Kid Einsteins · Multi-Digit Multiplication · Grades 3–5 · NYC Math Class

Multi-Digit Multiplication — 2×1 and 2×2, With & Without Carrying (30 Questions with Hidden Answers) | Kid Einsteins Class 5

The class where students go from “I can multiply” to “I can multiply anything.” This lesson teaches 2-digit × 1-digit and 2-digit × 2-digit multiplication in four escalating levels — no carrying, then carrying — using the box method and the standard algorithm side by side. Finishes with 30 practice questions (16 easy, 6 medium, 4 hard, 4 word problems), each with click-to-reveal step-by-step answers. Built for the SOMATH Kid Einsteins program (grades 3–5) on the Upper West Side of Manhattan.

· By the SOMATH team · 226 W 79th St, UWS · (646) 668-6151

This is Class 5 of the SOMATH Kid Einsteins multiplication arc. In Class 4 we learned that any hard multiplication can be split into easier pieces using the distributive property and partial products. In this class we take that same idea and run it through the four cases students will meet every week for the rest of grade school:

  1. 2-digit × 1-digit, no carrying — the confidence builder
  2. 2-digit × 1-digit, with carrying — where regrouping enters
  3. 2-digit × 2-digit, no carrying — where the second partial product appears
  4. 2-digit × 2-digit, with carrying — the full standard algorithm

Every case is the same math — area of a rectangle, split into pieces. The difference between the cases is only whether a piece happens to be bigger than the column it lives in. When it is, we carry. Written by the same team that teaches Kid Einsteins at SOMATH, run by cofounders Marcelo Ambrozio (Northwestern) and Vivianne Wright (Harvard).

How to use this in a class (45–55 min):
  1. Warm-up: 3 quick 2×1 no-carry problems on the board (5 min).
  2. Work through the four theory sections aloud, drawing the box for each (15 min).
  3. Students attempt Q1–Q16 with answers hidden (15 min).
  4. Reveal answers together; students continue to Q17–Q30 (15 min).
  5. Wrap-up: pick 1 word problem, solve it as a class using the box method (5 min).

What’s in this lesson

  1. The big idea in one sentence
  2. Case 1 — 2-digit × 1-digit, no carrying
  3. Case 2 — 2-digit × 1-digit, with carrying
  4. Case 3 — 2-digit × 2-digit, no carrying
  5. Case 4 — 2-digit × 2-digit, with carrying
  6. Six mistakes that cost points on every test
  7. 30 practice questions (hidden answers)
  8. Answer key summary
  9. About SOMATH & Kid Einsteins

The big idea in one sentence

Every multi-digit multiplication is a stack of one-digit multiplications, kept in the right columns by place value. Carrying is what you do when one of those one-digit multiplications overflows its column.

That’s the whole lesson. Everything below is just how to keep things in the right columns.

Case 1 — 2-digit × 1-digit, no carrying

What to do

Stack the numbers so the ones digits line up. Multiply the ones column. Then multiply the tens column. Write each answer directly below its column. If both partial answers are one digit, you are done — no carrying is needed.

Example: 34 × 2

3 4 × 2 ----- 6 8 Ones: 4 × 2 = 8 → write 8 in the ones column Tens: 3 × 2 = 6 → write 6 in the tens column

Both partial answers (8 and 6) fit in a single column. No carrying. Answer: 68.

Rule of thumb: if the ones digit of the top number is 4 or less, and the multiplier is small (2 or 3), you will almost always avoid carrying. It’s a great starting point for building confidence.

Case 2 — 2-digit × 1-digit, with carrying

What to do

Same stack, same left-to-right idea — but this time a single-digit multiplication produces a two-digit result. The ones digit of that result stays in the current column; the tens digit is written above the next column to the left and added into that multiplication.

Example: 27 × 3

2 2 7 × 3 ----- 8 1 Ones: 7 × 3 = 21 → write 1 in the ones column, carry the 2 above the tens column. Tens: 2 × 3 = 6, then add the carry: 6 + 2 = 8 → write 8 in the tens column.

