Practice · Compare numbers with an unknown digit · Sensim Math

Variant 1 easy

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$4\,\square\,8\,9 < 4789 \qquad 5\,3\,\square\,1 > 5341$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $4\,\square\,8\,9 < 4789$ ( $\square$ is the hundreds digit).
Second inequality: $5\,3\,\square\,1 > 5341$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 6. Allowed digits: 0, 1, 2, 3, 4, 5, 6.

4\,\square\,8\,9 < 4789 \Rightarrow \square \in \{0,1,2,3,4,5,6\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 5 to 9. Allowed digits: 5, 6, 7, 8, 9.

5\,3\,\square\,1 > 5341 \Rightarrow \square \in \{5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5,6} and the second allows {5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5,6\} \cap \{5,6,7,8,9\} = \{5,6\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 5, 6

4 · Reviewdoes it hold up?

Check 5 directly: 4589 < 4789 is true and 5351 > 5341 is true, so 5 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 2 easy

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$5\,\square\,2\,7 < 5832 \qquad 3\,9\,\square\,4 > 3945$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $5\,\square\,2\,7 < 5832$ ( $\square$ is the hundreds digit).
Second inequality: $3\,9\,\square\,4 > 3945$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 8. Allowed digits: 0, 1, 2, 3, 4, 5, 6, 7, 8.

5\,\square\,2\,7 < 5832 \Rightarrow \square \in \{0,1,2,3,4,5,6,7,8\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 5 to 9. Allowed digits: 5, 6, 7, 8, 9.

3\,9\,\square\,4 > 3945 \Rightarrow \square \in \{5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5,6,7,8} and the second allows {5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5,6,7,8\} \cap \{5,6,7,8,9\} = \{5,6,7,8\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 5, 6, 7, 8

4 · Reviewdoes it hold up?

Check 5 directly: 5527 < 5832 is true and 3954 > 3945 is true, so 5 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 3 easy

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$6\,\square\,0\,5 < 6805 \qquad 4\,7\,\square\,9 > 4749$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $6\,\square\,0\,5 < 6805$ ( $\square$ is the hundreds digit).
Second inequality: $4\,7\,\square\,9 > 4749$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 7. Allowed digits: 0, 1, 2, 3, 4, 5, 6, 7.

6\,\square\,0\,5 < 6805 \Rightarrow \square \in \{0,1,2,3,4,5,6,7\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 5 to 9. Allowed digits: 5, 6, 7, 8, 9.

4\,7\,\square\,9 > 4749 \Rightarrow \square \in \{5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5,6,7} and the second allows {5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5,6,7\} \cap \{5,6,7,8,9\} = \{5,6,7\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 5, 6, 7

4 · Reviewdoes it hold up?

Check 5 directly: 6505 < 6805 is true and 4759 > 4749 is true, so 5 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 4 medium

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$1\,\square\,6\,4 < 1364 \qquad 6\,8\,\square\,7 > 6817$

Your answer

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $1\,\square\,6\,4 < 1364$ ( $\square$ is the hundreds digit).
Second inequality: $6\,8\,\square\,7 > 6817$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 2. Allowed digits: 0, 1, 2.

1\,\square\,6\,4 < 1364 \Rightarrow \square \in \{0,1,2\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 2 to 9. Allowed digits: 2, 3, 4, 5, 6, 7, 8, 9.

6\,8\,\square\,7 > 6817 \Rightarrow \square \in \{2,3,4,5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2} and the second allows {2,3,4,5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2\} \cap \{2,3,4,5,6,7,8,9\} = \{2\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 2

4 · Reviewdoes it hold up?

Check 2 directly: 1264 < 1364 is true and 6827 > 6817 is true, so 2 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 5 medium

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$4\,\square\,6\,3 < 4759 \qquad 7\,6\,\square\,8 > 7665$

Your answer

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $4\,\square\,6\,3 < 4759$ ( $\square$ is the hundreds digit).
Second inequality: $7\,6\,\square\,8 > 7665$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 6. Allowed digits: 0, 1, 2, 3, 4, 5, 6.

4\,\square\,6\,3 < 4759 \Rightarrow \square \in \{0,1,2,3,4,5,6\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 6 to 9. Allowed digits: 6, 7, 8, 9.

7\,6\,\square\,8 > 7665 \Rightarrow \square \in \{6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5,6} and the second allows {6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5,6\} \cap \{6,7,8,9\} = \{6\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 6

4 · Reviewdoes it hold up?

Check 6 directly: 4663 < 4759 is true and 7668 > 7665 is true, so 6 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 6 medium

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$7\,\square\,1\,8 < 7918 \qquad 1\,5\,\square\,3 > 1553$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $7\,\square\,1\,8 < 7918$ ( $\square$ is the hundreds digit).
Second inequality: $1\,5\,\square\,3 > 1553$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 8. Allowed digits: 0, 1, 2, 3, 4, 5, 6, 7, 8.

7\,\square\,1\,8 < 7918 \Rightarrow \square \in \{0,1,2,3,4,5,6,7,8\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 6 to 9. Allowed digits: 6, 7, 8, 9.

1\,5\,\square\,3 > 1553 \Rightarrow \square \in \{6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5,6,7,8} and the second allows {6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5,6,7,8\} \cap \{6,7,8,9\} = \{6,7,8\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 6, 7, 8

4 · Reviewdoes it hold up?

