← Find the rule linking position to term · Generalize a Growing Pattern into a Rule

Find the rule linking position to term · 12 practice problems

4.OA.C.55.OA.B.35.OA.A.2

Generated variants — 12

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

Variant 1 easy answer: 97

The numbers below are arranged according to a rule. Find the 2020th number.

2, 7, 12, 17, 22, ⋯2,\ 7,\ 12,\ 17,\ 22,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 2 and each number after that is bigger than the one before by the same amount. We need the 20th number in the list.

Givens
  • The list begins 2, 7, 12, 17, 22, and it keeps going the same way.
  • We want the number sitting in position 20.
Unknowns
  • The value of the 20th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 20th number would take too long, so find the size of each jump, count how many jumps happen before position 20, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 2, 7, 12, 17, 22 are all the same.
7−2=5,12−7=57 - 2 = 5,\quad 12 - 7 = 5
Every step forward adds 5, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 20th number takes one fewer jump than its position.
20−1=19 jumps20 - 1 = 19\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 2, then add 5 once for every jump. That turns position into value in a single step.
value=2+(position−1)×5\text{value} = 2 + (\text{position} - 1) \times 5
One expression replaces writing out the whole list.

4Put 20 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 19 jumps of 5 from the starting number 2.
19×5=95,2+95=9719 \times 5 = 95,\quad 2 + 95 = 97
The 20th number is 97.
Answer: 97
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 2 + 2 x 5 = 12, which matches the list. So position 20 giving 97 is trustworthy.

Another way: Notice every number is 5 times its position plus -3: 5 x 20 + (-3) = 97, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 5 each time and extending it to position 20.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 2 easy answer: 148

The numbers below are arranged according to a rule. Find the 2525th number.

4, 10, 16, 22, 28, ⋯4,\ 10,\ 16,\ 22,\ 28,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 4 and each number after that is bigger than the one before by the same amount. We need the 25th number in the list.

Givens
  • The list begins 4, 10, 16, 22, 28, and it keeps going the same way.
  • We want the number sitting in position 25.
Unknowns
  • The value of the 25th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 25th number would take too long, so find the size of each jump, count how many jumps happen before position 25, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 4, 10, 16, 22, 28 are all the same.
10−4=6,16−10=610 - 4 = 6,\quad 16 - 10 = 6
Every step forward adds 6, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 25th number takes one fewer jump than its position.
25−1=24 jumps25 - 1 = 24\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 4, then add 6 once for every jump. That turns position into value in a single step.
value=4+(position−1)×6\text{value} = 4 + (\text{position} - 1) \times 6
One expression replaces writing out the whole list.

4Put 25 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 24 jumps of 6 from the starting number 4.
24×6=144,4+144=14824 \times 6 = 144,\quad 4 + 144 = 148
The 25th number is 148.
Answer: 148
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 4 + 2 x 6 = 16, which matches the list. So position 25 giving 148 is trustworthy.

Another way: Notice every number is 6 times its position plus -2: 6 x 25 + (-2) = 148, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 6 each time and extending it to position 25.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 3 easy answer: 88

The numbers below are arranged according to a rule. Find the 3030th number.

1, 4, 7, 10, 13, ⋯1,\ 4,\ 7,\ 10,\ 13,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 1 and each number after that is bigger than the one before by the same amount. We need the 30th number in the list.

Givens
  • The list begins 1, 4, 7, 10, 13, and it keeps going the same way.
  • We want the number sitting in position 30.
Unknowns
  • The value of the 30th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 30th number would take too long, so find the size of each jump, count how many jumps happen before position 30, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 1, 4, 7, 10, 13 are all the same.
4−1=3,7−4=34 - 1 = 3,\quad 7 - 4 = 3
Every step forward adds 3, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 30th number takes one fewer jump than its position.
30−1=29 jumps30 - 1 = 29\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 1, then add 3 once for every jump. That turns position into value in a single step.
value=1+(position−1)×3\text{value} = 1 + (\text{position} - 1) \times 3
One expression replaces writing out the whole list.

4Put 30 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 29 jumps of 3 from the starting number 1.
29×3=87,1+87=8829 \times 3 = 87,\quad 1 + 87 = 88
The 30th number is 88.
Answer: 88
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 1 + 2 x 3 = 7, which matches the list. So position 30 giving 88 is trustworthy.

Another way: Notice every number is 3 times its position plus -2: 3 x 30 + (-2) = 88, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 3 each time and extending it to position 30.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 4 easy answer: 101

The numbers below are arranged according to a rule. Find the 1515th number.

