← Knowing the multiple aids factoring into products · Divisibility and Remainder Reasoning

Knowing the multiple aids factoring into products · 12 practice problems

4.OA.B.44.OA.C.54.NBT.B.6

Generated variants — 12

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

Variant 1 easy answer: 1 digit

When the 4-digit number $160\blacksquare$ is a multiple of $18$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 160 with one more digit on the end, and the whole thing has to be a multiple of 18. We must count how many digits can go in that last place.

Givens

The number reads 160 then the hidden digit.
It must be a multiple of 18.

Unknowns

How many digits make it a multiple of 18.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 18 ten times is slow. Since 18 is 2 times 9 and those share no factor, a number is a multiple of 18 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 18 into 2 and 9

#2 Make a Systematic List 4.OA.B.4

2 and 9 have no common factor, so being a multiple of both is the same as being a multiple of 18.

18 = 2 \times 9

One test becomes two easier ones.

2Apply the multiple-of-2 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 2 rule narrows the field before the harder test.

\square \in \{0, 2, 4, 6, 8\}

5 digits still in the running.

3Check each survivor against the multiple-of-9 rule

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

The digits of 160 already add to 7, so the hidden digit has to bring the total to a multiple of 9 as well.

1602 \div 18 = 89

Only 2 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 18 come every 18 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 1600 through 1609 by 18 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 18 into 2 and 9 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 2.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 2 easy answer: 1 digit

When the 4-digit number $218\blacksquare$ is a multiple of $12$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 218 with one more digit on the end, and the whole thing has to be a multiple of 12. We must count how many digits can go in that last place.

Givens

The number reads 218 then the hidden digit.
It must be a multiple of 12.

Unknowns

How many digits make it a multiple of 12.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 12 ten times is slow. Since 12 is 4 times 3 and those share no factor, a number is a multiple of 12 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 12 into 4 and 3

#2 Make a Systematic List 4.OA.B.4

4 and 3 have no common factor, so being a multiple of both is the same as being a multiple of 12.

12 = 4 \times 3

One test becomes two easier ones.

2Apply the multiple-of-4 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 4 rule narrows the field before the harder test.

\square \in \{0, 4, 8\}

3 digits still in the running.

3Check each survivor against the multiple-of-3 rule

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

The digits of 218 already add to 11, so the hidden digit has to bring the total to a multiple of 3 as well.

2184 \div 12 = 182

Only 4 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 12 come every 12 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 2180 through 2189 by 12 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 12 into 4 and 3 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 4.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 3 easy answer: 1 digit

When the 4-digit number $284\blacksquare$ is a multiple of $6$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 284 with one more digit on the end, and the whole thing has to be a multiple of 6. We must count how many digits can go in that last place.

Givens

The number reads 284 then the hidden digit.
It must be a multiple of 6.

Unknowns

How many digits make it a multiple of 6.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 6 ten times is slow. Since 6 is 2 times 3 and those share no factor, a number is a multiple of 6 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 6 into 2 and 3

#2 Make a Systematic List 4.OA.B.4

2 and 3 have no common factor, so being a multiple of both is the same as being a multiple of 6.

6 = 2 \times 3

One test becomes two easier ones.

2Apply the multiple-of-2 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 2 rule narrows the field before the harder test.

\square \in \{0, 2, 4, 6, 8\}

5 digits still in the running.

3Check each survivor against the multiple-of-3 rule

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

The digits of 284 already add to 14, so the hidden digit has to bring the total to a multiple of 3 as well.

2844 \div 6 = 474

Only 4 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 6 come every 6 numbers, and ten consecutive numbers hold 1 or 2 of them -- 1 is in that range.

Another way: Dividing all ten candidates 2840 through 2849 by 6 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 6 into 2 and 3 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 2.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 4 easy answer: 1 digit

When the 4-digit number $352\blacksquare$ is a multiple of $18$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 352 with one more digit on the end, and the whole thing has to be a multiple of 18. We must count how many digits can go in that last place.

Givens

The number reads 352 then the hidden digit.
It must be a multiple of 18.

Unknowns

How many digits make it a multiple of 18.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 18 ten times is slow. Since 18 is 2 times 9 and those share no factor, a number is a multiple of 18 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 18 into 2 and 9

#2 Make a Systematic List 4.OA.B.4

2 and 9 have no common factor, so being a multiple of both is the same as being a multiple of 18.

18 = 2 \times 9

One test becomes two easier ones.

2Apply the multiple-of-2 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 2 rule narrows the field before the harder test.

\square \in \{0, 2, 4, 6, 8\}

5 digits still in the running.

3Check each survivor against the multiple-of-9 rule

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

The digits of 352 already add to 10, so the hidden digit has to bring the total to a multiple of 9 as well.

3528 \div 18 = 196

Only 8 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 18 come every 18 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 3520 through 3529 by 18 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 18 into 2 and 9 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 2.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 5 medium answer: 1 digit

When the 4-digit number $372\blacksquare$ is a multiple of $15$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 372 with one more digit on the end, and the whole thing has to be a multiple of 15. We must count how many digits can go in that last place.