Answer: 81.

What the carry really is. 7 × 3 = 21 is really 20 + 1. The 1 belongs in the ones column. The 20 belongs one column to the left — that’s the tens column. Instead of writing “20” underneath, we just note the “2 tens” at the top of the tens column and add them into the next multiplication. Carrying is bookkeeping, not magic.

Order matters: multiply first, then add the carry. Students who do it in the wrong order (add carry, then multiply) get wrong answers every time.

Case 3 — 2-digit × 2-digit, no carrying

What to do

Now the multiplier has two digits, which means there will be two partial products. The first partial product is the top number times the ones digit of the bottom. The second partial product is the top number times the tens digit of the bottom — but the tens digit is really a multiple of 10, so the second partial product must be shifted one column to the left. We usually write a 0 in the ones column of the second line as a placeholder.

Example: 23 × 12

2 3 × 1 2 ----- 4 6 ← 23 × 2 (first partial product) 2 3 0 ← 23 × 10 (second partial product, shift left) ----- 2 7 6 Step 1: 23 × 2 = 46 → write 46 Step 2: 23 × 1 (of the tens) = 23, but that "1" is 10, so the real product is 23 × 10 = 230. Write 230, or equivalently write 23 shifted one column left. Step 3: Add the two partial products: 46 + 230 = 276

Answer: 276.

The shift is the whole point. The reason 2-digit × 2-digit is harder than 2-digit × 1-digit is not the multiplication — it’s the alignment. The 2 in 12 is not 2, it’s 20. Every wrong answer on a 2-digit × 2-digit problem where the multiplications themselves are correct is either (a) the student forgot to shift, or (b) the student forgot to add. Practice the shift until it becomes automatic.

Case 4 — 2-digit × 2-digit, with carrying

What to do

The full standard algorithm. Combine everything so far: two partial products, and each partial product may itself require carrying. Also — and this trips up almost every student — the carry from the first partial product must be erased or crossed out before starting the second partial product. Otherwise it gets added into the wrong multiplication.

Example: 47 × 26

4 4 7 × 2 6 ----- 2 8 2 ← 47 × 6 (first partial product, with carry) ----- Now erase the "4" carry — we're done with it. 1 4 7 × 2 6 ----- 2 8 2 9 4 0 ← 47 × 20 (second partial product, with new carry) ----- 1 2 2 2 First partial product (47 × 6): Ones: 7 × 6 = 42 → write 2, carry 4 Tens: 4 × 6 = 24, + carried 4 = 28 → write 8 in tens, 2 in hundreds First partial product = 282 Second partial product (47 × 20): Ones: put a 0 as the placeholder (we're multiplying by 20, not 2) Then compute 47 × 2: Ones (of 47) × 2: 7 × 2 = 14 → write 4, carry 1 Tens (of 47) × 2: 4 × 2 = 8, + carried 1 = 9 Second partial product = 940 Add: 282 + 940 ----- 1,222

Answer: 1,222.

Two carries, two purposes. The first carry (the 4) belongs to 47 × 6. The second carry (the 1) belongs to 47 × 2. They live in different multiplications; they never interact. Kids who mix them up write things like “1,262” instead of “1,222”. Cross out the first carry before starting the second partial product.

Six mistakes that cost points on every test

  1. Forgetting to add the carry. You wrote the “2” above the tens column of 27 × 3, then computed 2 × 3 = 6 and wrote 6 — forgetting to add the 2. Correct: 6 + 2 = 8.
  2. Adding the carry before multiplying. Some students do (2 + 2) × 3 = 12. Wrong. Multiply first, add carry after.
  3. Forgetting the 0 (shift) on the second line of 2×2. Writing 282 + 94 = 376 instead of 282 + 940 = 1,222. The zero is not decoration — it’s the difference between 2 and 20.
  4. Using the first carry inside the second partial product. After finishing 47 × 6, the “4” on top belongs to that multiplication. When you start 47 × 2, cross it out.
  5. Lining up the columns wrong when you add. The ones of the second partial product must sit under the tens of the first. If you write the shift as a placeholder 0, this alignment is automatic. If you skip the 0, you have to eyeball it — and eyeballing loses points.
  6. Multiplying only the ones digit of the top by the tens digit of the bottom. The top number is whole: you must multiply the entire top number by each digit of the bottom. So in 47 × 26, the second partial product is 47 × 2 = 94 (then shift), not 7 × 2 = 14.