3, 10, 17, 24, 31, ⋯3,\ 10,\ 17,\ 24,\ 31,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 3 and each number after that is bigger than the one before by the same amount. We need the 15th number in the list.

Givens
  • The list begins 3, 10, 17, 24, 31, and it keeps going the same way.
  • We want the number sitting in position 15.
Unknowns
  • The value of the 15th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 15th number would take too long, so find the size of each jump, count how many jumps happen before position 15, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 3, 10, 17, 24, 31 are all the same.
10−3=7,17−10=710 - 3 = 7,\quad 17 - 10 = 7
Every step forward adds 7, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 15th number takes one fewer jump than its position.
15−1=14 jumps15 - 1 = 14\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 3, then add 7 once for every jump. That turns position into value in a single step.
value=3+(position−1)×7\text{value} = 3 + (\text{position} - 1) \times 7
One expression replaces writing out the whole list.

4Put 15 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 14 jumps of 7 from the starting number 3.
14×7=98,3+98=10114 \times 7 = 98,\quad 3 + 98 = 101
The 15th number is 101.
Answer: 101
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 3 + 2 x 7 = 17, which matches the list. So position 15 giving 101 is trustworthy.

Another way: Notice every number is 7 times its position plus -4: 7 x 15 + (-4) = 101, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 7 each time and extending it to position 15.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 5 medium answer: 135

The numbers below are arranged according to a rule. Find the 2222th number.

9, 15, 21, 27, 33, ⋯9,\ 15,\ 21,\ 27,\ 33,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 9 and each number after that is bigger than the one before by the same amount. We need the 22nd number in the list.

Givens
  • The list begins 9, 15, 21, 27, 33, and it keeps going the same way.
  • We want the number sitting in position 22.
Unknowns
  • The value of the 22nd number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 22nd number would take too long, so find the size of each jump, count how many jumps happen before position 22, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 9, 15, 21, 27, 33 are all the same.
15−9=6,21−15=615 - 9 = 6,\quad 21 - 15 = 6
Every step forward adds 6, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 22nd number takes one fewer jump than its position.
22−1=21 jumps22 - 1 = 21\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 9, then add 6 once for every jump. That turns position into value in a single step.
value=9+(position−1)×6\text{value} = 9 + (\text{position} - 1) \times 6
One expression replaces writing out the whole list.

4Put 22 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 21 jumps of 6 from the starting number 9.
21×6=126,9+126=13521 \times 6 = 126,\quad 9 + 126 = 135
The 22nd number is 135.
Answer: 135
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 9 + 2 x 6 = 21, which matches the list. So position 22 giving 135 is trustworthy.

Another way: Notice every number is 6 times its position plus 3: 6 x 22 + (3) = 135, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 6 each time and extending it to position 22.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 6 medium answer: 166

The numbers below are arranged according to a rule. Find the 3333th number.

6, 11, 16, 21, 26, ⋯6,\ 11,\ 16,\ 21,\ 26,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 6 and each number after that is bigger than the one before by the same amount. We need the 33rd number in the list.

Givens
  • The list begins 6, 11, 16, 21, 26, and it keeps going the same way.
  • We want the number sitting in position 33.
Unknowns
  • The value of the 33rd number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 33rd number would take too long, so find the size of each jump, count how many jumps happen before position 33, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 6, 11, 16, 21, 26 are all the same.
11−6=5,16−11=511 - 6 = 5,\quad 16 - 11 = 5
Every step forward adds 5, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 33rd number takes one fewer jump than its position.
33−1=32 jumps33 - 1 = 32\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 6, then add 5 once for every jump. That turns position into value in a single step.
value=6+(position−1)×5\text{value} = 6 + (\text{position} - 1) \times 5
One expression replaces writing out the whole list.

4Put 33 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 32 jumps of 5 from the starting number 6.
32×5=160,6+160=16632 \times 5 = 160,\quad 6 + 160 = 166
The 33rd number is 166.
Answer: 166
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 6 + 2 x 5 = 16, which matches the list. So position 33 giving 166 is trustworthy.

Another way: Notice every number is 5 times its position plus 1: 5 x 33 + (1) = 166, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 5 each time and extending it to position 33.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 7 medium answer: 138

The numbers below are arranged according to a rule. Find the 1818th number.

2, 10, 18, 26, 34, ⋯2,\ 10,\ 18,\ 26,\ 34,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 2 and each number after that is bigger than the one before by the same amount. We need the 18th number in the list.

