← Subtract the remainder to find a divisor · Divisibility and Remainder Reasoning

Subtract the remainder to find a divisor · 12 practice problems

4.NBT.B.64.OA.B.4

Generated variants — 12

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

Variant 1 easy answer: 12

Find the largest number that satisfies both of the following conditions.

When $99$ is divided by the number, the remainder is $3$ .
When $137$ is divided by the number, the remainder is $5$ .

Show solution

1 · Understandwhat's really being asked

One number divides 99 leaving 3 left over, and divides 137 leaving 5 left over. Among all numbers that do both, we want the largest.

Givens

99 divided by the number leaves a remainder of 3.
137 divided by the number leaves a remainder of 5.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 5.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

99 leaves 3 over, so removing those 3 leaves an amount the number divides exactly.

99 - 3 = 96

The number is a factor of 96.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 5 that is left over.

137 - 5 = 132

The number is a factor of 132 as well.

3Find the common factors of 96 and 132

#6 Guess and Check 4.OA.B.4

A number that divides both 96 and 132 is a common factor, and every common factor divides their greatest common factor.

\gcd(96, 132) = 12

The candidates are 1, 2, 3, 4, 6, 12.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 5 is out: that removes 1, 2, 3, 4. The largest survivor is 12.

99 \div 12 = 8 \cdots 3,\quad 137 \div 12 = 11 \cdots 5

Both divisions give exactly the promised remainders, so 12 is the answer.

Answer: 12

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 12 is really the biggest: no common factor of 96 and 132 exceeds 12, since every common factor divides 12.

Another way: List the factors of 96 and of 132 separately and mark the ones that appear on both lists; the largest mark, above 5, is 12.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 96 and 132 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 2 easy answer: 16

Find the largest number that satisfies both of the following conditions.

When $148$ is divided by the number, the remainder is $4$ .
When $165$ is divided by the number, the remainder is $5$ .

Show solution

1 · Understandwhat's really being asked

One number divides 148 leaving 4 left over, and divides 165 leaving 5 left over. Among all numbers that do both, we want the largest.

Givens

148 divided by the number leaves a remainder of 4.
165 divided by the number leaves a remainder of 5.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 5.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

148 leaves 4 over, so removing those 4 leaves an amount the number divides exactly.

148 - 4 = 144

The number is a factor of 144.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 5 that is left over.

165 - 5 = 160

The number is a factor of 160 as well.

3Find the common factors of 144 and 160

#6 Guess and Check 4.OA.B.4

A number that divides both 144 and 160 is a common factor, and every common factor divides their greatest common factor.

\gcd(144, 160) = 16

The candidates are 1, 2, 4, 8, 16.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 5 is out: that removes 1, 2, 4. The largest survivor is 16.

148 \div 16 = 9 \cdots 4,\quad 165 \div 16 = 10 \cdots 5

Both divisions give exactly the promised remainders, so 16 is the answer.

Answer: 16

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 16 is really the biggest: no common factor of 144 and 160 exceeds 16, since every common factor divides 16.

Another way: List the factors of 144 and of 160 separately and mark the ones that appear on both lists; the largest mark, above 5, is 16.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 144 and 160 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 3 easy answer: 18

Find the largest number that satisfies both of the following conditions.

When $132$ is divided by the number, the remainder is $6$ .
When $166$ is divided by the number, the remainder is $4$ .

Show solution

1 · Understandwhat's really being asked

One number divides 132 leaving 6 left over, and divides 166 leaving 4 left over. Among all numbers that do both, we want the largest.

Givens

132 divided by the number leaves a remainder of 6.
166 divided by the number leaves a remainder of 4.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 6.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

132 leaves 6 over, so removing those 6 leaves an amount the number divides exactly.

132 - 6 = 126

The number is a factor of 126.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 4 that is left over.

166 - 4 = 162

The number is a factor of 162 as well.

3Find the common factors of 126 and 162

#6 Guess and Check 4.OA.B.4

A number that divides both 126 and 162 is a common factor, and every common factor divides their greatest common factor.

\gcd(126, 162) = 18

The candidates are 1, 2, 3, 6, 9, 18.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 6 is out: that removes 1, 2, 3, 6. The largest survivor is 18.

132 \div 18 = 7 \cdots 6,\quad 166 \div 18 = 9 \cdots 4

Both divisions give exactly the promised remainders, so 18 is the answer.

Answer: 18

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 18 is really the biggest: no common factor of 126 and 162 exceeds 18, since every common factor divides 18.

Another way: List the factors of 126 and of 162 separately and mark the ones that appear on both lists; the largest mark, above 6, is 18.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 126 and 162 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 4 easy answer: 24

Find the largest number that satisfies both of the following conditions.