30 practice questions

Answers are hidden. Click “Show answer & solution” to reveal each one. Questions 1–16 are 2-digit × 1-digit (with and without carrying). Questions 17–22 are 2-digit × 2-digit without carrying. Questions 23–26 are 2-digit × 2-digit with carrying. Questions 27–30 are word problems.

Part A — 2-digit × 1-digit, NO carrying (Q1–Q8)

Build confidence. Every partial product fits in one column.

Question 1 Easy

Compute 12 × 3.

Answer: 36

1 2 × 3 ----- 3 6 Ones: 2 × 3 = 6 Tens: 1 × 3 = 3

No carrying: both partial answers are one digit.

Question 2 Easy

Compute 21 × 4.

Answer: 84

2 1 × 4 ----- 8 4 Ones: 1 × 4 = 4 Tens: 2 × 4 = 8

Ones digit 1 makes this a no-carry problem for any single-digit multiplier.

Question 3 Easy

Compute 34 × 2.

Answer: 68

3 4 × 2 ----- 6 8 Ones: 4 × 2 = 8 Tens: 3 × 2 = 6

The example from the theory section. Because 4 × 2 = 8 (single digit), no carry.

Question 4 Easy

Compute 23 × 3.

Answer: 69

2 3 × 3 ----- 6 9 Ones: 3 × 3 = 9 Tens: 2 × 3 = 6

3 × 3 = 9, which just barely fits in one column. Any bigger and we’d be carrying.

Question 5 Easy

Compute 32 × 3.

Answer: 96

3 2 × 3 ----- 9 6 Ones: 2 × 3 = 6 Tens: 3 × 3 = 9

Both partial products are single digits. Notice the tens digit of the answer is exactly the top tens digit times the multiplier.

Question 6 Easy

Compute 43 × 2.

Answer: 86

4 3 × 2 ----- 8 6 Ones: 3 × 2 = 6 Tens: 4 × 2 = 8

No carry — the biggest partial product is 4 × 2 = 8, still single digit.

Question 7 Easy

Compute 24 × 2.

Answer: 48

2 4 × 2 ----- 4 8 Ones: 4 × 2 = 8 Tens: 2 × 2 = 4

Simple check: doubling 24 should give 48. It does.

Question 8 Easy

Compute 11 × 9.

Answer: 99

1 1 × 9 ----- 9 9 Ones: 1 × 9 = 9 Tens: 1 × 9 = 9

Even with a large multiplier (9), 1 × 9 = 9 stays in one column. Whenever both digits of the top number are 1, no carry is possible.

Part B — 2-digit × 1-digit, WITH carrying (Q9–Q16)

Now a single-digit multiplication overflows its column. Multiply first, then add the carry.

Question 9 Easy

Compute 27 × 3.

Answer: 81

2 2 7 × 3 ----- 8 1 Ones: 7 × 3 = 21 → write 1, carry 2. Tens: 2 × 3 = 6, + carry 2 = 8.

The example from the theory section. Note the carry is written above the tens column so we don’t forget to add it.

Question 10 Easy

Compute 46 × 2.

Answer: 92

1 4 6 × 2 ----- 9 2 Ones: 6 × 2 = 12 → write 2, carry 1. Tens: 4 × 2 = 8, + carry 1 = 9.

A small carry (1) — but skip it and you’d write 82, which is wrong by 10. That’s exactly what the carried 1 represents: one whole ten.

