← 2-2 · Bound an inequality at equality · Pin Down a Number from Digit and Range Conditions

Bound an inequality at equality · 12 practice problems

3.OA.A.43.OA.C.7

Generated variants — 12

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

Variant 1 answer: 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$3\times9<4\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 4 times the box is larger than 3 times 9.

Givens

The inequality is 3 x 9 < 4 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 4 x box greater than 27.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 4 x box first passes 27; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 3 times 9 so the inequality becomes a comparison with a single number.

3 \times 9 = 27

3 times 9 is a basic multiplication fact, so the left side is just the number 27.

#6 Guess and Check 3.OA.A.4

Test multiples of 4 in order: 4x6 = 24 is not greater than 27, but 4x7 = 28 is greater than 27. So the box must be 7 or larger.

4 \times 6 = 24,\ 4 \times 7 = 28

Checking the 4-times facts shows exactly when the right side becomes bigger than 27.

#2 Make a Systematic List 3.OA.A.4

Every box value 7 and above (and at most 9) makes the right side larger than 27, so collect them.

\square \in \{7, 8, 9\}

Since bigger box values only make 4 x box larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

Review

Checking the smallest answer: 4x7 = 28 > 27, true; and the value just below, 4x6 = 24, is not greater than 27, so 7 is correctly the cutoff and 7, 8, 9 all work.

You could find where the two sides are equal: 27 divided by 4 is between 6 and 7, so the box must be more than 6, giving 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 4 x box greater than 27.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 3 x 9 and the 4-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 2 answer: 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$6\times9<8\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 8 times the box is larger than 6 times 9.

Givens

The inequality is 6 x 9 < 8 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 8 x box greater than 54.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 8 x box first passes 54; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 6 times 9 so the inequality becomes a comparison with a single number.

6 \times 9 = 54

6 times 9 is a basic multiplication fact, so the left side is just the number 54.

#6 Guess and Check 3.OA.A.4

Test multiples of 8 in order: 8x6 = 48 is not greater than 54, but 8x7 = 56 is greater than 54. So the box must be 7 or larger.

8 \times 6 = 48,\ 8 \times 7 = 56

Checking the 8-times facts shows exactly when the right side becomes bigger than 54.

#2 Make a Systematic List 3.OA.A.4

Every box value 7 and above (and at most 9) makes the right side larger than 54, so collect them.

\square \in \{7, 8, 9\}

Since bigger box values only make 8 x box larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

Review

Checking the smallest answer: 8x7 = 56 > 54, true; and the value just below, 8x6 = 48, is not greater than 54, so 7 is correctly the cutoff and 7, 8, 9 all work.

You could find where the two sides are equal: 54 divided by 8 is between 6 and 7, so the box must be more than 6, giving 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 8 x box greater than 54.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 6 x 9 and the 8-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 3 answer: 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$7\times7<9\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 9 times the box is larger than 7 times 7.

Givens

The inequality is 7 x 7 < 9 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x box greater than 49.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 9 x box first passes 49; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 7 times 7 so the inequality becomes a comparison with a single number.

7 \times 7 = 49

7 times 7 is a basic multiplication fact, so the left side is just the number 49.

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x5 = 45 is not greater than 49, but 9x6 = 54 is greater than 49. So the box must be 6 or larger.

9 \times 5 = 45,\ 9 \times 6 = 54

Checking the 9-times facts shows exactly when the right side becomes bigger than 49.

#2 Make a Systematic List 3.OA.A.4

Every box value 6 and above (and at most 9) makes the right side larger than 49, so collect them.

\square \in \{6, 7, 8, 9\}

Since bigger box values only make 9 x box larger, all of 6, 7, 8, 9 keep the inequality true.

Answer: 6, 7, 8, 9

Review

Checking the smallest answer: 9x6 = 54 > 49, true; and the value just below, 9x5 = 45, is not greater than 49, so 6 is correctly the cutoff and 6, 7, 8, 9 all work.

