Practice · Use neighbor differences to find rule · Sensim Math

Variant 1 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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 11 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

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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

Review

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.

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.

Standards · min 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.

💡 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!

Use neighbor differences to find rule · 11 practice problems

Generated variants — 11

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Standards · min grade 4

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Standards · min grade 4

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Standards · min grade 4

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Standards · min grade 4

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Standards · min grade 4

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Standards · min grade 4

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Standards · min grade 4