Question 11 Medium

Compute 38 × 4.

Answer: 152

3 3 8 × 4 ----- 1 5 2 Ones: 8 × 4 = 32 → write 2, carry 3. Tens: 3 × 4 = 12, + carry 3 = 15. → write 5 in tens, 1 in hundreds.

Here the tens multiplication itself gives a two-digit number (15), so the answer has a hundreds digit. The single-digit multiplier didn’t prevent a three-digit answer — that’s allowed.

Question 12 Medium

Compute 56 × 7.

Answer: 392

4 5 6 × 7 ----- 3 9 2 Ones: 6 × 7 = 42 → write 2, carry 4. Tens: 5 × 7 = 35, + carry 4 = 39. → write 9 in tens, 3 in hundreds.

Sanity check: 56 × 7 should be a little less than 60 × 7 = 420. 392 is 28 less than 420, and 4 × 7 = 28. Checks out.

Question 13 Medium

Compute 89 × 6.

Answer: 534

5 8 9 × 6 ----- 5 3 4 Ones: 9 × 6 = 54 → write 4, carry 5. Tens: 8 × 6 = 48, + carry 5 = 53. → write 3 in tens, 5 in hundreds.

Bigger carry (5). Sanity check: 89 × 6 ≈ 90 × 6 = 540, and we get 534, which is 6 less than 540 — exactly right because 1 × 6 = 6.

Question 14 Medium

Compute 47 × 8.

Answer: 376

5 4 7 × 8 ----- 3 7 6 Ones: 7 × 8 = 56 → write 6, carry 5. Tens: 4 × 8 = 32, + carry 5 = 37. → write 7 in tens, 3 in hundreds.

Both the ones and tens multiplications produce two-digit numbers, so we carry once and produce a three-digit answer.

Question 15 Medium

Compute 63 × 9.

Answer: 567

2 6 3 × 9 ----- 5 6 7 Ones: 3 × 9 = 27 → write 7, carry 2. Tens: 6 × 9 = 54, + carry 2 = 56. → write 6 in tens, 5 in hundreds.

63 × 9 = 63 × 10 − 63 = 630 − 63 = 567. Same answer, different route — a great mental-math cross-check.

Question 16 Medium

Compute 78 × 5.

Answer: 390

4 7 8 × 5 ----- 3 9 0 Ones: 8 × 5 = 40 → write 0, carry 4. Tens: 7 × 5 = 35, + carry 4 = 39. → write 9 in tens, 3 in hundreds.

Multiplying by 5 — the ones digit of the answer is always 0 or 5. Here we get 0 because 8 × 5 = 40 has zero ones.

Part C — 2-digit × 2-digit, NO carrying (Q17–Q22)

Now the multiplier has two digits, so there are two partial products. Shift the second one left.

Question 17 Medium

Compute 12 × 13.

Answer: 156

1 2 × 1 3 ----- 3 6 ← 12 × 3 1 2 0 ← 12 × 10 ----- 1 5 6 12 × 3 = 36 (no carry: 2×3=6, 1×3=3) 12 × 10 = 120 (shift left, add a 0) Total: 36 + 120 = 156

Classic starter 2×2 problem. Both partial products are one-carry-free, so the whole answer is arithmetic-clean.

Question 18 Medium

Compute 21 × 14.

Answer: 294

2 1 × 1 4 ----- 8 4 ← 21 × 4 2 1 0 ← 21 × 10 ----- 2 9 4 21 × 4 = 84 21 × 10 = 210 Total: 84 + 210 = 294

Both partial products are clean single-carry-free multiplications. Adding them is straightforward — no borrowing, no regrouping.

Question 19 Medium

Compute 31 × 22.

Answer: 682

3 1 × 2 2 ----- 6 2 ← 31 × 2 6 2 0 ← 31 × 20 ----- 6 8 2 31 × 2 = 62 31 × 20 = 620 Total: 62 + 620 = 682

Doubling twice: 31 × 2 = 62, and 62 × 11 = 682 (since 22 = 2 × 11). Or just: 31 × 22 = 682.