You could find where the two sides are equal: 49 divided by 9 is between 5 and 6, so the box must be more than 5, giving 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 9 x box greater than 49.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 7 x 7 and the 9-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 4 answer: 5, 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$5\times8<9\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 9 times the box is larger than 5 times 8.

Givens

The inequality is 5 x 8 < 9 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x box greater than 40.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 9 x box first passes 40; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 5 times 8 so the inequality becomes a comparison with a single number.

5 \times 8 = 40

5 times 8 is a basic multiplication fact, so the left side is just the number 40.

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x4 = 36 is not greater than 40, but 9x5 = 45 is greater than 40. So the box must be 5 or larger.

9 \times 4 = 36,\ 9 \times 5 = 45

Checking the 9-times facts shows exactly when the right side becomes bigger than 40.

#2 Make a Systematic List 3.OA.A.4

Every box value 5 and above (and at most 9) makes the right side larger than 40, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger box values only make 9 x box larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

Review

Checking the smallest answer: 9x5 = 45 > 40, true; and the value just below, 9x4 = 36, is not greater than 40, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

You could find where the two sides are equal: 40 divided by 9 is between 4 and 5, so the box must be more than 4, giving 5, 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 9 x box greater than 40.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 5 x 8 and the 9-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 5 answer: 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$4\times9<5\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 5 times the box is larger than 4 times 9.

Givens

The inequality is 4 x 9 < 5 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 5 x box greater than 36.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 5 x box first passes 36; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 4 times 9 so the inequality becomes a comparison with a single number.

4 \times 9 = 36

4 times 9 is a basic multiplication fact, so the left side is just the number 36.

#6 Guess and Check 3.OA.A.4

Test multiples of 5 in order: 5x7 = 35 is not greater than 36, but 5x8 = 40 is greater than 36. So the box must be 8 or larger.

5 \times 7 = 35,\ 5 \times 8 = 40

Checking the 5-times facts shows exactly when the right side becomes bigger than 36.

#2 Make a Systematic List 3.OA.A.4

Every box value 8 and above (and at most 9) makes the right side larger than 36, so collect them.

\square \in \{8, 9\}

Since bigger box values only make 5 x box larger, all of 8, 9 keep the inequality true.

Answer: 8, 9

Review

Checking the smallest answer: 5x8 = 40 > 36, true; and the value just below, 5x7 = 35, is not greater than 36, so 8 is correctly the cutoff and 8, 9 all work.

You could find where the two sides are equal: 36 divided by 5 is between 7 and 8, so the box must be more than 7, giving 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 5 x box greater than 36.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 9 and the 5-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 6 answer: 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$5\times9<7\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 7 times the box is larger than 5 times 9.

Givens

The inequality is 5 x 9 < 7 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 7 x box greater than 45.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 7 x box first passes 45; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 5 times 9 so the inequality becomes a comparison with a single number.

5 \times 9 = 45

5 times 9 is a basic multiplication fact, so the left side is just the number 45.

#6 Guess and Check 3.OA.A.4

Test multiples of 7 in order: 7x6 = 42 is not greater than 45, but 7x7 = 49 is greater than 45. So the box must be 7 or larger.

7 \times 6 = 42,\ 7 \times 7 = 49

Checking the 7-times facts shows exactly when the right side becomes bigger than 45.

#2 Make a Systematic List 3.OA.A.4

Every box value 7 and above (and at most 9) makes the right side larger than 45, so collect them.

\square \in \{7, 8, 9\}

Since bigger box values only make 7 x box larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

Review

Checking the smallest answer: 7x7 = 49 > 45, true; and the value just below, 7x6 = 42, is not greater than 45, so 7 is correctly the cutoff and 7, 8, 9 all work.

You could find where the two sides are equal: 45 divided by 7 is between 6 and 7, so the box must be more than 6, giving 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 7 x box greater than 45.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 5 x 9 and the 7-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 7 answer: 4, 5, 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$4\times7<8\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 8 times the box is larger than 4 times 7.

