← Estimate a product to solve an inequality · Pin Down a Number from Digit and Range Conditions

Estimate a product to solve an inequality · 10 practice problems

3.NBT.A.33.OA.D.8

Generated variants — 10

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

Variant 1 easy answer: 4

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

19×□>9519 \times \square > 95

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 19 times the box greater than 95.

Givens
  • The inequality is 19 times the box > 95.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 95 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 19 is close to 20, so 19 times a digit is a bit under 20 times that digit. That points to a boundary, and because 19 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 19 up to 20. Multiplying 20 by small digits gives a quick size estimate that locates roughly where the product first passes 95.
20×6=120≈19×620 \times 6 = 120 \approx 19 \times 6
Rounding 19 to 20 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 19 times 5 to confirm that digit fails.
19×5=95≯9519 \times 5 = 95 \not> 95
95 is not greater than 95, so 5 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 19 times 6 to confirm the first digit that works.
19×6=114>9519 \times 6 = 114 > 95
114 is greater than 95, so 6 works; any digit bigger than 6 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 6 up makes it true. The working digits are 6, 7, 8, 9, giving 4.
{6,7,8,9}⇒4 numbers\{6, 7, 8, 9\} \Rightarrow 4 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 4
4 · Reviewdoes it hold up?

95 divided by 19 is about 5.00, so the box must be at least 6. That leaves 4 single digit(s) that reach it. Spot check: 19 times 6 = 114 (true).

Another way: Convert to division (tool 11): solve 95 divided by 19 is about 5.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 4.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 19 times the digit by rounding 19 to 20 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 19 to 20 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 2 easy answer: 4

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

26×□>13026 \times \square > 130

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 26 times the box greater than 130.

Givens
  • The inequality is 26 times the box > 130.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 130 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 26 is close to 30, so 26 times a digit is a bit under 30 times that digit. That points to a boundary, and because 26 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 26 up to 30. Multiplying 30 by small digits gives a quick size estimate that locates roughly where the product first passes 130.
30×6=180≈26×630 \times 6 = 180 \approx 26 \times 6
Rounding 26 to 30 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 26 times 5 to confirm that digit fails.
26×5=130≯13026 \times 5 = 130 \not> 130
130 is not greater than 130, so 5 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 26 times 6 to confirm the first digit that works.
26×6=156>13026 \times 6 = 156 > 130
156 is greater than 130, so 6 works; any digit bigger than 6 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 6 up makes it true. The working digits are 6, 7, 8, 9, giving 4.
{6,7,8,9}⇒4 numbers\{6, 7, 8, 9\} \Rightarrow 4 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 4
4 · Reviewdoes it hold up?

130 divided by 26 is about 5.00, so the box must be at least 6. That leaves 4 single digit(s) that reach it. Spot check: 26 times 6 = 156 (true).

Another way: Convert to division (tool 11): solve 130 divided by 26 is about 5.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 4.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 26 times the digit by rounding 26 to 30 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 26 to 30 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 3 easy answer: 4

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

34×□>17034 \times \square > 170

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 34 times the box greater than 170.

Givens
  • The inequality is 34 times the box > 170.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 170 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 34 is close to 40, so 34 times a digit is a bit under 40 times that digit. That points to a boundary, and because 34 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 34 up to 40. Multiplying 40 by small digits gives a quick size estimate that locates roughly where the product first passes 170.
40×6=240≈34×640 \times 6 = 240 \approx 34 \times 6
Rounding 34 to 40 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 34 times 5 to confirm that digit fails.
34×5=170≯17034 \times 5 = 170 \not> 170
170 is not greater than 170, so 5 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 34 times 6 to confirm the first digit that works.
34×6=204>17034 \times 6 = 204 > 170
204 is greater than 170, so 6 works; any digit bigger than 6 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 6 up makes it true. The working digits are 6, 7, 8, 9, giving 4.
{6,7,8,9}⇒4 numbers\{6, 7, 8, 9\} \Rightarrow 4 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 4
4 · Reviewdoes it hold up?

170 divided by 34 is about 5.00, so the box must be at least 6. That leaves 4 single digit(s) that reach it. Spot check: 34 times 6 = 204 (true).

Another way: Convert to division (tool 11): solve 170 divided by 34 is about 5.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 4.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 34 times the digit by rounding 34 to 40 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 34 to 40 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 4 medium answer: 4

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

38×□>19038 \times \square > 190

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 38 times the box greater than 190.

Givens
  • The inequality is 38 times the box > 190.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 190 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 38 is close to 40, so 38 times a digit is a bit under 40 times that digit. That points to a boundary, and because 38 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 38 up to 40. Multiplying 40 by small digits gives a quick size estimate that locates roughly where the product first passes 190.
40×6=240≈38×640 \times 6 = 240 \approx 38 \times 6
Rounding 38 to 40 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 38 times 5 to confirm that digit fails.
38×5=190≯19038 \times 5 = 190 \not> 190
190 is not greater than 190, so 5 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 38 times 6 to confirm the first digit that works.
38×6=228>19038 \times 6 = 228 > 190
228 is greater than 190, so 6 works; any digit bigger than 6 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 6 up makes it true. The working digits are 6, 7, 8, 9, giving 4.
{6,7,8,9}⇒4 numbers\{6, 7, 8, 9\} \Rightarrow 4 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 4
4 · Reviewdoes it hold up?

