Practice · Use neighbor differences to find rule · Sensim Math

Variant 1 easy answer: 51

Find the pattern in the number arrangement below, then find the first number in the sixth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the sixth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9.

Unknowns

The first number in the sixth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the sixth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the sixth row.

$2,\ 6,\ 10,\ 14,\ 18$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the sixth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 6.

1 + 2 + 6 + 10 + 14 + 18 = 51

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 6.

Answer: 51

4 · Reviewdoes it hold up?

Count check: Rows 1 through 5 hold 25 odd numbers in total, so Row 6 begins with the 26th odd number, which is 2 x 26 - 1 = 51. This matches, and 51 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 6 starts at odd number (5 x 5) + 1 = 26th, equal to 51.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the sixth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the sixth row starts at 51!

Variant 2 easy answer: 99

Find the pattern in the number arrangement below, then find the first number in the eighth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the eighth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9.

Unknowns

The first number in the eighth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the eighth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the eighth row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the eighth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 8.

1 + 2 + 6 + 10 + 14 + 18 + 22 + 26 = 99

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 8.

Answer: 99

4 · Reviewdoes it hold up?

Count check: Rows 1 through 7 hold 49 odd numbers in total, so Row 8 begins with the 50th odd number, which is 2 x 50 - 1 = 99. This matches, and 99 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 8 starts at odd number (7 x 7) + 1 = 50th, equal to 99.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the eighth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the eighth row starts at 99!

Variant 3 medium answer: 51

Find the pattern in the number arrangement below, then find the first number in the sixth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the sixth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the sixth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the sixth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the sixth row.

$2,\ 6,\ 10,\ 14,\ 18$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the sixth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 6.

1 + 2 + 6 + 10 + 14 + 18 = 51

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 6.

Answer: 51

4 · Reviewdoes it hold up?

Count check: Rows 1 through 5 hold 25 odd numbers in total, so Row 6 begins with the 26th odd number, which is 2 x 26 - 1 = 51. This matches, and 51 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 6 starts at odd number (5 x 5) + 1 = 26th, equal to 51.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the sixth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the sixth row starts at 51!

Variant 4 easy answer: 99

Find the pattern in the number arrangement below, then find the first number in the eighth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the eighth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the eighth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the eighth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the eighth row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the eighth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 8.

1 + 2 + 6 + 10 + 14 + 18 + 22 + 26 = 99

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 8.

Answer: 99

4 · Reviewdoes it hold up?

Count check: Rows 1 through 7 hold 49 odd numbers in total, so Row 8 begins with the 50th odd number, which is 2 x 50 - 1 = 99. This matches, and 99 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 8 starts at odd number (7 x 7) + 1 = 50th, equal to 99.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the eighth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the eighth row starts at 99!

Variant 5 medium answer: 33

Find the pattern in the number arrangement below, then find the first number in the fifth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the fifth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the fifth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the fifth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the fifth row.

$2,\ 6,\ 10,\ 14$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the fifth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 5.

1 + 2 + 6 + 10 + 14 = 33

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 5.

Answer: 33

4 · Reviewdoes it hold up?

Count check: Rows 1 through 4 hold 16 odd numbers in total, so Row 5 begins with the 17th odd number, which is 2 x 17 - 1 = 33. This matches, and 33 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 5 starts at odd number (4 x 4) + 1 = 17th, equal to 33.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the fifth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the fifth row starts at 33!

Variant 6 medium answer: 163

Find the pattern in the number arrangement below, then find the first number in the tenth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the tenth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the tenth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the tenth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the tenth row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26,\ 30,\ 34$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the tenth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 10.

1 + 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 = 163

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 10.

Answer: 163

4 · Reviewdoes it hold up?

Count check: Rows 1 through 9 hold 81 odd numbers in total, so Row 10 begins with the 82th odd number, which is 2 x 82 - 1 = 163. This matches, and 163 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 10 starts at odd number (9 x 9) + 1 = 82th, equal to 163.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the tenth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the tenth row starts at 163!

Variant 7 medium answer: 73

Find the pattern in the number arrangement below, then find the first number in the seventh row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the seventh row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the seventh row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the seventh row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the seventh row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the seventh row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 7.

1 + 2 + 6 + 10 + 14 + 18 + 22 = 73

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 7.

Answer: 73

4 · Reviewdoes it hold up?

Count check: Rows 1 through 6 hold 36 odd numbers in total, so Row 7 begins with the 37th odd number, which is 2 x 37 - 1 = 73. This matches, and 73 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 7 starts at odd number (6 x 6) + 1 = 37th, equal to 73.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the seventh row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the seventh row starts at 73!

Variant 8 hard answer: 201

Find the pattern in the number arrangement below, then find the first number in the eleventh row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the eleventh row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the eleventh row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the eleventh row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the eleventh row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26,\ 30,\ 34,\ 38$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the eleventh row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 11.

1 + 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38 = 201

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 11.

Answer: 201

4 · Reviewdoes it hold up?

