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← 2-2 · 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 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 2 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 3 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 4 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 5 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 6 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 7 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 8 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 9 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

Standards · min 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.
💡 Compare numbers one place at a time from the left, find each rule's range, then keep the digit both rules share!
Variant 10 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

Understand

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

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

Plan

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

The box 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.

Execute

#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the hundreds digit (the box) decides the comparison. To stay less than the right side, the box 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.
#3 Eliminate Possibilities 4.NBT.A.2
Compare place by place from the left. The higher places already match, so the tens digit (the box) decides the comparison. To stay greater than the right side, the box 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.
#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

Review

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.

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

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