Question 20 Medium

Compute 23 × 12.

Answer: 276

2 3 × 1 2 ----- 4 6 ← 23 × 2 2 3 0 ← 23 × 10 ----- 2 7 6 23 × 2 = 46 23 × 10 = 230 Total: 46 + 230 = 276

The theory example. Note the placeholder 0 on the second line — it’s what makes the tens of 46 line up under the tens of 23.

Question 21 Medium

Compute 41 × 21.

Answer: 861

4 1 × 2 1 ----- 4 1 ← 41 × 1 8 2 0 ← 41 × 20 ----- 8 6 1 41 × 1 = 41 41 × 20 = 820 Total: 41 + 820 = 861

Multiplying by 1 is copying — the first partial product is just the top number. Notice how easy the addition becomes when both partial products are carry-free.

Question 22 Medium

Compute 22 × 34.

Answer: 748

2 2 × 3 4 ----- 8 8 ← 22 × 4 6 6 0 ← 22 × 30 ----- 7 4 8 22 × 4 = 88 22 × 30 = 660 Total: 88 + 660 = 748

Every digit of the top is 2, so both partial products have the same digit pattern (88 and 66). Beautiful test of the shift: 660 is 66 shifted one column left.

Part D — 2-digit × 2-digit, WITH carrying (Q23–Q26)

The full standard algorithm. Two partial products, each with its own carry. Cross out the first carry before starting the second.

Question 23 Hard

Compute 47 × 26.

Answer: 1,222

4 7 × 2 6 ----- 2 8 2 ← 47 × 6 (7×6=42 write 2 carry 4; 4×6=24+4=28) 9 4 0 ← 47 × 20 (7×2=14 write 4 carry 1; 4×2=8+1=9; then shift) ----- 1 2 2 2 First partial product 47 × 6 = 282 Second partial product 47 × 20 = 940 Sum: 282 + 940 = 1,222

The full theory-section example. The two carries (4 and 1) belong to different multiplications — don’t let them get confused.

Question 24 Hard

Compute 68 × 39.

Answer: 2,652

6 8 × 3 9 ----- 6 1 2 ← 68 × 9 (8×9=72 write 2 carry 7; 6×9=54+7=61) 2 0 4 0 ← 68 × 30 (8×3=24 write 4 carry 2; 6×3=18+2=20; then shift) ----- 2 6 5 2 First partial product 68 × 9 = 612 Second partial product 68 × 30 = 2,040 Sum: 612 + 2,040 = 2,652

Big carries (7 and 2). Sanity check: 68 × 39 is close to 70 × 40 = 2,800, and we’re at 2,652 — about 150 less. That’s reasonable because we overestimated both factors.

Question 25 Hard

Compute 84 × 57.

Answer: 4,788

8 4 × 5 7 ----- 5 8 8 ← 84 × 7 (4×7=28 write 8 carry 2; 8×7=56+2=58) 4 2 0 0 ← 84 × 50 (4×5=20 write 0 carry 2; 8×5=40+2=42; then shift) ----- 4 7 8 8 First partial product 84 × 7 = 588 Second partial product 84 × 50 = 4,200 Sum: 588 + 4,200 = 4,788

Notice the second partial product 84 × 50 = 4,200 has a zero from the ones-digit multiplication (4 × 5 = 20 → write 0 carry 2) and the placeholder zero from the shift. Two different zeros with two different meanings.

Question 26 Hard

Compute 96 × 48.

Answer: 4,608

9 6 × 4 8 ----- 7 6 8 ← 96 × 8 (6×8=48 write 8 carry 4; 9×8=72+4=76) 3 8 4 0 ← 96 × 40 (6×4=24 write 4 carry 2; 9×4=36+2=38; then shift) ----- 4 6 0 8 First partial product 96 × 8 = 768 Second partial product 96 × 40 = 3,840 Sum: 768 + 3,840 = 4,608

Both factors are close to 100 and 50, so 96 × 48 ≈ 100 × 50 = 5,000 — and we’re at 4,608. Perfect ballpark check. Always estimate before you multiply, then compare when you’re done.

