← Sum consecutive numbers via the middle · Sum of Evenly Spaced Numbers via the Middle

Sum consecutive numbers via the middle · 11 practice problems

3.OA.D.93.NBT.A.22.NBT.A.2

Generated variants — 11

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 25

Find the sum of the two consecutive whole numbers below.

12+1312 + 13

Show solution
1 · Understandwhat's really being asked

We must add two numbers that increase by 1 each time: 12, 13, and report their total.

Givens
  • The two consecutive whole numbers are 12, 13
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all two numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly two of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the first with the last, the second with the second-to-last, and so on. Each pair adds to the same total.
12+13=2512+13 = 25
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 2 numbers split into 1 pairs, and every pair equals 25. So the total is 1 copies of 25.
25×125 \times 1
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
25×1=2525 \times 1 = 25
A small multiplication finishes the job that many separate additions would have done.
Answer: 25
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 12.5, so the sum should be near 2 x 12.5 = 25. Our answer 25 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the two numbers is 12.5, and multiplying it by 2 gives 25, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 2 easy answer: 15

Find the sum of the five consecutive whole numbers below.

1+2+3+4+51 + 2 + 3 + 4 + 5

Show solution
1 · Understandwhat's really being asked

We must add five numbers that increase by 1 each time: 1, 2, 3, 4, 5, and report their total.

Givens
  • The five consecutive whole numbers are 1, 2, 3, 4, 5
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all five numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly five of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the smallest with the largest, working inward. Each pair adds to the same total, and the middle number 3 is left over.
1+5=6, 2+4=61+5 = 6,\ 2+4 = 6
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 5 numbers split into 2 equal pairs of 6, plus the leftover middle 3.
6×2+36 \times 2 + 3
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
6×2+3=12+3=156 \times 2 + 3 = 12 + 3 = 15
A small multiplication finishes the job that many separate additions would have done.
Answer: 15
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 3, so the sum should be near 5 x 3 = 15. Our answer 15 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the five numbers is 3, and multiplying it by 5 gives 15, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 3 easy answer: 46

Find the sum of the four consecutive whole numbers below.

10+11+12+1310 + 11 + 12 + 13

Show solution
1 · Understandwhat's really being asked

We must add four numbers that increase by 1 each time: 10, 11, 12, 13, and report their total.

Givens
  • The four consecutive whole numbers are 10, 11, 12, 13
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all four numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly four of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the first with the last, the second with the second-to-last, and so on. Each pair adds to the same total.
10+13=23, 11+12=2310+13 = 23,\ 11+12 = 23
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 4 numbers split into 2 pairs, and every pair equals 23. So the total is 2 copies of 23.
23×223 \times 2
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
23×2=4623 \times 2 = 46
A small multiplication finishes the job that many separate additions would have done.
Answer: 46
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 11.5, so the sum should be near 4 x 11.5 = 46. Our answer 46 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the four numbers is 11.5, and multiplying it by 4 gives 46, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 4 medium answer: 102

Find the sum of the three consecutive whole numbers below.

33+34+3533 + 34 + 35

Show solution
1 · Understandwhat's really being asked

We must add three numbers that increase by 1 each time: 33, 34, 35, and report their total.

Givens
  • The three consecutive whole numbers are 33, 34, 35
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all three numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly three of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the smallest with the largest, working inward. Each pair adds to the same total, and the middle number 34 is left over.
33+35=6833+35 = 68
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 3 numbers split into 1 equal pairs of 68, plus the leftover middle 34.
68×1+3468 \times 1 + 34
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
68×1+34=68+34=10268 \times 1 + 34 = 68 + 34 = 102
A small multiplication finishes the job that many separate additions would have done.
Answer: 102
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 34, so the sum should be near 3 x 34 = 102. Our answer 102 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the three numbers is 34, and multiplying it by 3 gives 102, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 5 medium answer: 385

Find the sum of the five consecutive whole numbers below.

75+76+77+78+7975 + 76 + 77 + 78 + 79

Show solution
1 · Understandwhat's really being asked

We must add five numbers that increase by 1 each time: 75, 76, 77, 78, 79, and report their total.