190 divided by 38 is about 5.00, so the box must be at least 6. That leaves 4 single digit(s) that reach it. Spot check: 38 times 6 = 228 (true).

Another way: Convert to division (tool 11): solve 190 divided by 38 is about 5.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 4.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 38 times the digit by rounding 38 to 40 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 38 to 40 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 5 medium answer: 2

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

29×□>20829 \times \square > 208

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 29 times the box greater than 208.

Givens
  • The inequality is 29 times the box > 208.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 208 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 29 is close to 30, so 29 times a digit is a bit under 30 times that digit. That points to a boundary, and because 29 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 29 up to 30. Multiplying 30 by small digits gives a quick size estimate that locates roughly where the product first passes 208.
30×8=240≈29×830 \times 8 = 240 \approx 29 \times 8
Rounding 29 to 30 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 29 times 7 to confirm that digit fails.
29×7=203≯20829 \times 7 = 203 \not> 208
203 is not greater than 208, so 7 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 29 times 8 to confirm the first digit that works.
29×8=232>20829 \times 8 = 232 > 208
232 is greater than 208, so 8 works; any digit bigger than 8 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 8 up makes it true. The working digits are 8, 9, giving 2.
{8,9}⇒2 numbers\{8, 9\} \Rightarrow 2 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 2
4 · Reviewdoes it hold up?

208 divided by 29 is about 7.17, so the box must be at least 8. That leaves 2 single digit(s) that reach it. Spot check: 29 times 8 = 232 (true).

Another way: Convert to division (tool 11): solve 208 divided by 29 is about 7.17, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 2.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 29 times the digit by rounding 29 to 30 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 29 to 30 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 6 medium answer: 2

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

31×□>21731 \times \square > 217

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 31 times the box greater than 217.

Givens
  • The inequality is 31 times the box > 217.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 217 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 31 is close to 40, so 31 times a digit is a bit under 40 times that digit. That points to a boundary, and because 31 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 31 up to 40. Multiplying 40 by small digits gives a quick size estimate that locates roughly where the product first passes 217.
40×8=320≈31×840 \times 8 = 320 \approx 31 \times 8
Rounding 31 to 40 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 31 times 7 to confirm that digit fails.
31×7=217≯21731 \times 7 = 217 \not> 217
217 is not greater than 217, so 7 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 31 times 8 to confirm the first digit that works.
31×8=248>21731 \times 8 = 248 > 217
248 is greater than 217, so 8 works; any digit bigger than 8 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 8 up makes it true. The working digits are 8, 9, giving 2.
{8,9}⇒2 numbers\{8, 9\} \Rightarrow 2 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 2
4 · Reviewdoes it hold up?

217 divided by 31 is about 7.00, so the box must be at least 8. That leaves 2 single digit(s) that reach it. Spot check: 31 times 8 = 248 (true).

Another way: Convert to division (tool 11): solve 217 divided by 31 is about 7.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 2.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 31 times the digit by rounding 31 to 40 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 31 to 40 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 7 medium answer: 3

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

42×□>25242 \times \square > 252

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 42 times the box greater than 252.

Givens
  • The inequality is 42 times the box > 252.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 252 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 42 is close to 50, so 42 times a digit is a bit under 50 times that digit. That points to a boundary, and because 42 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 42 up to 50. Multiplying 50 by small digits gives a quick size estimate that locates roughly where the product first passes 252.
50×7=350≈42×750 \times 7 = 350 \approx 42 \times 7
Rounding 42 to 50 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 42 times 6 to confirm that digit fails.
42×6=252≯25242 \times 6 = 252 \not> 252
252 is not greater than 252, so 6 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 42 times 7 to confirm the first digit that works.
42×7=294>25242 \times 7 = 294 > 252
294 is greater than 252, so 7 works; any digit bigger than 7 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 7 up makes it true. The working digits are 7, 8, 9, giving 3.
{7,8,9}⇒3 numbers\{7, 8, 9\} \Rightarrow 3 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 3
4 · Reviewdoes it hold up?

252 divided by 42 is about 6.00, so the box must be at least 7. That leaves 3 single digit(s) that reach it. Spot check: 42 times 7 = 294 (true).

Another way: Convert to division (tool 11): solve 252 divided by 42 is about 6.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 3.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 42 times the digit by rounding 42 to 50 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 42 to 50 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 8 hard answer: 4

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

47×□>28047 \times \square > 280

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 47 times the box greater than 280.

