← Numbers divisible by several are common multiples · Divisibility and Remainder Reasoning

Numbers divisible by several are common multiples · 11 practice problems

3.OA.B.63.OA.C.7

Generated variants — 11

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

Variant 1 easy answer: 24

From the numbers below, find the one that can be divided evenly by $4$ and also by $6$ .

$8 \quad 9 \quad 24 \quad 15$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 8, 9, 24, 15, we must find the single number that can be divided evenly (no remainder) by both 4 and 6.

Givens

The candidate numbers are 8, 9, 24, 15.
The answer must divide evenly by 4.
The answer must also divide evenly by 6.

Unknowns

Which of the numbers is divisible by both 4 and 6.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 4, divisible by 6). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 6

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 6: 8 leaves a remainder; 9 leaves a remainder; 24 is 6 times 4; 15 leaves a remainder. So 24 pass the 'divisible by 6' clue; 8, 9, 15 eliminated.

24 = 6 \times 4

Grade 3 times-tables: a number is divisible by 6 only if it appears in the 6 times table.

2Of those, keep the multiples of 4

#3 Eliminate Possibilities 3.OA.B.6

. Only 24 is in the 4 times table, so 24 survives.

24 = 4 \times 6

Grade 3 unknown-factor sense: 24 divided by 4 is 6 with nothing left over, so 24 is a multiple of 4.

Answer: 24

4 · Reviewdoes it hold up?

Check 24 against both clues directly: 24 divided by 4 is 6 (exact) and 24 divided by 6 is 4 (exact). Both are whole numbers, so 24 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 4 and 6 is a multiple of 12. List the multiples of 12 (12, 24, 36, ...) and notice only 24 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 6 times table to test divisibility by 6.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 4 by finding the missing factor.

💡Takeaway. Divisible by both 4 and 6 just means it shows up in both times tables, and 24 is the one that does!

Variant 2 easy answer: 24

From the numbers below, find the one that can be divided evenly by $3$ and also by $4$ .

$18 \quad 16 \quad 24 \quad 21$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 18, 16, 24, 21, we must find the single number that can be divided evenly (no remainder) by both 3 and 4.

Givens

The candidate numbers are 18, 16, 24, 21.
The answer must divide evenly by 3.
The answer must also divide evenly by 4.

Unknowns

Which of the numbers is divisible by both 3 and 4.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 3, divisible by 4). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 4

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 4: 18 leaves a remainder; 16 is 4 times 4; 24 is 4 times 6; 21 leaves a remainder. So 16, 24 pass the 'divisible by 4' clue; 18, 21 eliminated.

16 = 4 \times 4, \quad 24 = 4 \times 6

Grade 3 times-tables: a number is divisible by 4 only if it appears in the 4 times table.

2Of those, keep the multiples of 3

#3 Eliminate Possibilities 3.OA.B.6

Now test 16, 24 against the 'divisible by 3' clue: 16 is not in the 3 times table; 24 is 3 times 8. So 16 eliminated and 24 survives.

24 = 3 \times 8

Grade 3 unknown-factor sense: 24 divided by 3 is 8 with nothing left over, so 24 is a multiple of 3.

Answer: 24

4 · Reviewdoes it hold up?

Check 24 against both clues directly: 24 divided by 3 is 8 (exact) and 24 divided by 4 is 6 (exact). Both are whole numbers, so 24 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 3 and 4 is a multiple of 12. List the multiples of 12 (12, 24, 36, ...) and notice only 24 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 4 times table to test divisibility by 4.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 3 by finding the missing factor.

💡Takeaway. Divisible by both 3 and 4 just means it shows up in both times tables, and 24 is the one that does!

Variant 3 easy answer: 18

From the numbers below, find the one that can be divided evenly by $2$ and also by $3$ .

$14 \quad 9 \quad 18 \quad 25$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 14, 9, 18, 25, we must find the single number that can be divided evenly (no remainder) by both 2 and 3.

Givens

The candidate numbers are 14, 9, 18, 25.
The answer must divide evenly by 2.
The answer must also divide evenly by 3.