Givens
  • The five consecutive whole numbers are 75, 76, 77, 78, 79
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all five numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly five of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the smallest with the largest, working inward. Each pair adds to the same total, and the middle number 77 is left over.
75+79=154, 76+78=15475+79 = 154,\ 76+78 = 154
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 5 numbers split into 2 equal pairs of 154, plus the leftover middle 77.
154×2+77154 \times 2 + 77
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
154×2+77=308+77=385154 \times 2 + 77 = 308 + 77 = 385
A small multiplication finishes the job that many separate additions would have done.
Answer: 385
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 77, so the sum should be near 5 x 77 = 385. Our answer 385 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the five numbers is 77, and multiplying it by 5 gives 385, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 6 medium answer: 297

Find the sum of the six consecutive whole numbers below.

47+48+49+50+51+5247 + 48 + 49 + 50 + 51 + 52

Show solution
1 · Understandwhat's really being asked

We must add six numbers that increase by 1 each time: 47, 48, 49, 50, 51, 52, and report their total.

Givens
  • The six consecutive whole numbers are 47, 48, 49, 50, 51, 52
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all six numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly six of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the first with the last, the second with the second-to-last, and so on. Each pair adds to the same total.
47+52=99, 48+51=99, 49+50=9947+52 = 99,\ 48+51 = 99,\ 49+50 = 99
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 6 numbers split into 3 pairs, and every pair equals 99. So the total is 3 copies of 99.
99×399 \times 3
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
99×3=29799 \times 3 = 297
A small multiplication finishes the job that many separate additions would have done.
Answer: 297
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 49.5, so the sum should be near 6 x 49.5 = 297. Our answer 297 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the six numbers is 49.5, and multiplying it by 6 gives 297, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 7 medium answer: 721

Find the sum of the seven consecutive whole numbers below.

100+101+102+103+104+105+106100 + 101 + 102 + 103 + 104 + 105 + 106

Show solution
1 · Understandwhat's really being asked

We must add seven numbers that increase by 1 each time: 100, 101, 102, 103, 104, 105, 106, and report their total.

Givens
  • The seven consecutive whole numbers are 100, 101, 102, 103, 104, 105, 106
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all seven numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly seven of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the smallest with the largest, working inward. Each pair adds to the same total, and the middle number 103 is left over.
100+106=206, 101+105=206, 102+104=206100+106 = 206,\ 101+105 = 206,\ 102+104 = 206
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 7 numbers split into 3 equal pairs of 206, plus the leftover middle 103.
206×3+103206 \times 3 + 103
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
206×3+103=618+103=721206 \times 3 + 103 = 618 + 103 = 721
A small multiplication finishes the job that many separate additions would have done.
Answer: 721
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 103, so the sum should be near 7 x 103 = 721. Our answer 721 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the seven numbers is 103, and multiplying it by 7 gives 721, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 8 medium answer: 1521

Find the sum of the six consecutive whole numbers below.

251+252+253+254+255+256251 + 252 + 253 + 254 + 255 + 256

Show solution
1 · Understandwhat's really being asked

We must add six numbers that increase by 1 each time: 251, 252, 253, 254, 255, 256, and report their total.

Givens
  • The six consecutive whole numbers are 251, 252, 253, 254, 255, 256
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all six numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly six of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the first with the last, the second with the second-to-last, and so on. Each pair adds to the same total.
251+256=507, 252+255=507, 253+254=507251+256 = 507,\ 252+255 = 507,\ 253+254 = 507
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 6 numbers split into 3 pairs, and every pair equals 507. So the total is 3 copies of 507.
507×3507 \times 3
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
507×3=1521507 \times 3 = 1521
A small multiplication finishes the job that many separate additions would have done.
Answer: 1521
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 253.5, so the sum should be near 6 x 253.5 = 1521. Our answer 1521 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the six numbers is 253.5, and multiplying it by 6 gives 1521, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 9 hard answer: 1628

Find the sum of the eight consecutive whole numbers below.

200+201+202+203+204+205+206+207200 + 201 + 202 + 203 + 204 + 205 + 206 + 207

Show solution
1 · Understandwhat's really being asked

We must add eight numbers that increase by 1 each time: 200, 201, 202, 203, 204, 205, 206, 207, and report their total.