When $125$ is divided by the number, the remainder is $5$ .
When $179$ is divided by the number, the remainder is $11$ .

Show solution

1 · Understandwhat's really being asked

One number divides 125 leaving 5 left over, and divides 179 leaving 11 left over. Among all numbers that do both, we want the largest.

Givens

125 divided by the number leaves a remainder of 5.
179 divided by the number leaves a remainder of 11.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 11.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

125 leaves 5 over, so removing those 5 leaves an amount the number divides exactly.

125 - 5 = 120

The number is a factor of 120.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 11 that is left over.

179 - 11 = 168

The number is a factor of 168 as well.

3Find the common factors of 120 and 168

#6 Guess and Check 4.OA.B.4

A number that divides both 120 and 168 is a common factor, and every common factor divides their greatest common factor.

\gcd(120, 168) = 24

The candidates are 1, 2, 3, 4, 6, 8, 12, 24.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 11 is out: that removes 1, 2, 3, 4, 6, 8. The largest survivor is 24.

125 \div 24 = 5 \cdots 5,\quad 179 \div 24 = 7 \cdots 11

Both divisions give exactly the promised remainders, so 24 is the answer.

Answer: 24

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 24 is really the biggest: no common factor of 120 and 168 exceeds 24, since every common factor divides 24.

Another way: List the factors of 120 and of 168 separately and mark the ones that appear on both lists; the largest mark, above 11, is 24.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 120 and 168 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 5 medium answer: 14

Find the largest number that satisfies both of the following conditions.

When $128$ is divided by the number, the remainder is $2$ .
When $191$ is divided by the number, the remainder is $9$ .

Show solution

1 · Understandwhat's really being asked

One number divides 128 leaving 2 left over, and divides 191 leaving 9 left over. Among all numbers that do both, we want the largest.

Givens

128 divided by the number leaves a remainder of 2.
191 divided by the number leaves a remainder of 9.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 9.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

128 leaves 2 over, so removing those 2 leaves an amount the number divides exactly.

128 - 2 = 126

The number is a factor of 126.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 9 that is left over.

191 - 9 = 182

The number is a factor of 182 as well.

3Find the common factors of 126 and 182

#6 Guess and Check 4.OA.B.4

A number that divides both 126 and 182 is a common factor, and every common factor divides their greatest common factor.

\gcd(126, 182) = 14

The candidates are 1, 2, 7, 14.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 9 is out: that removes 1, 2, 7. The largest survivor is 14.

128 \div 14 = 9 \cdots 2,\quad 191 \div 14 = 13 \cdots 9

Both divisions give exactly the promised remainders, so 14 is the answer.

Answer: 14

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 14 is really the biggest: no common factor of 126 and 182 exceeds 14, since every common factor divides 14.

Another way: List the factors of 126 and of 182 separately and mark the ones that appear on both lists; the largest mark, above 9, is 14.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 126 and 182 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 6 medium answer: 15

Find the largest number that satisfies both of the following conditions.

When $172$ is divided by the number, the remainder is $7$ .
When $198$ is divided by the number, the remainder is $3$ .

Show solution

1 · Understandwhat's really being asked

One number divides 172 leaving 7 left over, and divides 198 leaving 3 left over. Among all numbers that do both, we want the largest.

Givens

172 divided by the number leaves a remainder of 7.
198 divided by the number leaves a remainder of 3.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 7.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

172 leaves 7 over, so removing those 7 leaves an amount the number divides exactly.

172 - 7 = 165

The number is a factor of 165.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 3 that is left over.

198 - 3 = 195

The number is a factor of 195 as well.

3Find the common factors of 165 and 195

#6 Guess and Check 4.OA.B.4

A number that divides both 165 and 195 is a common factor, and every common factor divides their greatest common factor.

\gcd(165, 195) = 15

The candidates are 1, 3, 5, 15.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 7 is out: that removes 1, 3, 5. The largest survivor is 15.

172 \div 15 = 11 \cdots 7,\quad 198 \div 15 = 13 \cdots 3

Both divisions give exactly the promised remainders, so 15 is the answer.

Answer: 15

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 15 is really the biggest: no common factor of 165 and 195 exceeds 15, since every common factor divides 15.

Another way: List the factors of 165 and of 195 separately and mark the ones that appear on both lists; the largest mark, above 7, is 15.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 165 and 195 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 7 medium answer: 20

Find the largest number that satisfies both of the following conditions.

When $149$ is divided by the number, the remainder is $9$ .
When $226$ is divided by the number, the remainder is $6$ .

Show solution

1 · Understandwhat's really being asked

One number divides 149 leaving 9 left over, and divides 226 leaving 6 left over. Among all numbers that do both, we want the largest.