Count check: Rows 1 through 10 hold 100 odd numbers in total, so Row 11 begins with the 101th odd number, which is 2 x 101 - 1 = 201. This matches, and 201 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 11 starts at odd number (10 x 10) + 1 = 101th, equal to 201.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the eleventh row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the eleventh row starts at 201!

Variant 9 hard answer: 243

Find the pattern in the number arrangement below, then find the first number in the twelfth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the twelfth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the twelfth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the twelfth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the twelfth row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26,\ 30,\ 34,\ 38,\ 42$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the twelfth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 12.

1 + 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38 + 42 = 243

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 12.

Answer: 243

4 · Reviewdoes it hold up?

Count check: Rows 1 through 11 hold 121 odd numbers in total, so Row 12 begins with the 122th odd number, which is 2 x 122 - 1 = 243. This matches, and 243 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 12 starts at odd number (11 x 11) + 1 = 122th, equal to 243.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the twelfth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the twelfth row starts at 243!

Variant 10 medium answer: 129

Find the pattern in the number arrangement below, then find the first number in the ninth row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the ninth row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19.

Unknowns

The first number in the ninth row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the ninth row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the ninth row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26,\ 30$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the ninth row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 9.

1 + 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 = 129

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 9.

Answer: 129

4 · Reviewdoes it hold up?

Count check: Rows 1 through 8 hold 64 odd numbers in total, so Row 9 begins with the 65th odd number, which is 2 x 65 - 1 = 129. This matches, and 129 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 9 starts at odd number (8 x 8) + 1 = 65th, equal to 129.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the ninth row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the ninth row starts at 129!

Variant 11 hard answer: 73

Find the pattern in the number arrangement below, then find the first number in the seventh row.

Row	Numbers
Row 1	$1$
Row 2	$3\ \ 5\ \ 7$
Row 3	$9\ \ 11\ \ 13\ \ 15\ \ 17$
Row 4	$19\ \ 21\ \ 23\ \ 25\ \ 27\ \ 29\ \ 31$
Row 5	$33\ \ 35\ \ 37\ \ 39\ \ 41\ \ 43\ \ 45\ \ 47\ \ 49$

Each row lists consecutive odd numbers: Row 1 has $1$ number, Row 2 has $3$ , Row 3 has $5$ , Row 4 has $7$ , Row 5 has $9$ . The count of numbers in each row increases by $2$ from one row to the next.

Show solution

1 · Understandwhat's really being asked

Consecutive odd numbers are written in rows. Row 1 holds 1 number, Row 2 holds 3 numbers, and each row holds 2 more numbers than the row above it. We must find the first number in the seventh row.

Givens

Row k holds 2k - 1 odd numbers (1, 3, 5, 7, ...).
Every entry is an odd number, listed in order with no odd numbers skipped.
The first numbers of the shown rows are 1, 3, 9, 19, 33.

Unknowns

The first number in the seventh row.

Constraints

Numbers are consecutive odd numbers (1, 3, 5, 7, ...) flowing row to row.
Row n contains 2n - 1 numbers.

2 · Planchoose the strategy

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

List the first number of each early row and the gaps between them. The gaps grow by a steady amount, which lets us extend the rule down to the seventh row without writing out every odd number.

3 · Execute3 carry out the plan

1List the first number of each row

#9 Solve an Easier Related Problem 4.OA.C.5

Read off the first entry of each given row.

1,\ 3,\ 9,\ 19,\ 33,\ \ldots

Just looking at the first number of each row turns a big table into a short list.

2Find how the first numbers grow

#5 Look for a Pattern 3.OA.D.9

Find the jump from one row's start to the next. Each jump is 4 more than the one before, because each new row adds 2 more odd numbers and skipping 2 odd numbers means moving ahead by 4. Continue the jumps down to the seventh row.

$2,\ 6,\ 10,\ 14,\ 18,\ 22$ (each is 4 more than the last)

The jump grows by 4 every step, a steady arithmetic pattern.

3Add the jumps up to the seventh row

#5 Look for a Pattern 4.OA.C.5

Start at Row 1's first number, 1, then add the jumps that lead into Rows 2 through 7.

1 + 2 + 6 + 10 + 14 + 18 + 22 = 73

Each jump carries us into the next row, so adding the jumps lands us on the start of Row 7.

Answer: 73

4 · Reviewdoes it hold up?

Count check: Rows 1 through 6 hold 36 odd numbers in total, so Row 7 begins with the 37th odd number, which is 2 x 37 - 1 = 73. This matches, and 73 is odd as required.

Another way: Use the count rule directly: the numbers before Row n total (n-1) x (n-1) odd numbers, so Row 7 starts at odd number (6 x 6) + 1 = 37th, equal to 73.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the row-start sequence down to the seventh row.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Spotting that the jumps between row-starts grow by 4 each time.

💡Takeaway. Track just the first number of each row and how far it jumps each time, then add the jumps: the seventh row starts at 73!

Use neighbor differences to find rule · 11 practice problems

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