Givens
  • The list begins 2, 10, 18, 26, 34, and it keeps going the same way.
  • We want the number sitting in position 18.
Unknowns
  • The value of the 18th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 18th number would take too long, so find the size of each jump, count how many jumps happen before position 18, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 2, 10, 18, 26, 34 are all the same.
10−2=8,18−10=810 - 2 = 8,\quad 18 - 10 = 8
Every step forward adds 8, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 18th number takes one fewer jump than its position.
18−1=17 jumps18 - 1 = 17\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 2, then add 8 once for every jump. That turns position into value in a single step.
value=2+(position−1)×8\text{value} = 2 + (\text{position} - 1) \times 8
One expression replaces writing out the whole list.

4Put 18 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 17 jumps of 8 from the starting number 2.
17×8=136,2+136=13817 \times 8 = 136,\quad 2 + 136 = 138
The 18th number is 138.
Answer: 138
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 2 + 2 x 8 = 18, which matches the list. So position 18 giving 138 is trustworthy.

Another way: Notice every number is 8 times its position plus -6: 8 x 18 + (-6) = 138, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 8 each time and extending it to position 18.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 8 medium answer: 197

The numbers below are arranged according to a rule. Find the 2828th number.

8, 15, 22, 29, 36, ⋯8,\ 15,\ 22,\ 29,\ 36,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 8 and each number after that is bigger than the one before by the same amount. We need the 28th number in the list.

Givens
  • The list begins 8, 15, 22, 29, 36, and it keeps going the same way.
  • We want the number sitting in position 28.
Unknowns
  • The value of the 28th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 28th number would take too long, so find the size of each jump, count how many jumps happen before position 28, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 8, 15, 22, 29, 36 are all the same.
15−8=7,22−15=715 - 8 = 7,\quad 22 - 15 = 7
Every step forward adds 7, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 28th number takes one fewer jump than its position.
28−1=27 jumps28 - 1 = 27\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 8, then add 7 once for every jump. That turns position into value in a single step.
value=8+(position−1)×7\text{value} = 8 + (\text{position} - 1) \times 7
One expression replaces writing out the whole list.

4Put 28 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 27 jumps of 7 from the starting number 8.
27×7=189,8+189=19727 \times 7 = 189,\quad 8 + 189 = 197
The 28th number is 197.
Answer: 197
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 8 + 2 x 7 = 22, which matches the list. So position 28 giving 197 is trustworthy.

Another way: Notice every number is 7 times its position plus 1: 7 x 28 + (1) = 197, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 7 each time and extending it to position 28.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 9 hard answer: 161

The numbers below are arranged according to a rule. Find the 4040th number.

5, 9, 13, 17, 21, ⋯5,\ 9,\ 13,\ 17,\ 21,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 5 and each number after that is bigger than the one before by the same amount. We need the 40th number in the list.

Givens
  • The list begins 5, 9, 13, 17, 21, and it keeps going the same way.
  • We want the number sitting in position 40.
Unknowns
  • The value of the 40th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 40th number would take too long, so find the size of each jump, count how many jumps happen before position 40, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 5, 9, 13, 17, 21 are all the same.
9−5=4,13−9=49 - 5 = 4,\quad 13 - 9 = 4
Every step forward adds 4, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 40th number takes one fewer jump than its position.
40−1=39 jumps40 - 1 = 39\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 5, then add 4 once for every jump. That turns position into value in a single step.
value=5+(position−1)×4\text{value} = 5 + (\text{position} - 1) \times 4
One expression replaces writing out the whole list.

4Put 40 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 39 jumps of 4 from the starting number 5.
39×4=156,5+156=16139 \times 4 = 156,\quad 5 + 156 = 161
The 40th number is 161.
Answer: 161
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 5 + 2 x 4 = 13, which matches the list. So position 40 giving 161 is trustworthy.

Another way: Notice every number is 4 times its position plus 1: 4 x 40 + (1) = 161, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 4 each time and extending it to position 40.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 10 hard answer: 106

The numbers below are arranged according to a rule. Find the 1212th number.

7, 16, 25, 34, 43, ⋯7,\ 16,\ 25,\ 34,\ 43,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 7 and each number after that is bigger than the one before by the same amount. We need the 12th number in the list.

Givens
  • The list begins 7, 16, 25, 34, 43, and it keeps going the same way.
  • We want the number sitting in position 12.
Unknowns
  • The value of the 12th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 12th number would take too long, so find the size of each jump, count how many jumps happen before position 12, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 7, 16, 25, 34, 43 are all the same.
16−7=9,25−16=916 - 7 = 9,\quad 25 - 16 = 9
Every step forward adds 9, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 12th number takes one fewer jump than its position.
12−1=11 jumps12 - 1 = 11\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 7, then add 9 once for every jump. That turns position into value in a single step.
value=7+(position−1)×9\text{value} = 7 + (\text{position} - 1) \times 9
One expression replaces writing out the whole list.