Givens

149 divided by the number leaves a remainder of 9.
226 divided by the number leaves a remainder of 6.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 9.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

149 leaves 9 over, so removing those 9 leaves an amount the number divides exactly.

149 - 9 = 140

The number is a factor of 140.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 6 that is left over.

226 - 6 = 220

The number is a factor of 220 as well.

3Find the common factors of 140 and 220

#6 Guess and Check 4.OA.B.4

A number that divides both 140 and 220 is a common factor, and every common factor divides their greatest common factor.

\gcd(140, 220) = 20

The candidates are 1, 2, 4, 5, 10, 20.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 9 is out: that removes 1, 2, 4, 5. The largest survivor is 20.

149 \div 20 = 7 \cdots 9,\quad 226 \div 20 = 11 \cdots 6

Both divisions give exactly the promised remainders, so 20 is the answer.

Answer: 20

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 20 is really the biggest: no common factor of 140 and 220 exceeds 20, since every common factor divides 20.

Another way: List the factors of 140 and of 220 separately and mark the ones that appear on both lists; the largest mark, above 9, is 20.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 140 and 220 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 8 medium answer: 25

Find the largest number that satisfies both of the following conditions.

When $187$ is divided by the number, the remainder is $12$ .
When $232$ is divided by the number, the remainder is $7$ .

Show solution

1 · Understandwhat's really being asked

One number divides 187 leaving 12 left over, and divides 232 leaving 7 left over. Among all numbers that do both, we want the largest.

Givens

187 divided by the number leaves a remainder of 12.
232 divided by the number leaves a remainder of 7.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 12.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

187 leaves 12 over, so removing those 12 leaves an amount the number divides exactly.

187 - 12 = 175

The number is a factor of 175.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 7 that is left over.

232 - 7 = 225

The number is a factor of 225 as well.

3Find the common factors of 175 and 225

#6 Guess and Check 4.OA.B.4

A number that divides both 175 and 225 is a common factor, and every common factor divides their greatest common factor.

\gcd(175, 225) = 25

The candidates are 1, 5, 25.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 12 is out: that removes 1, 5. The largest survivor is 25.

187 \div 25 = 7 \cdots 12,\quad 232 \div 25 = 9 \cdots 7

Both divisions give exactly the promised remainders, so 25 is the answer.

Answer: 25

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 25 is really the biggest: no common factor of 175 and 225 exceeds 25, since every common factor divides 25.

Another way: List the factors of 175 and of 225 separately and mark the ones that appear on both lists; the largest mark, above 12, is 25.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 175 and 225 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 9 hard answer: 28

Find the largest number that satisfies both of the following conditions.

When $151$ is divided by the number, the remainder is $11$ .
When $241$ is divided by the number, the remainder is $17$ .

Show solution

1 · Understandwhat's really being asked

One number divides 151 leaving 11 left over, and divides 241 leaving 17 left over. Among all numbers that do both, we want the largest.

Givens

151 divided by the number leaves a remainder of 11.
241 divided by the number leaves a remainder of 17.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 17.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

151 leaves 11 over, so removing those 11 leaves an amount the number divides exactly.

151 - 11 = 140

The number is a factor of 140.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 17 that is left over.

241 - 17 = 224

The number is a factor of 224 as well.

3Find the common factors of 140 and 224

#6 Guess and Check 4.OA.B.4

A number that divides both 140 and 224 is a common factor, and every common factor divides their greatest common factor.

\gcd(140, 224) = 28

The candidates are 1, 2, 4, 7, 14, 28.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 17 is out: that removes 1, 2, 4, 7, 14. The largest survivor is 28.

151 \div 28 = 5 \cdots 11,\quad 241 \div 28 = 8 \cdots 17

Both divisions give exactly the promised remainders, so 28 is the answer.

Answer: 28

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 28 is really the biggest: no common factor of 140 and 224 exceeds 28, since every common factor divides 28.

Another way: List the factors of 140 and of 224 separately and mark the ones that appear on both lists; the largest mark, above 17, is 28.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 140 and 224 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 10 hard answer: 21

Find the largest number that satisfies both of the following conditions.

When $174$ is divided by the number, the remainder is $6$ .
When $244$ is divided by the number, the remainder is $13$ .

Show solution

1 · Understandwhat's really being asked

One number divides 174 leaving 6 left over, and divides 244 leaving 13 left over. Among all numbers that do both, we want the largest.

Givens

174 divided by the number leaves a remainder of 6.
244 divided by the number leaves a remainder of 13.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 13.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

174 leaves 6 over, so removing those 6 leaves an amount the number divides exactly.

174 - 6 = 168

The number is a factor of 168.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 13 that is left over.

244 - 13 = 231

The number is a factor of 231 as well.