Givens
  • The inequality is 47 times the box > 280.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 280 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 47 is close to 50, so 47 times a digit is a bit under 50 times that digit. That points to a boundary, and because 47 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 47 up to 50. Multiplying 50 by small digits gives a quick size estimate that locates roughly where the product first passes 280.
50×6=300≈47×650 \times 6 = 300 \approx 47 \times 6
Rounding 47 to 50 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 47 times 5 to confirm that digit fails.
47×5=235≯28047 \times 5 = 235 \not> 280
235 is not greater than 280, so 5 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 47 times 6 to confirm the first digit that works.
47×6=282>28047 \times 6 = 282 > 280
282 is greater than 280, so 6 works; any digit bigger than 6 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 6 up makes it true. The working digits are 6, 7, 8, 9, giving 4.
{6,7,8,9}⇒4 numbers\{6, 7, 8, 9\} \Rightarrow 4 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 4
4 · Reviewdoes it hold up?

280 divided by 47 is about 5.96, so the box must be at least 6. That leaves 4 single digit(s) that reach it. Spot check: 47 times 6 = 282 (true).

Another way: Convert to division (tool 11): solve 280 divided by 47 is about 5.96, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 4.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 47 times the digit by rounding 47 to 50 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 47 to 50 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 9 hard answer: 3

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

53×□>31853 \times \square > 318

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 53 times the box greater than 318.

Givens
  • The inequality is 53 times the box > 318.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 318 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 53 is close to 60, so 53 times a digit is a bit under 60 times that digit. That points to a boundary, and because 53 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 53 up to 60. Multiplying 60 by small digits gives a quick size estimate that locates roughly where the product first passes 318.
60×7=420≈53×760 \times 7 = 420 \approx 53 \times 7
Rounding 53 to 60 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 53 times 6 to confirm that digit fails.
53×6=318≯31853 \times 6 = 318 \not> 318
318 is not greater than 318, so 6 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 53 times 7 to confirm the first digit that works.
53×7=371>31853 \times 7 = 371 > 318
371 is greater than 318, so 7 works; any digit bigger than 7 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 7 up makes it true. The working digits are 7, 8, 9, giving 3.
{7,8,9}⇒3 numbers\{7, 8, 9\} \Rightarrow 3 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 3
4 · Reviewdoes it hold up?

318 divided by 53 is about 6.00, so the box must be at least 7. That leaves 3 single digit(s) that reach it. Spot check: 53 times 7 = 371 (true).

Another way: Convert to division (tool 11): solve 318 divided by 53 is about 6.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 3.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 53 times the digit by rounding 53 to 60 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 53 to 60 to guess the cutoff, then check around it: Grade 3 estimation does the job!
Variant 10 hard answer: 3

Among the numbers from 00 to 99, how many different numbers can go in the □\square to make the statement true?

58×□>34858 \times \square > 348

Show solution
1 · Understandwhat's really being asked

Using a single digit from 0 to 9 in the box, I need to count how many of those digits make 58 times the box greater than 348.

Givens
  • The inequality is 58 times the box > 348.
  • The box can hold any whole number from 0 to 9.
Unknowns
  • How many of the digits 0 through 9 make the inequality true.
Constraints
  • Only single digits 0,1,2,...,9 are allowed in the box.
  • The statement must be strictly greater than 348 (equal does not count).
2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Estimate first: 58 is close to 60, so 58 times a digit is a bit under 60 times that digit. That points to a boundary, and because 58 times the box grows steadily as the box grows, once a digit works every larger digit works too. So I just check the digits around the boundary and count the ones that pass.

3 · Execute4 carry out the plan

1Estimate the boundary digit

#6 Guess and Check 3.NBT.A.3
Round 58 up to 60. Multiplying 60 by small digits gives a quick size estimate that locates roughly where the product first passes 348.
60×7=420≈58×760 \times 7 = 420 \approx 58 \times 7
Rounding 58 to 60 gives a quick size estimate of the product without exact multiplying.

2Check the boundary exactly

#6 Guess and Check 3.OA.D.8
Compute 58 times 6 to confirm that digit fails.
58×6=348≯34858 \times 6 = 348 \not> 348
348 is not greater than 348, so 6 does not work.

3Check the first working digit

#6 Guess and Check 3.OA.D.8
Compute 58 times 7 to confirm the first digit that works.
58×7=406>34858 \times 7 = 406 > 348
406 is greater than 348, so 7 works; any digit bigger than 7 gives an even larger product.

4Count the digits that work

#5 Look for a Pattern 3.OA.D.8
Because the product grows as the box grows, every digit from 7 up makes it true. The working digits are 7, 8, 9, giving 3.
{7,8,9}⇒3 numbers\{7, 8, 9\} \Rightarrow 3 \text{ numbers}
Once the inequality first holds, it keeps holding for all larger digits, so I only count from the boundary up.
Answer: 3
4 · Reviewdoes it hold up?

348 divided by 58 is about 6.00, so the box must be at least 7. That leaves 3 single digit(s) that reach it. Spot check: 58 times 7 = 406 (true).

Another way: Convert to division (tool 11): solve 348 divided by 58 is about 6.00, so the box must be that much or more, and counting the qualifying digits in 0 to 9 gives 3.

Standardsmin grade 3
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Estimating 58 times the digit by rounding 58 to 60 to locate the boundary.
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 — Checking exact products and counting which digits satisfy the inequality.
💡Takeaway. Round 58 to 60 to guess the cutoff, then check around it: Grade 3 estimation does the job!