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← 3-2 · Estimate a product to bound an unknown · Pin Down a Number from Digit and Range Conditions

Estimate a product to bound an unknown · 11 practice problems

3.NBT.A.33.OA.C.7

Generated variants — 11

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

Variant 1 answer: 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×95>20×30\square \times 95 > 20 \times 30

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 95 is greater than 20 times 30.

Givens
  • The inequality is (square) x 95 > 20 x 30.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 30 = 600). Then estimate: each step up in the square adds about 95. Test the boundary digit to find where the product first passes 600, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 20 by 30 to get the number the left side must beat.
20×30=60020 \times 30 = 600
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 95 is close to 100, (square) x 95 is roughly (square) times 100. Test near the boundary: 6 x 95 = 570 (not over 600) and 7 x 95 = 665 (over 600).
6×95=570<600,7×95=665>6006 \times 95 = 570 < 600,\quad 7 \times 95 = 665 > 600
Estimating 95 as about 100 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 570 or less, which is not over 600.
7×95=665, 8×95=760, 9×95=8557 \times 95 = 665,\ 8 \times 95 = 760,\ 9 \times 95 = 855
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 7, 8, and 9

Review

The boundary 6 x 95 = 570 is just under 600 and 7 x 95 = 665 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Divide instead: 600 / 95 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 30 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 95 against 600.
💡 Treat 95 as almost 100 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!
Variant 2 answer: 6, 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×85>10×50\square \times 85 > 10 \times 50

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 85 is greater than 10 times 50.

Givens
  • The inequality is (square) x 85 > 10 x 50.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (10 x 50 = 500). Then estimate: each step up in the square adds about 85. Test the boundary digit to find where the product first passes 500, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 10 by 50 to get the number the left side must beat.
10×50=50010 \times 50 = 500
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 85 is close to 80, (square) x 85 is roughly (square) times 80. Test near the boundary: 5 x 85 = 425 (not over 500) and 6 x 85 = 510 (over 500).
5×85=425<500,6×85=510>5005 \times 85 = 425 < 500,\quad 6 \times 85 = 510 > 500
Estimating 85 as about 80 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 6 up, the product only grows, so 6, 7, 8, and 9 all satisfy the inequality. Digits 5 and below give 425 or less, which is not over 500.
6×85=510, 7×85=595, 8×85=680, 9×85=7656 \times 85 = 510,\ 7 \times 85 = 595,\ 8 \times 85 = 680,\ 9 \times 85 = 765
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 6, 7, 8, and 9

Review

The boundary 5 x 85 = 425 is just under 500 and 6 x 85 = 510 is just over, so 6 being the smallest that works makes sense; all one-digit values are 6, 7, 8, and 9.

Divide instead: 500 / 85 is between 5 and 6, so the square must be at least 6. The one-digit numbers 6, 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 10 x 50 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 85 against 500.
💡 Treat 85 as almost 80 and you can guess the boundary -- Grade 3 estimating points right to 6, 7, 8, and 9!
Variant 3 answer: 5, 6, 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×95>20×20\square \times 95 > 20 \times 20

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 95 is greater than 20 times 20.

Givens
  • The inequality is (square) x 95 > 20 x 20.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 20 = 400). Then estimate: each step up in the square adds about 95. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 20 by 20 to get the number the left side must beat.
20×20=40020 \times 20 = 400
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 95 is close to 100, (square) x 95 is roughly (square) times 100. Test near the boundary: 4 x 95 = 380 (not over 400) and 5 x 95 = 475 (over 400).
4×95=380<400,5×95=475>4004 \times 95 = 380 < 400,\quad 5 \times 95 = 475 > 400
Estimating 95 as about 100 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 380 or less, which is not over 400.
5×95=475, 6×95=570, 7×95=665, 8×95=760, 9×95=8555 \times 95 = 475,\ 6 \times 95 = 570,\ 7 \times 95 = 665,\ 8 \times 95 = 760,\ 9 \times 95 = 855
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 5, 6, 7, 8, and 9

Review

The boundary 4 x 95 = 380 is just under 400 and 5 x 95 = 475 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Divide instead: 400 / 95 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 20 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 95 against 400.
💡 Treat 95 as almost 100 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!
Variant 4 answer: 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×43>30×10\square \times 43 > 30 \times 10

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 43 is greater than 30 times 10.

Givens
  • The inequality is (square) x 43 > 30 x 10.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (30 x 10 = 300). Then estimate: each step up in the square adds about 43. Test the boundary digit to find where the product first passes 300, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 30 by 10 to get the number the left side must beat.
30×10=30030 \times 10 = 300
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 43 is close to 40, (square) x 43 is roughly (square) times 40. Test near the boundary: 6 x 43 = 258 (not over 300) and 7 x 43 = 301 (over 300).
6×43=258<300,7×43=301>3006 \times 43 = 258 < 300,\quad 7 \times 43 = 301 > 300
Estimating 43 as about 40 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 258 or less, which is not over 300.
7×43=301, 8×43=344, 9×43=3877 \times 43 = 301,\ 8 \times 43 = 344,\ 9 \times 43 = 387
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 7, 8, and 9

Review

The boundary 6 x 43 = 258 is just under 300 and 7 x 43 = 301 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Divide instead: 300 / 43 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 30 x 10 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 43 against 300.
💡 Treat 43 as almost 40 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!
Variant 5 answer: 9

Find every one-digit number that can go in the □\square.