Unknowns

Which of the numbers is divisible by both 2 and 3.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 2, divisible by 3). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 3

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 3: 14 leaves a remainder; 9 is 3 times 3; 18 is 3 times 6; 25 leaves a remainder. So 9, 18 pass the 'divisible by 3' clue; 14, 25 eliminated.

9 = 3 \times 3, \quad 18 = 3 \times 6

Grade 3 times-tables: a number is divisible by 3 only if it appears in the 3 times table.

2Of those, keep the multiples of 2

#3 Eliminate Possibilities 3.OA.B.6

Now test 9, 18 against the 'divisible by 2' clue: 9 is not in the 2 times table; 18 is 2 times 9. So 9 eliminated and 18 survives.

18 = 2 \times 9

Grade 3 unknown-factor sense: 18 divided by 2 is 9 with nothing left over, so 18 is a multiple of 2.

Answer: 18

4 · Reviewdoes it hold up?

Check 18 against both clues directly: 18 divided by 2 is 9 (exact) and 18 divided by 3 is 6 (exact). Both are whole numbers, so 18 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 2 and 3 is a multiple of 6. List the multiples of 6 (6, 12, 18, ...) and notice only 18 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 3 times table to test divisibility by 3.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 2 by finding the missing factor.

💡Takeaway. Divisible by both 2 and 3 just means it shows up in both times tables, and 18 is the one that does!

Variant 4 medium answer: 18

From the numbers below, find the one that can be divided evenly by $2$ and also by $9$ .

$27 \quad 14 \quad 18 \quad 21$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 27, 14, 18, 21, we must find the single number that can be divided evenly (no remainder) by both 2 and 9.

Givens

The candidate numbers are 27, 14, 18, 21.
The answer must divide evenly by 2.
The answer must also divide evenly by 9.

Unknowns

Which of the numbers is divisible by both 2 and 9.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 2, divisible by 9). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 9

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 9: 27 is 9 times 3; 14 leaves a remainder; 18 is 9 times 2; 21 leaves a remainder. So 27, 18 pass the 'divisible by 9' clue; 14, 21 eliminated.

27 = 9 \times 3, \quad 18 = 9 \times 2

Grade 3 times-tables: a number is divisible by 9 only if it appears in the 9 times table.

2Of those, keep the multiples of 2

#3 Eliminate Possibilities 3.OA.B.6

Now test 27, 18 against the 'divisible by 2' clue: 27 is not in the 2 times table; 18 is 2 times 9. So 27 eliminated and 18 survives.

18 = 2 \times 9

Grade 3 unknown-factor sense: 18 divided by 2 is 9 with nothing left over, so 18 is a multiple of 2.

Answer: 18

4 · Reviewdoes it hold up?

Check 18 against both clues directly: 18 divided by 2 is 9 (exact) and 18 divided by 9 is 2 (exact). Both are whole numbers, so 18 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 2 and 9 is a multiple of 18. List the multiples of 18 (18, 36, 54, ...) and notice only 18 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 9 times table to test divisibility by 9.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 2 by finding the missing factor.

💡Takeaway. Divisible by both 2 and 9 just means it shows up in both times tables, and 18 is the one that does!

Variant 5 medium answer: 30

From the numbers below, find the one that can be divided evenly by $5$ and also by $6$ .

$25 \quad 18 \quad 30 \quad 12$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 25, 18, 30, 12, we must find the single number that can be divided evenly (no remainder) by both 5 and 6.

Givens

The candidate numbers are 25, 18, 30, 12.
The answer must divide evenly by 5.
The answer must also divide evenly by 6.

Unknowns

Which of the numbers is divisible by both 5 and 6.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 5, divisible by 6). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 6

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 6: 25 leaves a remainder; 18 is 6 times 3; 30 is 6 times 5; 12 is 6 times 2. So 18, 30, 12 pass the 'divisible by 6' clue; 25 eliminated.

18 = 6 \times 3, \quad 30 = 6 \times 5, \quad 12 = 6 \times 2

Grade 3 times-tables: a number is divisible by 6 only if it appears in the 6 times table.

2Of those, keep the multiples of 5

#3 Eliminate Possibilities 3.OA.B.6

Now test 18, 30, 12 against the 'divisible by 5' clue: 18 is not in the 5 times table; 30 is 5 times 6; 12 is not in the 5 times table. So 18, 12 eliminated and 30 survives.