4Put 12 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 11 jumps of 9 from the starting number 7.
11×9=99,7+99=10611 \times 9 = 99,\quad 7 + 99 = 106
The 12th number is 106.
Answer: 106
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 7 + 2 x 9 = 25, which matches the list. So position 12 giving 106 is trustworthy.

Another way: Notice every number is 9 times its position plus -2: 9 x 12 + (-2) = 106, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 9 each time and extending it to position 12.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 11 hard answer: 144

The numbers below are arranged according to a rule. Find the 1414th number.

1, 12, 23, 34, 45, ⋯1,\ 12,\ 23,\ 34,\ 45,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 1 and each number after that is bigger than the one before by the same amount. We need the 14th number in the list.

Givens
  • The list begins 1, 12, 23, 34, 45, and it keeps going the same way.
  • We want the number sitting in position 14.
Unknowns
  • The value of the 14th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 14th number would take too long, so find the size of each jump, count how many jumps happen before position 14, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 1, 12, 23, 34, 45 are all the same.
12−1=11,23−12=1112 - 1 = 11,\quad 23 - 12 = 11
Every step forward adds 11, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 14th number takes one fewer jump than its position.
14−1=13 jumps14 - 1 = 13\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 1, then add 11 once for every jump. That turns position into value in a single step.
value=1+(position−1)×11\text{value} = 1 + (\text{position} - 1) \times 11
One expression replaces writing out the whole list.

4Put 14 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 13 jumps of 11 from the starting number 1.
13×11=143,1+143=14413 \times 11 = 143,\quad 1 + 143 = 144
The 14th number is 144.
Answer: 144
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 1 + 2 x 11 = 23, which matches the list. So position 14 giving 144 is trustworthy.

Another way: Notice every number is 11 times its position plus -10: 11 x 14 + (-10) = 144, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 11 each time and extending it to position 14.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.
Variant 12 hard answer: 183

The numbers below are arranged according to a rule. Find the 1616th number.

3, 15, 27, 39, 51, ⋯3,\ 15,\ 27,\ 39,\ 51,\ \cdots

Show solution
1 · Understandwhat's really being asked

A list starts at 3 and each number after that is bigger than the one before by the same amount. We need the 16th number in the list.

Givens
  • The list begins 3, 15, 27, 39, 51, and it keeps going the same way.
  • We want the number sitting in position 16.
Unknowns
  • The value of the 16th number.
Constraints
  • The same amount is added at every step, so the jumps never change.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

Writing out to the 16th number would take too long, so find the size of each jump, count how many jumps happen before position 16, and turn that into one calculation.

3 · Execute4 carry out the plan

1Find the size of each jump

#5 Look for a Pattern 4.OA.C.5
Subtract each number from the next one: the gaps between 3, 15, 27, 39, 51 are all the same.
15−3=12,27−15=1215 - 3 = 12,\quad 27 - 15 = 12
Every step forward adds 12, so the list climbs at a steady rate.

2Count the jumps, not the numbers

#9 Solve an Easier Related Problem 5.OA.B.3
The 1st number needs no jump, the 2nd needs one, the 3rd needs two. Reaching the 16th number takes one fewer jump than its position.
16−1=15 jumps16 - 1 = 15\ \text{jumps}
Fence posts and gaps: the first number is where you start, not a jump you take.

3Write the rule as one expression

#5 Look for a Pattern 5.OA.A.2
Start at 3, then add 12 once for every jump. That turns position into value in a single step.
value=3+(position−1)×12\text{value} = 3 + (\text{position} - 1) \times 12
One expression replaces writing out the whole list.

4Put 16 into the rule

#5 Look for a Pattern 4.OA.C.5
Take 15 jumps of 12 from the starting number 3.
15×12=180,3+180=18315 \times 12 = 180,\quad 3 + 180 = 183
The 16th number is 183.
Answer: 183
4 · Reviewdoes it hold up?

Test the rule on a number we can see: position 3 gives 3 + 2 x 12 = 27, which matches the list. So position 16 giving 183 is trustworthy.

Another way: Notice every number is 12 times its position plus -9: 12 x 16 + (-9) = 183, the same answer.

Standardsmin grade 5
  • 4.OA.C.5 Generate a number pattern that follows a given rule and identify features of the pattern — Spotting that the list grows by 12 each time and extending it to position 16.
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships between terms — Relating the position number to the count of jumps taken.
  • 5.OA.A.2 Write simple expressions that record calculations with numbers — Recording the position-to-value rule as a single expression.
💡Takeaway. Count the jumps between numbers, not the numbers themselves — reaching the 30th spot takes only 29 jumps.