Part E — Word problems (Q27–Q30)

Real-world uses of the four cases above. Set up the multiplication, then solve.

Question 27 Word

A pack of crayons contains 24 crayons. A kindergarten teacher orders 8 packs for her classroom. How many crayons will she receive in total?

Answer: 192 crayons

Setup: 24 × 8 3 2 4 × 8 ----- 1 9 2 Ones: 4 × 8 = 32 → write 2, carry 3. Tens: 2 × 8 = 16, + carry 3 = 19. → write 9 in tens, 1 in hundreds.

A 2-digit × 1-digit problem with carrying. Word problems that use “each” or “per” almost always translate to a multiplication.

Question 28 Word

A school orders 15 packs of pencils. Each pack contains 12 pencils. How many pencils total?

Answer: 180 pencils

Setup: 15 × 12 1 5 × 1 2 ----- 3 0 ← 15 × 2 1 5 0 ← 15 × 10 ----- 1 8 0 15 × 2 = 30 15 × 10 = 150 Total: 30 + 150 = 180

A 2-digit × 2-digit problem without carrying inside either partial product — a gentle intro to Case 3 in a real setting.

Question 29 Word

A field trip has 26 students. Each student needs to bring $17 for the museum entry fee. How much money should the class collect in total?

Answer: $442

Setup: 26 × 17 2 6 × 1 7 ----- 1 8 2 ← 26 × 7 (6×7=42 write 2 carry 4; 2×7=14+4=18) 2 6 0 ← 26 × 10 ----- 4 4 2 First partial product 26 × 7 = 182 Second partial product 26 × 10 = 260 Sum: 182 + 260 = 442

A 2-digit × 2-digit problem with carrying inside the first partial product but not the second (because the tens digit of 17 is 1). Real-world money problem — these show up on every 4th and 5th grade test.

Question 30 Word

A movie theater has 48 rows of seats. Each row has 32 seats. What is the total capacity of the theater?

Answer: 1,536 seats

Setup: 48 × 32 4 8 × 3 2 ----- 9 6 ← 48 × 2 (8×2=16 write 6 carry 1; 4×2=8+1=9) 1 4 4 0 ← 48 × 30 (8×3=24 write 4 carry 2; 4×3=12+2=14; then shift) ----- 1 5 3 6 First partial product 48 × 2 = 96 Second partial product 48 × 30 = 1,440 Sum: 96 + 1,440 = 1,536

The hardest word problem: a 2-digit × 2-digit multiplication with carrying in both partial products. Ballpark check: 50 × 30 = 1,500, and our answer 1,536 is right there. This is exactly the picture you use again in high school for “rectangle of area (a)(b)” problems in geometry and algebra.

Answer key summary

Q#AnswerQ#AnswerQ#Answer
1361115221861
2841239222748
36813534231,222
46914376242,652
59615567254,788
68616390264,608
7481715627192
8991829428180
9811968229442
109220276301,536

About SOMATH & Kid Einsteins

SOMATH — School of Math is a math-focused school on the Upper West Side of Manhattan for students in grades 1–12. The Kid Einsteins program (grades 3–5) drills multiplication and division fluency, fractions and decimals, area and perimeter, ratios, and structured word problems in small groups of 6–8 students. This is Class 5 of the Kid Einsteins arc — the natural sequel to Class 4 (Distributive Thinking & Partial Products).

Classes are taught by cofounder Marcelo Ambrozio (Northwestern-trained, 15+ years teaching math in NYC) and the SOMATH team.

Location: 226 W 79th St, 1st Floor, New York, NY 10024 (Upper West Side)
Phone: (646) 668-6151
Email: hello@schoolofmath.us

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