Givens

The inequality is 4 x 7 < 8 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 8 x box greater than 28.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 8 x box first passes 28; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 4 times 7 so the inequality becomes a comparison with a single number.

4 \times 7 = 28

4 times 7 is a basic multiplication fact, so the left side is just the number 28.

#6 Guess and Check 3.OA.A.4

Test multiples of 8 in order: 8x3 = 24 is not greater than 28, but 8x4 = 32 is greater than 28. So the box must be 4 or larger.

8 \times 3 = 24,\ 8 \times 4 = 32

Checking the 8-times facts shows exactly when the right side becomes bigger than 28.

#2 Make a Systematic List 3.OA.A.4

Every box value 4 and above (and at most 9) makes the right side larger than 28, so collect them.

\square \in \{4, 5, 6, 7, 8, 9\}

Since bigger box values only make 8 x box larger, all of 4, 5, 6, 7, 8, 9 keep the inequality true.

Answer: 4, 5, 6, 7, 8, 9

Review

Checking the smallest answer: 8x4 = 32 > 28, true; and the value just below, 8x3 = 24, is not greater than 28, so 4 is correctly the cutoff and 4, 5, 6, 7, 8, 9 all work.

You could find where the two sides are equal: 28 divided by 8 is between 3 and 4, so the box must be more than 3, giving 4, 5, 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 8 x box greater than 28.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 7 and the 8-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 8 answer: 5, 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$3\times8<6\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 6 times the box is larger than 3 times 8.

Givens

The inequality is 3 x 8 < 6 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 6 x box greater than 24.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 6 x box first passes 24; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 3 times 8 so the inequality becomes a comparison with a single number.

3 \times 8 = 24

3 times 8 is a basic multiplication fact, so the left side is just the number 24.

#6 Guess and Check 3.OA.A.4

Test multiples of 6 in order: 6x4 = 24 is not greater than 24, but 6x5 = 30 is greater than 24. So the box must be 5 or larger.

6 \times 4 = 24,\ 6 \times 5 = 30

Checking the 6-times facts shows exactly when the right side becomes bigger than 24.

#2 Make a Systematic List 3.OA.A.4

Every box value 5 and above (and at most 9) makes the right side larger than 24, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger box values only make 6 x box larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

Review

Checking the smallest answer: 6x5 = 30 > 24, true; and the value just below, 6x4 = 24, is not greater than 24, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

You could find where the two sides are equal: 24 divided by 6 is between 4 and 5, so the box must be more than 4, giving 5, 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 6 x box greater than 24.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 3 x 8 and the 6-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 9 answer: 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$7\times8<9\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 9 times the box is larger than 7 times 8.

Givens

The inequality is 7 x 8 < 9 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x box greater than 56.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 9 x box first passes 56; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 7 times 8 so the inequality becomes a comparison with a single number.

7 \times 8 = 56

7 times 8 is a basic multiplication fact, so the left side is just the number 56.

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x6 = 54 is not greater than 56, but 9x7 = 63 is greater than 56. So the box must be 7 or larger.

9 \times 6 = 54,\ 9 \times 7 = 63

Checking the 9-times facts shows exactly when the right side becomes bigger than 56.

#2 Make a Systematic List 3.OA.A.4

Every box value 7 and above (and at most 9) makes the right side larger than 56, so collect them.

\square \in \{7, 8, 9\}

Since bigger box values only make 9 x box larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

Review

Checking the smallest answer: 9x7 = 63 > 56, true; and the value just below, 9x6 = 54, is not greater than 56, so 7 is correctly the cutoff and 7, 8, 9 all work.

You could find where the two sides are equal: 56 divided by 9 is between 6 and 7, so the box must be more than 6, giving 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 9 x box greater than 56.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 7 x 8 and the 9-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 10 answer: 5, 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$6\times6<8\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 8 times the box is larger than 6 times 6.