30 = 5 \times 6

Grade 3 unknown-factor sense: 30 divided by 5 is 6 with nothing left over, so 30 is a multiple of 5.

Answer: 30

4 · Reviewdoes it hold up?

Check 30 against both clues directly: 30 divided by 5 is 6 (exact) and 30 divided by 6 is 5 (exact). Both are whole numbers, so 30 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 5 and 6 is a multiple of 30. List the multiples of 30 (30, 60, 90, ...) and notice only 30 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 6 times table to test divisibility by 6.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 5 by finding the missing factor.

💡Takeaway. Divisible by both 5 and 6 just means it shows up in both times tables, and 30 is the one that does!

Variant 6 medium answer: 30

From the numbers below, find the one that can be divided evenly by $3$ and also by $5$ .

$10 \quad 9 \quad 30 \quad 21$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 10, 9, 30, 21, we must find the single number that can be divided evenly (no remainder) by both 3 and 5.

Givens

The candidate numbers are 10, 9, 30, 21.
The answer must divide evenly by 3.
The answer must also divide evenly by 5.

Unknowns

Which of the numbers is divisible by both 3 and 5.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 3, divisible by 5). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 5

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 5: 10 is 5 times 2; 9 leaves a remainder; 30 is 5 times 6; 21 leaves a remainder. So 10, 30 pass the 'divisible by 5' clue; 9, 21 eliminated.

10 = 5 \times 2, \quad 30 = 5 \times 6

Grade 3 times-tables: a number is divisible by 5 only if it appears in the 5 times table.

2Of those, keep the multiples of 3

#3 Eliminate Possibilities 3.OA.B.6

Now test 10, 30 against the 'divisible by 3' clue: 10 is not in the 3 times table; 30 is 3 times 10. So 10 eliminated and 30 survives.

30 = 3 \times 10

Grade 3 unknown-factor sense: 30 divided by 3 is 10 with nothing left over, so 30 is a multiple of 3.

Answer: 30

4 · Reviewdoes it hold up?

Check 30 against both clues directly: 30 divided by 3 is 10 (exact) and 30 divided by 5 is 6 (exact). Both are whole numbers, so 30 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 3 and 5 is a multiple of 15. List the multiples of 15 (15, 30, 45, ...) and notice only 30 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 5 times table to test divisibility by 5.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 3 by finding the missing factor.

💡Takeaway. Divisible by both 3 and 5 just means it shows up in both times tables, and 30 is the one that does!

Variant 7 medium answer: 20

From the numbers below, find the one that can be divided evenly by $4$ and also by $5$ .

$20 \quad 12 \quad 15 \quad 35$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 20, 12, 15, 35, we must find the single number that can be divided evenly (no remainder) by both 4 and 5.

Givens

The candidate numbers are 20, 12, 15, 35.
The answer must divide evenly by 4.
The answer must also divide evenly by 5.

Unknowns

Which of the numbers is divisible by both 4 and 5.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 4, divisible by 5). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 5

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 5: 20 is 5 times 4; 12 leaves a remainder; 15 is 5 times 3; 35 is 5 times 7. So 20, 15, 35 pass the 'divisible by 5' clue; 12 eliminated.

20 = 5 \times 4, \quad 15 = 5 \times 3, \quad 35 = 5 \times 7

Grade 3 times-tables: a number is divisible by 5 only if it appears in the 5 times table.

2Of those, keep the multiples of 4

#3 Eliminate Possibilities 3.OA.B.6

Now test 20, 15, 35 against the 'divisible by 4' clue: 20 is 4 times 5; 15 is not in the 4 times table; 35 is not in the 4 times table. So 15, 35 eliminated and 20 survives.

20 = 4 \times 5

Grade 3 unknown-factor sense: 20 divided by 4 is 5 with nothing left over, so 20 is a multiple of 4.

Answer: 20

4 · Reviewdoes it hold up?