□×37>10×30\square \times 37 > 10 \times 30

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 37 is greater than 10 times 30.

Givens
  • The inequality is (square) x 37 > 10 x 30.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (10 x 30 = 300). Then estimate: each step up in the square adds about 37. Test the boundary digit to find where the product first passes 300, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 10 by 30 to get the number the left side must beat.
10×30=30010 \times 30 = 300
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 37 is close to 40, (square) x 37 is roughly (square) times 40. Test near the boundary: 8 x 37 = 296 (not over 300) and 9 x 37 = 333 (over 300).
8×37=296<300,9×37=333>3008 \times 37 = 296 < 300,\quad 9 \times 37 = 333 > 300
Estimating 37 as about 40 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 9 up, the product only grows, so 9 all satisfy the inequality. Digits 8 and below give 296 or less, which is not over 300.
9×37=3339 \times 37 = 333
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 9

Review

The boundary 8 x 37 = 296 is just under 300 and 9 x 37 = 333 is just over, so 9 being the smallest that works makes sense; all one-digit values are 9.

Divide instead: 300 / 37 is between 8 and 9, so the square must be at least 9. The one-digit numbers 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 10 x 30 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 37 against 300.
💡 Treat 37 as almost 40 and you can guess the boundary -- Grade 3 estimating points right to 9!
Variant 6 answer: 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×88>30×20\square \times 88 > 30 \times 20

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 88 is greater than 30 times 20.

Givens
  • The inequality is (square) x 88 > 30 x 20.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (30 x 20 = 600). Then estimate: each step up in the square adds about 88. Test the boundary digit to find where the product first passes 600, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 30 by 20 to get the number the left side must beat.
30×20=60030 \times 20 = 600
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 88 is close to 90, (square) x 88 is roughly (square) times 90. Test near the boundary: 6 x 88 = 528 (not over 600) and 7 x 88 = 616 (over 600).
6×88=528<600,7×88=616>6006 \times 88 = 528 < 600,\quad 7 \times 88 = 616 > 600
Estimating 88 as about 90 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 528 or less, which is not over 600.
7×88=616, 8×88=704, 9×88=7927 \times 88 = 616,\ 8 \times 88 = 704,\ 9 \times 88 = 792
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 7, 8, and 9

Review

The boundary 6 x 88 = 528 is just under 600 and 7 x 88 = 616 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Divide instead: 600 / 88 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 30 x 20 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 88 against 600.
💡 Treat 88 as almost 90 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!
Variant 7 answer: 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×66>20×20\square \times 66 > 20 \times 20

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 66 is greater than 20 times 20.

Givens
  • The inequality is (square) x 66 > 20 x 20.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 20 = 400). Then estimate: each step up in the square adds about 66. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 20 by 20 to get the number the left side must beat.
20×20=40020 \times 20 = 400
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 66 is close to 70, (square) x 66 is roughly (square) times 70. Test near the boundary: 6 x 66 = 396 (not over 400) and 7 x 66 = 462 (over 400).
6×66=396<400,7×66=462>4006 \times 66 = 396 < 400,\quad 7 \times 66 = 462 > 400
Estimating 66 as about 70 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 396 or less, which is not over 400.
7×66=462, 8×66=528, 9×66=5947 \times 66 = 462,\ 8 \times 66 = 528,\ 9 \times 66 = 594
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 7, 8, and 9

Review

The boundary 6 x 66 = 396 is just under 400 and 7 x 66 = 462 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Divide instead: 400 / 66 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 20 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 66 against 400.
💡 Treat 66 as almost 70 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!
Variant 8 answer: 5, 6, 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×91>40×10\square \times 91 > 40 \times 10

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 91 is greater than 40 times 10.

Givens
  • The inequality is (square) x 91 > 40 x 10.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (40 x 10 = 400). Then estimate: each step up in the square adds about 91. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 40 by 10 to get the number the left side must beat.
40×10=40040 \times 10 = 400
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 91 is close to 90, (square) x 91 is roughly (square) times 90. Test near the boundary: 4 x 91 = 364 (not over 400) and 5 x 91 = 455 (over 400).
4×91=364<400,5×91=455>4004 \times 91 = 364 < 400,\quad 5 \times 91 = 455 > 400
Estimating 91 as about 90 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 364 or less, which is not over 400.
5×91=455, 6×91=546, 7×91=637, 8×91=728, 9×91=8195 \times 91 = 455,\ 6 \times 91 = 546,\ 7 \times 91 = 637,\ 8 \times 91 = 728,\ 9 \times 91 = 819
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 5, 6, 7, 8, and 9

Review

The boundary 4 x 91 = 364 is just under 400 and 5 x 91 = 455 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Divide instead: 400 / 91 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 40 x 10 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 91 against 400.
💡 Treat 91 as almost 90 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!
Variant 9 answer: 6, 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×78>20×20\square \times 78 > 20 \times 20

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 78 is greater than 20 times 20.