Givens

The number reads 372 then the hidden digit.
It must be a multiple of 15.

Unknowns

How many digits make it a multiple of 15.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 15 ten times is slow. Since 15 is 3 times 5 and those share no factor, a number is a multiple of 15 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 15 into 3 and 5

#2 Make a Systematic List 4.OA.B.4

3 and 5 have no common factor, so being a multiple of both is the same as being a multiple of 15.

15 = 3 \times 5

One test becomes two easier ones.

2Apply the multiple-of-3 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 3 rule narrows the field before the harder test.

\square \in \{0, 3, 6, 9\}

4 digits still in the running.

3Check each survivor against the multiple-of-5 rule

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

Test each survivor against 5 directly.

3720 \div 15 = 248

Only 0 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 15 come every 15 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 3720 through 3729 by 15 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 15 into 3 and 5 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 3.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 6 medium answer: 1 digit

When the 4-digit number $497\blacksquare$ is a multiple of $6$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 497 with one more digit on the end, and the whole thing has to be a multiple of 6. We must count how many digits can go in that last place.

Givens

The number reads 497 then the hidden digit.
It must be a multiple of 6.

Unknowns

How many digits make it a multiple of 6.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 6 ten times is slow. Since 6 is 2 times 3 and those share no factor, a number is a multiple of 6 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 6 into 2 and 3

#2 Make a Systematic List 4.OA.B.4

2 and 3 have no common factor, so being a multiple of both is the same as being a multiple of 6.

6 = 2 \times 3

One test becomes two easier ones.

2Apply the multiple-of-2 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 2 rule narrows the field before the harder test.

\square \in \{0, 2, 4, 6, 8\}

5 digits still in the running.

3Check each survivor against the multiple-of-3 rule

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

The digits of 497 already add to 20, so the hidden digit has to bring the total to a multiple of 3 as well.

4974 \div 6 = 829

Only 4 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 6 come every 6 numbers, and ten consecutive numbers hold 1 or 2 of them -- 1 is in that range.

Another way: Dividing all ten candidates 4970 through 4979 by 6 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 6 into 2 and 3 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 2.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 7 medium answer: 1 digit

When the 4-digit number $526\blacksquare$ is a multiple of $10$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 526 with one more digit on the end, and the whole thing has to be a multiple of 10. We must count how many digits can go in that last place.

Givens

The number reads 526 then the hidden digit.
It must be a multiple of 10.

Unknowns

How many digits make it a multiple of 10.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 10 ten times is slow. Since 10 is 2 times 5 and those share no factor, a number is a multiple of 10 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 10 into 2 and 5

#2 Make a Systematic List 4.OA.B.4

2 and 5 have no common factor, so being a multiple of both is the same as being a multiple of 10.

10 = 2 \times 5

One test becomes two easier ones.

2Apply the multiple-of-2 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 2 rule narrows the field before the harder test.

\square \in \{0, 2, 4, 6, 8\}

5 digits still in the running.

3Check each survivor against the multiple-of-5 rule

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

Test each survivor against 5 directly.

5260 \div 10 = 526

Only 0 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 10 come every 10 numbers, and ten consecutive numbers hold 1 or 2 of them -- 1 is in that range.

Another way: Dividing all ten candidates 5260 through 5269 by 10 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 10 into 2 and 5 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 2.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 8 medium answer: 1 digit

When the 4-digit number $585\blacksquare$ is a multiple of $45$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 585 with one more digit on the end, and the whole thing has to be a multiple of 45. We must count how many digits can go in that last place.

Givens

The number reads 585 then the hidden digit.
It must be a multiple of 45.

Unknowns

How many digits make it a multiple of 45.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 45 ten times is slow. Since 45 is 9 times 5 and those share no factor, a number is a multiple of 45 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 45 into 9 and 5

#2 Make a Systematic List 4.OA.B.4

9 and 5 have no common factor, so being a multiple of both is the same as being a multiple of 45.

45 = 9 \times 5

One test becomes two easier ones.

2Apply the multiple-of-9 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 9 rule narrows the field before the harder test.

\square \in \{0, 9\}

2 digits still in the running.

3Check each survivor against the multiple-of-5 rule

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

Test each survivor against 5 directly.

5850 \div 45 = 130

Only 0 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 45 come every 45 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 5850 through 5859 by 45 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 45 into 9 and 5 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 9.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 9 hard answer: 1 digit

When the 4-digit number $621\blacksquare$ is a multiple of $45$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 621 with one more digit on the end, and the whole thing has to be a multiple of 45. We must count how many digits can go in that last place.

Givens

The number reads 621 then the hidden digit.
It must be a multiple of 45.

Unknowns

How many digits make it a multiple of 45.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 45 ten times is slow. Since 45 is 9 times 5 and those share no factor, a number is a multiple of 45 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 45 into 9 and 5

#2 Make a Systematic List 4.OA.B.4

9 and 5 have no common factor, so being a multiple of both is the same as being a multiple of 45.