Check 6 directly: 7618 < 7918 is true and 1563 > 1553 is true, so 6 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 7 medium

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$2\,\square\,4\,1 < 2651 \qquad 8\,\square\,1\,2 > 8312$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $2\,\square\,4\,1 < 2651$ ( $\square$ is the hundreds digit).
Second inequality: $8\,\square\,1\,2 > 8312$ ( $\square$ is the hundreds digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 6. Allowed digits: 0, 1, 2, 3, 4, 5, 6.

2\,\square\,4\,1 < 2651 \Rightarrow \square \in \{0,1,2,3,4,5,6\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 4 to 9. Allowed digits: 4, 5, 6, 7, 8, 9.

8\,\square\,1\,2 > 8312 \Rightarrow \square \in \{4,5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5,6} and the second allows {4,5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5,6\} \cap \{4,5,6,7,8,9\} = \{4,5,6\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 4, 5, 6

4 · Reviewdoes it hold up?

Check 4 directly: 2441 < 2651 is true and 8412 > 8312 is true, so 4 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 8 hard

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$8\,\square\,2\,3 < 8523 \qquad 7\,1\,\square\,5 > 7115$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $8\,\square\,2\,3 < 8523$ ( $\square$ is the hundreds digit).
Second inequality: $7\,1\,\square\,5 > 7115$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 4. Allowed digits: 0, 1, 2, 3, 4.

8\,\square\,2\,3 < 8523 \Rightarrow \square \in \{0,1,2,3,4\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 2 to 9. Allowed digits: 2, 3, 4, 5, 6, 7, 8, 9.

7\,1\,\square\,5 > 7115 \Rightarrow \square \in \{2,3,4,5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4} and the second allows {2,3,4,5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4\} \cap \{2,3,4,5,6,7,8,9\} = \{2,3,4\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 2, 3, 4

4 · Reviewdoes it hold up?

Check 2 directly: 8223 < 8523 is true and 7125 > 7115 is true, so 2 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 9 hard

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$3\,\square\,7\,6 < 3576 \qquad 9\,2\,\square\,1 > 9231$

Your answer

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $3\,\square\,7\,6 < 3576$ ( $\square$ is the hundreds digit).
Second inequality: $9\,2\,\square\,1 > 9231$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 4. Allowed digits: 0, 1, 2, 3, 4.

3\,\square\,7\,6 < 3576 \Rightarrow \square \in \{0,1,2,3,4\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 4 to 9. Allowed digits: 4, 5, 6, 7, 8, 9.

9\,2\,\square\,1 > 9231 \Rightarrow \square \in \{4,5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4} and the second allows {4,5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4\} \cap \{4,5,6,7,8,9\} = \{4\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 4

4 · Reviewdoes it hold up?

Check 4 directly: 3476 < 3576 is true and 9241 > 9231 is true, so 4 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Variant 10 hard

Among the digits from $0$ to $9$ , find every digit that can go in $\square$ to make both inequalities true.

$9\,\square\,5\,2 < 9652 \qquad 2\,4\,\square\,6 > 2436$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every digit from 0 to 9 that can be placed in the $\square$ so that both inequalities are true at the same time.

Givens

The $\square$ holds a single digit from 0 to 9.
First inequality: $9\,\square\,5\,2 < 9652$ ( $\square$ is the hundreds digit).
Second inequality: $2\,4\,\square\,6 > 2436$ ( $\square$ is the tens digit).

Unknowns

Which digit(s) make both inequalities true.

Constraints

The same digit goes in both boxes.
The digit is a whole number between 0 and 9.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #5 Look for a Pattern

The $\square$ has a finite set of candidates, 0 through 9. I compare place by place to find the range each inequality allows, then keep only the digits that satisfy both.

3 · Execute3 carry out the plan

1Solve inequality 1

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the hundreds digit (the

\square

) decides the comparison. To stay less than the right side, the

\square

may be 0 to 5. Allowed digits: 0, 1, 2, 3, 4, 5.

9\,\square\,5\,2 < 9652 \Rightarrow \square \in \{0,1,2,3,4,5\}

When the top places are tied, the next place down decides which number is larger.

2Solve inequality 2

#3 Eliminate Possibilities 4.NBT.A.2

Compare place by place from the left. The higher places already match, so the tens digit (the

\square

) decides the comparison. To stay greater than the right side, the

\square

may be 4 to 9. Allowed digits: 4, 5, 6, 7, 8, 9.

2\,4\,\square\,6 > 2436 \Rightarrow \square \in \{4,5,6,7,8,9\}

When the top places are tied, the next place down decides which number is larger.

3Keep only digits that satisfy both

#5 Look for a Pattern 4.NBT.A.2

The first inequality allows {0,1,2,3,4,5} and the second allows {4,5,6,7,8,9}. Keep the digits in both lists.

\{0,1,2,3,4,5\} \cap \{4,5,6,7,8,9\} = \{4,5\}

A digit must obey both rules, so only the overlap of the two allowed ranges survives.

Answer: 4, 5

4 · Reviewdoes it hold up?

Check 4 directly: 9452 < 9652 is true and 2446 > 2436 is true, so 4 works. The full answer is the overlap of the two ranges.

Another way: Test all ten digits 0 through 9 one at a time in both inequalities (guess and check) and keep the ones that pass both.

Standardsmin grade 4

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Comparing four-digit numbers place by place to bound the unknown digit.

💡Takeaway. Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!

Compare numbers with an unknown digit · 10 practice problems

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