    2. 2Multiply 312 by 3. The correct product is 936, but the shown first row is 636. The first partial product is incorrect.
    3. 3Check the second partial product. This row should show 312 multiplied by 20, and the shown value of 6240 is correct.
    4. 4The first partial product is wrong (636 instead of 936), so the final sum is also affected. Since the error begins at the first row, the first partial product is the first incorrect part.

    Common mistake. A common mistake is choosing the final sum just because 6876 looks wrong. In first-error problems, stop at the earliest line that does not match the multiplication.

    The idea underneath. Check each line from top to bottom. The first partial product uses the ones digit. The 2 in 23 means 20, so the second partial product is 312 times 20. Its ones digit is 0. Stop at the first incorrect line.

  3. 3On level
    A standard-algorithm multiplication problem is partly completed below.
    Type the missing digits in the boxes to finish the partial products.
    Setup
    032×14‾\displaystyle \begin{array}{r} \phantom{0}32 \\ \underline{\times14} \end{array}
    Ones row
    1\displaystyle 1
    8\displaystyle 8
    Tens row
    3\displaystyle 3
    0\displaystyle 0
    Show the answer and how to solve it

    The missing digits are 2 and 2, so the partial products are 128 and 320.

    1. 1Multiply 32 by the ones digit, 4, to fill the first partial-product row.
    2. 2Multiply 32 by the 1 in the tens place to fill the second row. Because that digit means 10, the second partial product is 320. The multiplication by 10 already includes the tens-place value.
    3. 3Put those results back into the vertical work.

    Common mistake. The 1 in 14 means 10, not 1. The second partial product is 32×10=32032 \times 10 = 320, not 32.

    The idea underneath. Use the ones digit for the first partial product. For the second partial product, multiply by the value of the tens digit.

  4. 4Stretch
    Lena wrote the equation shown below.
    213×457=89,991\displaystyle 213 \times 457 = 89{,}991
    Select the statement that best describes Lena's claim.
    A
    The claim has the right size and the right ones digit, but an estimate using nearby hundreds still shows it cannot be right.
    B
    The claim is correct.
    C
    The claim has the wrong size.
    D
    The ones digit of the claim does not match the ones-place multiplication.
    Show the answer and how to solve it

    The claim has the right ones digit and a plausible number of digits, but 89,991 falls below the nearby-hundreds interval of 91,400 to 137,100 for 213×457213 \times 457.

    1. 1Start with Lena's claim.
    2. 2Multiply the ones digits 3 and 7 to check the last digit the product must have.
    3. 3The product must end in 1. The claim 89,991 also ends in 1, so the ones-digit check passes.
    4. 4Round 213 down to 200 and up to 300, then multiply both by 457 to get a nearby-hundreds interval.
    5. 5The claim 89,991 is below 91,400, so it falls outside the interval [91,400, 137,100]. The nearby-hundreds estimate rules out the claim.

    Common mistake. A common mistake is choosing the ones-digit statement because 3×7=213 \times 7 = 21 ends in 1 and the claim ends in 1. The ones-digit check actually passes here; the nearby-hundreds estimate is the decisive check.

    The idea underneath. The last digit of a product depends only on the ones digits of the factors, so you can reject some products without doing the full multiplication.