Givens

The inequality is 6 x 6 < 8 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 8 x box greater than 36.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 8 x box first passes 36; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 6 times 6 so the inequality becomes a comparison with a single number.

6 \times 6 = 36

6 times 6 is a basic multiplication fact, so the left side is just the number 36.

#6 Guess and Check 3.OA.A.4

Test multiples of 8 in order: 8x4 = 32 is not greater than 36, but 8x5 = 40 is greater than 36. So the box must be 5 or larger.

8 \times 4 = 32,\ 8 \times 5 = 40

Checking the 8-times facts shows exactly when the right side becomes bigger than 36.

#2 Make a Systematic List 3.OA.A.4

Every box value 5 and above (and at most 9) makes the right side larger than 36, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger box values only make 8 x box larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

Review

Checking the smallest answer: 8x5 = 40 > 36, true; and the value just below, 8x4 = 32, is not greater than 36, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

You could find where the two sides are equal: 36 divided by 8 is between 4 and 5, so the box must be more than 4, giving 5, 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 8 x box greater than 36.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 6 x 6 and the 8-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 11 answer: 5, 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$8\times4<7\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 7 times the box is larger than 8 times 4.

Givens

The inequality is 8 x 4 < 7 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 7 x box greater than 32.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 7 x box first passes 32; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 8 times 4 so the inequality becomes a comparison with a single number.

8 \times 4 = 32

8 times 4 is a basic multiplication fact, so the left side is just the number 32.

#6 Guess and Check 3.OA.A.4

Test multiples of 7 in order: 7x4 = 28 is not greater than 32, but 7x5 = 35 is greater than 32. So the box must be 5 or larger.

7 \times 4 = 28,\ 7 \times 5 = 35

Checking the 7-times facts shows exactly when the right side becomes bigger than 32.

#2 Make a Systematic List 3.OA.A.4

Every box value 5 and above (and at most 9) makes the right side larger than 32, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger box values only make 7 x box larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

Review

Checking the smallest answer: 7x5 = 35 > 32, true; and the value just below, 7x4 = 28, is not greater than 32, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

You could find where the two sides are equal: 32 divided by 7 is between 4 and 5, so the box must be more than 4, giving 5, 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 7 x box greater than 32.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 8 x 4 and the 7-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 12 answer: 6, 7, 8, 9

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$6\times8<9\times\square$

Show solution

Understand

We need every digit from 1 to 9 that can fill the box so that 9 times the box is larger than 6 times 8.

Givens

The inequality is 6 x 8 < 9 x box.
The box must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x box greater than 48.

Constraints

The box is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

Plan

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the box values in order from small to large to find where 9 x box first passes 48; listing the candidates keeps every value from 1 to 9 accounted for.

Execute

#6 Guess and Check 3.OA.C.7

Work out 6 times 8 so the inequality becomes a comparison with a single number.

6 \times 8 = 48

6 times 8 is a basic multiplication fact, so the left side is just the number 48.

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x5 = 45 is not greater than 48, but 9x6 = 54 is greater than 48. So the box must be 6 or larger.

9 \times 5 = 45,\ 9 \times 6 = 54

Checking the 9-times facts shows exactly when the right side becomes bigger than 48.

#2 Make a Systematic List 3.OA.A.4

Every box value 6 and above (and at most 9) makes the right side larger than 48, so collect them.

\square \in \{6, 7, 8, 9\}

Since bigger box values only make 9 x box larger, all of 6, 7, 8, 9 keep the inequality true.

Answer: 6, 7, 8, 9

Review

Checking the smallest answer: 9x6 = 54 > 48, true; and the value just below, 9x5 = 45, is not greater than 48, so 6 is correctly the cutoff and 6, 7, 8, 9 all work.

You could find where the two sides are equal: 48 divided by 9 is between 5 and 6, so the box must be more than 5, giving 6, 7, 8, 9.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which box values keep 9 x box greater than 48.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 6 x 8 and the 9-times facts used in the comparison.

💡 Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!