3Find the common factors of 168 and 231

#6 Guess and Check 4.OA.B.4

A number that divides both 168 and 231 is a common factor, and every common factor divides their greatest common factor.

\gcd(168, 231) = 21

The candidates are 1, 3, 7, 21.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 13 is out: that removes 1, 3, 7. The largest survivor is 21.

174 \div 21 = 8 \cdots 6,\quad 244 \div 21 = 11 \cdots 13

Both divisions give exactly the promised remainders, so 21 is the answer.

Answer: 21

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 21 is really the biggest: no common factor of 168 and 231 exceeds 21, since every common factor divides 21.

Another way: List the factors of 168 and of 231 separately and mark the ones that appear on both lists; the largest mark, above 13, is 21.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 168 and 231 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 11 hard answer: 22

Find the largest number that satisfies both of the following conditions.

When $167$ is divided by the number, the remainder is $13$ .
When $268$ is divided by the number, the remainder is $4$ .

Show solution

1 · Understandwhat's really being asked

One number divides 167 leaving 13 left over, and divides 268 leaving 4 left over. Among all numbers that do both, we want the largest.

Givens

167 divided by the number leaves a remainder of 13.
268 divided by the number leaves a remainder of 4.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 13.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

167 leaves 13 over, so removing those 13 leaves an amount the number divides exactly.

167 - 13 = 154

The number is a factor of 154.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 4 that is left over.

268 - 4 = 264

The number is a factor of 264 as well.

3Find the common factors of 154 and 264

#6 Guess and Check 4.OA.B.4

A number that divides both 154 and 264 is a common factor, and every common factor divides their greatest common factor.

\gcd(154, 264) = 22

The candidates are 1, 2, 11, 22.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 13 is out: that removes 1, 2, 11. The largest survivor is 22.

167 \div 22 = 7 \cdots 13,\quad 268 \div 22 = 12 \cdots 4

Both divisions give exactly the promised remainders, so 22 is the answer.

Answer: 22

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 22 is really the biggest: no common factor of 154 and 264 exceeds 22, since every common factor divides 22.

Another way: List the factors of 154 and of 264 separately and mark the ones that appear on both lists; the largest mark, above 13, is 22.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 154 and 264 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.

Variant 12 hard answer: 27

Find the largest number that satisfies both of the following conditions.

When $197$ is divided by the number, the remainder is $8$ .
When $289$ is divided by the number, the remainder is $19$ .

Show solution

1 · Understandwhat's really being asked

One number divides 197 leaving 8 left over, and divides 289 leaving 19 left over. Among all numbers that do both, we want the largest.

Givens

197 divided by the number leaves a remainder of 8.
289 divided by the number leaves a remainder of 19.
The same number is used in both divisions.

Unknowns

The largest number that fits both conditions.

Constraints

A remainder is always smaller than the divisor, so the number is bigger than 19.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

A remainder is the part that does not fit. Take it away and each clue becomes an exact division, which turns the question into finding common factors -- something we can list.

3 · Execute4 carry out the plan

1Turn the first clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

197 leaves 8 over, so removing those 8 leaves an amount the number divides exactly.

197 - 8 = 189

The number is a factor of 189.

2Turn the second clue into an exact-division clue

#7 Identify Subproblems 4.NBT.B.6

The same move on the second clue: take away the 19 that is left over.

289 - 19 = 270

The number is a factor of 270 as well.

3Find the common factors of 189 and 270

#6 Guess and Check 4.OA.B.4

A number that divides both 189 and 270 is a common factor, and every common factor divides their greatest common factor.

\gcd(189, 270) = 27

The candidates are 1, 3, 9, 27.

4Check the size rule and confirm with the originals

#6 Guess and Check 4.NBT.B.6

A remainder must be smaller than the divisor, so anything not bigger than 19 is out: that removes 1, 3, 9. The largest survivor is 27.

197 \div 27 = 7 \cdots 8,\quad 289 \div 27 = 10 \cdots 19

Both divisions give exactly the promised remainders, so 27 is the answer.

Answer: 27

4 · Reviewdoes it hold up?

Test the next-largest candidate to be sure 27 is really the biggest: no common factor of 189 and 270 exceeds 27, since every common factor divides 27.

Another way: List the factors of 189 and of 270 separately and mark the ones that appear on both lists; the largest mark, above 19, is 27.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Reading each clue as a quotient-and-remainder and removing the remainder to get an exact division.
4.OA.B.4 Find all factor pairs and recognize multiples; determine whether a number is a multiple — Finding the common factors of 189 and 270 and picking the largest that beats the remainder.

💡Takeaway. A remainder is the part that would not fit. Take it away first, and a leftover puzzle turns into a factor puzzle.