Check 20 against both clues directly: 20 divided by 4 is 5 (exact) and 20 divided by 5 is 4 (exact). Both are whole numbers, so 20 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 4 and 5 is a multiple of 20. List the multiples of 20 (20, 40, 60, ...) and notice only 20 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 5 times table to test divisibility by 5.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 4 by finding the missing factor.

💡Takeaway. Divisible by both 4 and 5 just means it shows up in both times tables, and 20 is the one that does!

Variant 8 medium answer: 14

From the numbers below, find the one that can be divided evenly by $2$ and also by $7$ .

$21 \quad 14 \quad 9 \quad 35$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 21, 14, 9, 35, we must find the single number that can be divided evenly (no remainder) by both 2 and 7.

Givens

The candidate numbers are 21, 14, 9, 35.
The answer must divide evenly by 2.
The answer must also divide evenly by 7.

Unknowns

Which of the numbers is divisible by both 2 and 7.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 2, divisible by 7). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 7

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 7: 21 is 7 times 3; 14 is 7 times 2; 9 leaves a remainder; 35 is 7 times 5. So 21, 14, 35 pass the 'divisible by 7' clue; 9 eliminated.

21 = 7 \times 3, \quad 14 = 7 \times 2, \quad 35 = 7 \times 5

Grade 3 times-tables: a number is divisible by 7 only if it appears in the 7 times table.

2Of those, keep the multiples of 2

#3 Eliminate Possibilities 3.OA.B.6

Now test 21, 14, 35 against the 'divisible by 2' clue: 21 is not in the 2 times table; 14 is 2 times 7; 35 is not in the 2 times table. So 21, 35 eliminated and 14 survives.

14 = 2 \times 7

Grade 3 unknown-factor sense: 14 divided by 2 is 7 with nothing left over, so 14 is a multiple of 2.

Answer: 14

4 · Reviewdoes it hold up?

Check 14 against both clues directly: 14 divided by 2 is 7 (exact) and 14 divided by 7 is 2 (exact). Both are whole numbers, so 14 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 2 and 7 is a multiple of 14. List the multiples of 14 (14, 28, 42, ...) and notice only 14 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 7 times table to test divisibility by 7.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 2 by finding the missing factor.

💡Takeaway. Divisible by both 2 and 7 just means it shows up in both times tables, and 14 is the one that does!

Variant 9 hard answer: 36

From the numbers below, find the one that can be divided evenly by $4$ and also by $9$ .

$12 \quad 18 \quad 36 \quad 27$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 12, 18, 36, 27, we must find the single number that can be divided evenly (no remainder) by both 4 and 9.

Givens

The candidate numbers are 12, 18, 36, 27.
The answer must divide evenly by 4.
The answer must also divide evenly by 9.

Unknowns

Which of the numbers is divisible by both 4 and 9.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 4, divisible by 9). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 9

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 9: 12 leaves a remainder; 18 is 9 times 2; 36 is 9 times 4; 27 is 9 times 3. So 18, 36, 27 pass the 'divisible by 9' clue; 12 eliminated.

18 = 9 \times 2, \quad 36 = 9 \times 4, \quad 27 = 9 \times 3

Grade 3 times-tables: a number is divisible by 9 only if it appears in the 9 times table.

2Of those, keep the multiples of 4

#3 Eliminate Possibilities 3.OA.B.6

Now test 18, 36, 27 against the 'divisible by 4' clue: 18 is not in the 4 times table; 36 is 4 times 9; 27 is not in the 4 times table. So 18, 27 eliminated and 36 survives.

36 = 4 \times 9

Grade 3 unknown-factor sense: 36 divided by 4 is 9 with nothing left over, so 36 is a multiple of 4.

Answer: 36

4 · Reviewdoes it hold up?

Check 36 against both clues directly: 36 divided by 4 is 9 (exact) and 36 divided by 9 is 4 (exact). Both are whole numbers, so 36 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 4 and 9 is a multiple of 36. List the multiples of 36 (36, 72, 108, ...) and notice only 36 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 9 times table to test divisibility by 9.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 4 by finding the missing factor.

💡Takeaway. Divisible by both 4 and 9 just means it shows up in both times tables, and 36 is the one that does!

Variant 10 hard answer: 24

From the numbers below, find the one that can be divided evenly by $3$ and also by $8$ .

