Fluently multiply multi-digit whole numbers using the standard algorithm.
Official wording from the Common Core State Standards for Mathematics (© 2010 National Governors Association Center for Best Practices and Council of Chief State School Officers). View on thecorestandards.org
By the end of fifth grade, students are expected to multiply multi-digit whole numbers quickly and accurately using the standard algorithm, the vertical method that multiplies by one digit of the bottom factor at a time and records regrouped values above. Fluency here means efficient, accurate and flexible, not just fast.
The algorithm is a compressed form of the partial products students used in fourth grade. In 346 × 27, the first row is 346 × 7 = 2,422 and the second row is 346 × 20 = 6,920, which is why a zero is placed in the ones column before multiplying by the 2. Adding the rows gives 9,342. Students who understand this can explain each step in place value language, spot an answer that is far too small, and estimate first (about 350 × 30 = 10,500) to judge whether the result is reasonable. That habit of estimating first is what separates fluent multipliers from students who only follow steps.
When multiplying by the tens digit, students who omit the zero record 346 × 2 instead of 346 × 20, and the final product is far too small.
In 47 × 6, the carried 4 from 7 × 6 = 42 must be added after 4 × 6 is found (24 + 4 = 28), not added to the 4 before multiplying.
Regrouped digits from the first row confuse the second row. Crossing them out or using a fresh line for each partial product prevents this.
If digits drift out of their columns, the final addition is wrong even when every multiplication fact is right. Grid paper helps keep columns straight.
Multiply 346 × 27 using the standard algorithm and check with an estimate.
Answer: 346 × 27 = 9,342.
Put a partial products area model next to the standard algorithm for the same problem and ask students to match each row of the algorithm to regions of the area model. This keeps the place value meaning visible while students build speed.
Fluency is often assessed with straight computation items, such as 1,208 × 36, and through word problems where multiplication is one step of several. Short, frequent practice is more effective than long sets.
Original questions written for this standard. Choose an option or type your answer, then press Check. Every question has a worked explanation.
Answer: 2538 (also accepted: 2,538)
3 × 6 = 18 (write 8, carry 1). 2 × 6 + 1 = 13 (write 3, carry 1). 4 × 6 + 1 = 25. The product is 2,538.
Answer: 1972 (also accepted: 1,972)
58 × 4 = 232 and 58 × 30 = 1,740. Adding gives 232 + 1,740 = 1,972.
Answer: C) 27,365
2,105 × 3 = 6,315 and 2,105 × 10 = 21,050. 6,315 + 21,050 = 27,365.
Answer: B) 20,680
The second row is 517 × 40, because the 4 is in the tens place. 517 × 40 = 20,680.
Answer: 4060 (also accepted: 4,060, 4060 seats, 4,060 seats)
145 × 8 = 1,160 and 145 × 20 = 2,900. 1,160 + 2,900 = 4,060 seats.
Answer: D) 700 × 50 = 35,000
Rounding each factor to its nearest friendly number gives 700 × 50 = 35,000. The exact product, 33,216, is close to this.
A full lesson with slides, activities and an exit ticket on multi-digit multiplication (standard algorithm), pitched to grade 5 and editable in PowerPoint or Google Slides.
Make a lesson →A printable, differentiated worksheet on 5.NBT.B.5 with an answer key, ready in about a minute.
Make a worksheet →Turn multi-digit multiplication (standard algorithm) into a quiz students answer online that marks itself, with a class summary for you.
Build a test →It sets the standard algorithm as the fluency goal. Students may still use area models or partial products to understand or check, but by the end of the year they should be fluent with the algorithm.
The standard says multi-digit whole numbers. In practice that is usually up to four-digit by two-digit products, which are big enough to need the full algorithm.