Givens
  • The inequality is (square) x 78 > 20 x 20.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 20 = 400). Then estimate: each step up in the square adds about 78. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 20 by 20 to get the number the left side must beat.
20×20=40020 \times 20 = 400
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 78 is close to 80, (square) x 78 is roughly (square) times 80. Test near the boundary: 5 x 78 = 390 (not over 400) and 6 x 78 = 468 (over 400).
5×78=390<400,6×78=468>4005 \times 78 = 390 < 400,\quad 6 \times 78 = 468 > 400
Estimating 78 as about 80 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 6 up, the product only grows, so 6, 7, 8, and 9 all satisfy the inequality. Digits 5 and below give 390 or less, which is not over 400.
6×78=468, 7×78=546, 8×78=624, 9×78=7026 \times 78 = 468,\ 7 \times 78 = 546,\ 8 \times 78 = 624,\ 9 \times 78 = 702
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 6, 7, 8, and 9

Review

The boundary 5 x 78 = 390 is just under 400 and 6 x 78 = 468 is just over, so 6 being the smallest that works makes sense; all one-digit values are 6, 7, 8, and 9.

Divide instead: 400 / 78 is between 5 and 6, so the square must be at least 6. The one-digit numbers 6, 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 20 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 78 against 400.
💡 Treat 78 as almost 80 and you can guess the boundary -- Grade 3 estimating points right to 6, 7, 8, and 9!
Variant 10 answer: 5, 6, 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×62>30×10\square \times 62 > 30 \times 10

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 62 is greater than 30 times 10.

Givens
  • The inequality is (square) x 62 > 30 x 10.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (30 x 10 = 300). Then estimate: each step up in the square adds about 62. Test the boundary digit to find where the product first passes 300, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 30 by 10 to get the number the left side must beat.
30×10=30030 \times 10 = 300
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 62 is close to 60, (square) x 62 is roughly (square) times 60. Test near the boundary: 4 x 62 = 248 (not over 300) and 5 x 62 = 310 (over 300).
4×62=248<300,5×62=310>3004 \times 62 = 248 < 300,\quad 5 \times 62 = 310 > 300
Estimating 62 as about 60 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 248 or less, which is not over 300.
5×62=310, 6×62=372, 7×62=434, 8×62=496, 9×62=5585 \times 62 = 310,\ 6 \times 62 = 372,\ 7 \times 62 = 434,\ 8 \times 62 = 496,\ 9 \times 62 = 558
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 5, 6, 7, 8, and 9

Review

The boundary 4 x 62 = 248 is just under 300 and 5 x 62 = 310 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Divide instead: 300 / 62 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 30 x 10 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 62 against 300.
💡 Treat 62 as almost 60 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!
Variant 11 answer: 5, 6, 7, 8, and 9

Find every one-digit number that can go in the □\square.

□×48>10×20\square \times 48 > 10 \times 20

Show solution

Understand

Find every one-digit number that can replace the square so that (square) times 48 is greater than 10 times 20.

Givens
  • The inequality is (square) x 48 > 10 x 20.
  • The square must be a one-digit number.
Unknowns
  • All one-digit numbers that make the inequality true.
Constraints
  • The square is a single digit (0 through 9).

Plan

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (10 x 20 = 200). Then estimate: each step up in the square adds about 48. Test the boundary digit to find where the product first passes 200, then list every digit at or above it.

Execute

#8 Analyze the Units 3.NBT.A.3
Multiply 10 by 20 to get the number the left side must beat.
10×20=20010 \times 20 = 200
Multiplying two multiples of 10 is a small product with zeros tacked on.
#6 Guess and Check 3.OA.C.7
Since 48 is close to 50, (square) x 48 is roughly (square) times 50. Test near the boundary: 4 x 48 = 192 (not over 200) and 5 x 48 = 240 (over 200).
4×48=192<200,5×48=240>2004 \times 48 = 192 < 200,\quad 5 \times 48 = 240 > 200
Estimating 48 as about 50 quickly shows which digit is the turning point.
#6 Guess and Check 3.OA.C.7
From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 192 or less, which is not over 200.
5×48=240, 6×48=288, 7×48=336, 8×48=384, 9×48=4325 \times 48 = 240,\ 6 \times 48 = 288,\ 7 \times 48 = 336,\ 8 \times 48 = 384,\ 9 \times 48 = 432
Once a digit works, every larger digit works too, since the product keeps increasing.
Answer: 5, 6, 7, 8, and 9

Review

The boundary 4 x 48 = 192 is just under 200 and 5 x 48 = 240 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Divide instead: 200 / 48 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standards · min grade 3

  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 10 x 20 and estimating with multiples of 10.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 48 against 200.
💡 Treat 48 as almost 50 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!