45 = 9 \times 5

One test becomes two easier ones.

2Apply the multiple-of-9 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 9 rule narrows the field before the harder test.

\square \in \{0, 9\}

2 digits still in the running.

3Check each survivor against the multiple-of-5 rule

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

Test each survivor against 5 directly.

6210 \div 45 = 138

Only 0 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 45 come every 45 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 6210 through 6219 by 45 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 45 into 9 and 5 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 9.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 10 hard answer: 1 digit

When the 4-digit number $735\blacksquare$ is a multiple of $15$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 735 with one more digit on the end, and the whole thing has to be a multiple of 15. We must count how many digits can go in that last place.

Givens

The number reads 735 then the hidden digit.
It must be a multiple of 15.

Unknowns

How many digits make it a multiple of 15.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 15 ten times is slow. Since 15 is 3 times 5 and those share no factor, a number is a multiple of 15 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 15 into 3 and 5

#2 Make a Systematic List 4.OA.B.4

3 and 5 have no common factor, so being a multiple of both is the same as being a multiple of 15.

15 = 3 \times 5

One test becomes two easier ones.

2Apply the multiple-of-3 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 3 rule narrows the field before the harder test.

\square \in \{0, 3, 6, 9\}

4 digits still in the running.

3Check each survivor against the multiple-of-5 rule

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

Test each survivor against 5 directly.

7350 \div 15 = 490

Only 0 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 15 come every 15 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 7350 through 7359 by 15 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 15 into 3 and 5 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 3.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 11 hard answer: 1 digit

When the 4-digit number $843\blacksquare$ is a multiple of $12$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 843 with one more digit on the end, and the whole thing has to be a multiple of 12. We must count how many digits can go in that last place.

Givens

The number reads 843 then the hidden digit.
It must be a multiple of 12.

Unknowns

How many digits make it a multiple of 12.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 12 ten times is slow. Since 12 is 4 times 3 and those share no factor, a number is a multiple of 12 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 12 into 4 and 3

#2 Make a Systematic List 4.OA.B.4

4 and 3 have no common factor, so being a multiple of both is the same as being a multiple of 12.

12 = 4 \times 3

One test becomes two easier ones.

2Apply the multiple-of-4 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 4 rule narrows the field before the harder test.

\square \in \{2, 6\}

2 digits still in the running.

3Check each survivor against the multiple-of-3 rule

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

The digits of 843 already add to 15, so the hidden digit has to bring the total to a multiple of 3 as well.

8436 \div 12 = 703

Only 6 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 12 come every 12 numbers, and ten consecutive numbers hold 0 or 1 of them -- 1 is in that range.

Another way: Dividing all ten candidates 8430 through 8439 by 12 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 12 into 4 and 3 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 4.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.

Variant 12 hard answer: 1 digit

When the 4-digit number $964\blacksquare$ is a multiple of $10$ , how many digits can go in the $\blacksquare$ ?

Show solution

1 · Understandwhat's really being asked

The number is 964 with one more digit on the end, and the whole thing has to be a multiple of 10. We must count how many digits can go in that last place.

Givens

The number reads 964 then the hidden digit.
It must be a multiple of 10.

Unknowns

How many digits make it a multiple of 10.

Constraints

The hidden digit is one of 0 through 9.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

Dividing by 10 ten times is slow. Since 10 is 2 times 5 and those share no factor, a number is a multiple of 10 exactly when it is a multiple of both -- and each has an easy digit test.

3 · Execute4 carry out the plan

1Split 10 into 2 and 5

#2 Make a Systematic List 4.OA.B.4

2 and 5 have no common factor, so being a multiple of both is the same as being a multiple of 10.

10 = 2 \times 5

One test becomes two easier ones.

2Apply the multiple-of-2 test

#2 Make a Systematic List 4.OA.C.5

Checking each digit for the 2 rule narrows the field before the harder test.

\square \in \{0, 2, 4, 6, 8\}

5 digits still in the running.

3Check each survivor against the multiple-of-5 rule

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

Test each survivor against 5 directly.

9640 \div 10 = 964

Only 0 pass both.

4Count the digits that work

#2 Make a Systematic List 4.NBT.B.6

The survivors are the answer.

1

1 digit work.

Answer: 1 digit

4 · Reviewdoes it hold up?

Multiples of 10 come every 10 numbers, and ten consecutive numbers hold 1 or 2 of them -- 1 is in that range.

Another way: Dividing all ten candidates 9640 through 9649 by 10 finds the same 1, at ten long divisions instead of two digit rules.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Splitting 10 into 2 and 5 and testing both.
4.OA.C.5 Generate a number or shape pattern following a given rule — Using the digit pattern for multiples of 2.
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Confirming each surviving digit by division.

💡Takeaway. Break the divisor into two pieces that share no factor, and one hard check becomes two easy ones.