Givens
  • The eight consecutive whole numbers are 200, 201, 202, 203, 204, 205, 206, 207
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all eight numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly eight of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the first with the last, the second with the second-to-last, and so on. Each pair adds to the same total.
200+207=407, 201+206=407, 202+205=407, 203+204=407200+207 = 407,\ 201+206 = 407,\ 202+205 = 407,\ 203+204 = 407
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 8 numbers split into 4 pairs, and every pair equals 407. So the total is 4 copies of 407.
407×4407 \times 4
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
407×4=1628407 \times 4 = 1628
A small multiplication finishes the job that many separate additions would have done.
Answer: 1628
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 203.5, so the sum should be near 8 x 203.5 = 1628. Our answer 1628 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the eight numbers is 203.5, and multiplying it by 8 gives 1628, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 10 hard answer: 1055

Find the sum of the 10 consecutive whole numbers below.

101+102+103+104+105+106+107+108+109+110101 + 102 + 103 + 104 + 105 + 106 + 107 + 108 + 109 + 110

Show solution
1 · Understandwhat's really being asked

We must add 10 numbers that increase by 1 each time: 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, and report their total.

Givens
  • The 10 consecutive whole numbers are 101, 102, 103, 104, 105, 106, 107, 108, 109, 110
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all 10 numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly 10 of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the first with the last, the second with the second-to-last, and so on. Each pair adds to the same total.
101+110=211, 102+109=211, 103+108=211, 104+107=211, 105+106=211101+110 = 211,\ 102+109 = 211,\ 103+108 = 211,\ 104+107 = 211,\ 105+106 = 211
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 10 numbers split into 5 pairs, and every pair equals 211. So the total is 5 copies of 211.
211×5211 \times 5
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
211×5=1055211 \times 5 = 1055
A small multiplication finishes the job that many separate additions would have done.
Answer: 1055
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 105.5, so the sum should be near 10 x 105.5 = 1055. Our answer 1055 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the 10 numbers is 105.5, and multiplying it by 10 gives 1055, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!
Variant 11 hard answer: 4536

Find the sum of the nine consecutive whole numbers below.

500+501+502+503+504+505+506+507+508500 + 501 + 502 + 503 + 504 + 505 + 506 + 507 + 508

Show solution
1 · Understandwhat's really being asked

We must add nine numbers that increase by 1 each time: 500, 501, 502, 503, 504, 505, 506, 507, 508, and report their total.

Givens
  • The nine consecutive whole numbers are 500, 501, 502, 503, 504, 505, 506, 507, 508
  • Each number is exactly 1 more than the one before it
Unknowns
  • The sum of all nine numbers
Constraints
  • The numbers are consecutive (step of 1)
  • There are exactly nine of them
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems

Consecutive numbers have a balancing pattern: pairing the smallest with the largest gives equal partial sums. Spotting this turns many additions into a simple multiplication.

3 · Execute3 carry out the plan

1Pair from the outside in

#5 Look for a Pattern 3.OA.D.9
Pair the smallest with the largest, working inward. Each pair adds to the same total, and the middle number 504 is left over.
500+508=1008, 501+507=1008, 502+506=1008, 503+505=1008500+508 = 1008,\ 501+507 = 1008,\ 502+506 = 1008,\ 503+505 = 1008
When numbers go up by 1, what you add to one end you take from the other, so the outside-in pairs stay equal.

2Count the equal pairs

#7 Identify Subproblems 3.OA.D.9
The 9 numbers split into 4 equal pairs of 1008, plus the leftover middle 504.
1008×4+5041008 \times 4 + 504
Turning repeated equal sums into a multiplication is faster and less error-prone than adding one by one.

3Add it all up

#7 Identify Subproblems 3.NBT.A.2
Finish the arithmetic to get the final total.
1008×4+504=4032+504=45361008 \times 4 + 504 = 4032 + 504 = 4536
A small multiplication finishes the job that many separate additions would have done.
Answer: 4536
4 · Reviewdoes it hold up?

The numbers cluster around their middle value 504, so the sum should be near 9 x 504 = 4536. Our answer 4536 matches exactly, so the magnitude is sensible.

Another way: Use the middle-value idea (tool 9, easier related problem): the average of the nine numbers is 504, and multiplying it by 9 gives 4536, confirming the pairing result.

Standardsmin grade 3
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing that outside-in pairs of consecutive numbers are equal
  • 3.NBT.A.2 Fluently add and subtract within 1000 — Adding each pair and the final total
  • 2.NBT.A.2 Count within 1000, skip-count by 5s, 10s, and 100s — Recognizing the consecutive counting pattern of the numbers
💡Takeaway. Pair the ends of consecutive numbers - they all match, so you just multiply one pair by how many pairs there are!