← Compare numbers with an unknown digit · Pin Down a Number from Digit and Range Conditions

Compare numbers with an unknown digit · 10 practice problems

4.NBT.A.2

Generated variants — 10

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

Variant 1 easy answer: 5, 6

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

4 □ 8 9<47895 3 □ 1>53414\,\square\,8\,9 < 4789 \qquad 5\,3\,\square\,1 > 5341

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 □ 8 9<47894\,\square\,8\,9 < 4789 (□\square is the hundreds digit).
  • Second inequality: 5 3 □ 1>53415\,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 □ 8 9<4789⇒□∈{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 □ 1>5341⇒□∈{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}∩{5,6,7,8,9}={5,6}\{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 answer: 5, 6, 7, 8

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

5 □ 2 7<58323 9 □ 4>39455\,\square\,2\,7 < 5832 \qquad 3\,9\,\square\,4 > 3945

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 □ 2 7<58325\,\square\,2\,7 < 5832 (□\square is the hundreds digit).
  • Second inequality: 3 9 □ 4>39453\,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 □ 2 7<5832⇒□∈{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 □ 4>3945⇒□∈{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}∩{5,6,7,8,9}={5,6,7,8}\{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 answer: 5, 6, 7

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

6 □ 0 5<68054 7 □ 9>47496\,\square\,0\,5 < 6805 \qquad 4\,7\,\square\,9 > 4749

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 □ 0 5<68056\,\square\,0\,5 < 6805 (□\square is the hundreds digit).
  • Second inequality: 4 7 □ 9>47494\,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 □ 0 5<6805⇒□∈{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 □ 9>4749⇒□∈{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}∩{5,6,7,8,9}={5,6,7}\{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 answer: 2

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

1 □ 6 4<13646 8 □ 7>68171\,\square\,6\,4 < 1364 \qquad 6\,8\,\square\,7 > 6817

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 □ 6 4<13641\,\square\,6\,4 < 1364 (□\square is the hundreds digit).
  • Second inequality: 6 8 □ 7>68176\,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 □ 6 4<1364⇒□∈{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 □ 7>6817⇒□∈{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}∩{2,3,4,5,6,7,8,9}={2}\{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 answer: 6

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

4 □ 6 3<47597 6 □ 8>76654\,\square\,6\,3 < 4759 \qquad 7\,6\,\square\,8 > 7665

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 □ 6 3<47594\,\square\,6\,3 < 4759 (□\square is the hundreds digit).
  • Second inequality: 7 6 □ 8>76657\,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 □ 6 3<4759⇒□∈{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 □ 8>7665⇒□∈{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}∩{6,7,8,9}={6}\{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 answer: 6, 7, 8

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

7 □ 1 8<79181 5 □ 3>15537\,\square\,1\,8 < 7918 \qquad 1\,5\,\square\,3 > 1553

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 □ 1 8<79187\,\square\,1\,8 < 7918 (□\square is the hundreds digit).
  • Second inequality: 1 5 □ 3>15531\,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 □ 1 8<7918⇒□∈{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 □ 3>1553⇒□∈{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}∩{6,7,8,9}={6,7,8}\{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 answer: 4, 5, 6

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

2 □ 4 1<26518 □ 1 2>83122\,\square\,4\,1 < 2651 \qquad 8\,\square\,1\,2 > 8312

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 □ 4 1<26512\,\square\,4\,1 < 2651 (□\square is the hundreds digit).
  • Second inequality: 8 □ 1 2>83128\,\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 □ 4 1<2651⇒□∈{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 □ 1 2>8312⇒□∈{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}∩{4,5,6,7,8,9}={4,5,6}\{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 answer: 2, 3, 4

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

8 □ 2 3<85237 1 □ 5>71158\,\square\,2\,3 < 8523 \qquad 7\,1\,\square\,5 > 7115

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 □ 2 3<85238\,\square\,2\,3 < 8523 (□\square is the hundreds digit).
  • Second inequality: 7 1 □ 5>71157\,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 □ 2 3<8523⇒□∈{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 □ 5>7115⇒□∈{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}∩{2,3,4,5,6,7,8,9}={2,3,4}\{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 answer: 4

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

3 □ 7 6<35769 2 □ 1>92313\,\square\,7\,6 < 3576 \qquad 9\,2\,\square\,1 > 9231

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 □ 7 6<35763\,\square\,7\,6 < 3576 (□\square is the hundreds digit).
  • Second inequality: 9 2 □ 1>92319\,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 □ 7 6<3576⇒□∈{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 □ 1>9231⇒□∈{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}∩{4,5,6,7,8,9}={4}\{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 answer: 4, 5

Among the digits from 00 to 99, find every digit that can go in □\square to make both inequalities true.

9 □ 5 2<96522 4 □ 6>24369\,\square\,5\,2 < 9652 \qquad 2\,4\,\square\,6 > 2436

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 □ 5 2<96529\,\square\,5\,2 < 9652 (□\square is the hundreds digit).
  • Second inequality: 2 4 □ 6>24362\,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 □ 5 2<9652⇒□∈{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 □ 6>2436⇒□∈{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}∩{4,5,6,7,8,9}={4,5}\{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!