$16 \quad 24 \quad 9 \quad 40$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 16, 24, 9, 40, we must find the single number that can be divided evenly (no remainder) by both 3 and 8.

Givens

The candidate numbers are 16, 24, 9, 40.
The answer must divide evenly by 3.
The answer must also divide evenly by 8.

Unknowns

Which of the numbers is divisible by both 3 and 8.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 3, divisible by 8). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 8

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 8: 16 is 8 times 2; 24 is 8 times 3; 9 leaves a remainder; 40 is 8 times 5. So 16, 24, 40 pass the 'divisible by 8' clue; 9 eliminated.

16 = 8 \times 2, \quad 24 = 8 \times 3, \quad 40 = 8 \times 5

Grade 3 times-tables: a number is divisible by 8 only if it appears in the 8 times table.

2Of those, keep the multiples of 3

#3 Eliminate Possibilities 3.OA.B.6

Now test 16, 24, 40 against the 'divisible by 3' clue: 16 is not in the 3 times table; 24 is 3 times 8; 40 is not in the 3 times table. So 16, 40 eliminated and 24 survives.

24 = 3 \times 8

Grade 3 unknown-factor sense: 24 divided by 3 is 8 with nothing left over, so 24 is a multiple of 3.

Answer: 24

4 · Reviewdoes it hold up?

Check 24 against both clues directly: 24 divided by 3 is 8 (exact) and 24 divided by 8 is 3 (exact). Both are whole numbers, so 24 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 3 and 8 is a multiple of 24. List the multiples of 24 (24, 48, 72, ...) and notice only 24 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 8 times table to test divisibility by 8.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 3 by finding the missing factor.

💡Takeaway. Divisible by both 3 and 8 just means it shows up in both times tables, and 24 is the one that does!

Variant 11 hard answer: 42

From the numbers below, find the one that can be divided evenly by $6$ and also by $7$ .

$14 \quad 12 \quad 42 \quad 21$

Show solution

1 · Understandwhat's really being asked

Out of the numbers 14, 12, 42, 21, we must find the single number that can be divided evenly (no remainder) by both 6 and 7.

Givens

The candidate numbers are 14, 12, 42, 21.
The answer must divide evenly by 6.
The answer must also divide evenly by 7.

Unknowns

Which of the numbers is divisible by both 6 and 7.

Constraints

Even division means a remainder of 0.
Exactly one of the candidate numbers should satisfy both conditions.

2 · Planchoose the strategy

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, finite set of numbers and two clear logic clues (divisible by 6, divisible by 7). Testing each clue lets us cross off numbers until one survives both.

3 · Execute2 carry out the plan

1Keep only the multiples of 7

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 7: 14 is 7 times 2; 12 leaves a remainder; 42 is 7 times 6; 21 is 7 times 3. So 14, 42, 21 pass the 'divisible by 7' clue; 12 eliminated.

14 = 7 \times 2, \quad 42 = 7 \times 6, \quad 21 = 7 \times 3

Grade 3 times-tables: a number is divisible by 7 only if it appears in the 7 times table.

2Of those, keep the multiples of 6

#3 Eliminate Possibilities 3.OA.B.6

Now test 14, 42, 21 against the 'divisible by 6' clue: 14 is not in the 6 times table; 42 is 6 times 7; 21 is not in the 6 times table. So 14, 21 eliminated and 42 survives.

42 = 6 \times 7

Grade 3 unknown-factor sense: 42 divided by 6 is 7 with nothing left over, so 42 is a multiple of 6.

Answer: 42

4 · Reviewdoes it hold up?

Check 42 against both clues directly: 42 divided by 6 is 7 (exact) and 42 divided by 7 is 6 (exact). Both are whole numbers, so 42 genuinely meets both conditions, and it is the only survivor.

Another way: A number divisible by both 6 and 7 is a multiple of 42. List the multiples of 42 (42, 84, 126, ...) and notice only 42 appears among the candidates.

Standardsmin grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 7 times table to test divisibility by 7.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 6 by finding the missing factor.

💡Takeaway. Divisible by both 6 and 7 just means it shows up in both times tables, and 42 is the one that does!