Practice · Common factor over 1 means not in lowest terms · Sensim Math

Variant 1 easy

Among the proper fractions with denominator 45,
$\frac{1}{45},\ \frac{2}{45},\ \frac{3}{45},\ \cdots,\ \frac{42}{45},\ \frac{43}{45},\ \frac{44}{45},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 45 has a numerator from 1 to 44. We must count how many of them cannot be reduced.

Givens

The denominator is always 45.
The numerators run 1 through 44.

Unknowns

How many of the 44 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 45's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 45 into primes to see which numerators can share a factor with it.

45 = 3 \times 3 \times 5

Only multiples of 3 or 5 can reduce.

2Count the total

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

The numerators run from 1 to 44.

45 - 1 = 44

44 fractions to sort through.

3Count numerators that are multiples of 3

#16 Count the Complement 4.OA.B.4

Every 3th numerator shares the factor 3. Since 3 divides 45 exactly, there are 45 / 3 multiples up to 45, and one of them is 45 itself.

45 \div 3 - 1 = 14

14 of them reduce because of 3.

4Count numerators that are multiples of 5

#16 Count the Complement 4.OA.B.4

Likewise every 5th numerator shares the factor 5.

45 \div 5 - 1 = 8

8 of them reduce because of 5.

5Fix the double-counting (multiples of 15)

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

A multiple of 15 was counted in both lists, so it has been counted twice.

45 \div 15 - 1 = 2

2 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

14 + 8 - 2 = 20

20 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

44 - 20 = 24

24 fractions are already in lowest terms.

Answer: 24

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 45 shares nothing with 45 and counts, while 3 over 45 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 44 for a shared factor; it gives the same 24, at 44 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 45 and counting multiples of 3, 5 and 15.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 2 easy

Among the proper fractions with denominator 50,
$\frac{1}{50},\ \frac{2}{50},\ \frac{3}{50},\ \cdots,\ \frac{47}{50},\ \frac{48}{50},\ \frac{49}{50},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 50 has a numerator from 1 to 49. We must count how many of them cannot be reduced.

Givens

The denominator is always 50.
The numerators run 1 through 49.

Unknowns

How many of the 49 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 50's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 50 into primes to see which numerators can share a factor with it.

50 = 2 \times 5 \times 5

Only multiples of 2 or 5 can reduce.

2Count the total

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

The numerators run from 1 to 49.

50 - 1 = 49

49 fractions to sort through.

3Count numerators that are multiples of 2

#16 Count the Complement 4.OA.B.4

Every 2th numerator shares the factor 2. Since 2 divides 50 exactly, there are 50 / 2 multiples up to 50, and one of them is 50 itself.

50 \div 2 - 1 = 24

24 of them reduce because of 2.

4Count numerators that are multiples of 5

#16 Count the Complement 4.OA.B.4

Likewise every 5th numerator shares the factor 5.

50 \div 5 - 1 = 9

9 of them reduce because of 5.

5Fix the double-counting (multiples of 10)

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

A multiple of 10 was counted in both lists, so it has been counted twice.

50 \div 10 - 1 = 4

4 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

24 + 9 - 4 = 29

29 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

49 - 29 = 20

20 fractions are already in lowest terms.

Answer: 20

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 50 shares nothing with 50 and counts, while 2 over 50 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 49 for a shared factor; it gives the same 20, at 49 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 50 and counting multiples of 2, 5 and 10.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 3 easy

Among the proper fractions with denominator 55,
$\frac{1}{55},\ \frac{2}{55},\ \frac{3}{55},\ \cdots,\ \frac{52}{55},\ \frac{53}{55},\ \frac{54}{55},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 55 has a numerator from 1 to 54. We must count how many of them cannot be reduced.

Givens

The denominator is always 55.
The numerators run 1 through 54.

Unknowns

How many of the 54 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 55's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 55 into primes to see which numerators can share a factor with it.

55 = 5 \times 11

Only multiples of 5 or 11 can reduce.

2Count the total

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

The numerators run from 1 to 54.

55 - 1 = 54

54 fractions to sort through.

3Count numerators that are multiples of 5

#16 Count the Complement 4.OA.B.4

Every 5th numerator shares the factor 5. Since 5 divides 55 exactly, there are 55 / 5 multiples up to 55, and one of them is 55 itself.

55 \div 5 - 1 = 10

10 of them reduce because of 5.

4Count numerators that are multiples of 11

#16 Count the Complement 4.OA.B.4

Likewise every 11th numerator shares the factor 11.

55 \div 11 - 1 = 4

4 of them reduce because of 11.

5Fix the double-counting (multiples of 55)

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

A multiple of 55 was counted in both lists, so it has been counted twice.

55 \div 55 - 1 = 0

0 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

10 + 4 - 0 = 14

14 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

54 - 14 = 40

40 fractions are already in lowest terms.

Answer: 40

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 55 shares nothing with 55 and counts, while 5 over 55 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 54 for a shared factor; it gives the same 40, at 54 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 55 and counting multiples of 5, 11 and 55.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 4 easy

Among the proper fractions with denominator 63,
$\frac{1}{63},\ \frac{2}{63},\ \frac{3}{63},\ \cdots,\ \frac{60}{63},\ \frac{61}{63},\ \frac{62}{63},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 63 has a numerator from 1 to 62. We must count how many of them cannot be reduced.

Givens

The denominator is always 63.
The numerators run 1 through 62.

Unknowns

How many of the 62 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 63's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 63 into primes to see which numerators can share a factor with it.

63 = 3 \times 3 \times 7

Only multiples of 3 or 7 can reduce.

2Count the total

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

The numerators run from 1 to 62.

63 - 1 = 62

62 fractions to sort through.

3Count numerators that are multiples of 3

#16 Count the Complement 4.OA.B.4

Every 3th numerator shares the factor 3. Since 3 divides 63 exactly, there are 63 / 3 multiples up to 63, and one of them is 63 itself.

63 \div 3 - 1 = 20

20 of them reduce because of 3.

4Count numerators that are multiples of 7

#16 Count the Complement 4.OA.B.4

Likewise every 7th numerator shares the factor 7.

63 \div 7 - 1 = 8

8 of them reduce because of 7.

5Fix the double-counting (multiples of 21)

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

A multiple of 21 was counted in both lists, so it has been counted twice.

63 \div 21 - 1 = 2

2 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

20 + 8 - 2 = 26

26 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

62 - 26 = 36

36 fractions are already in lowest terms.

Answer: 36

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 63 shares nothing with 63 and counts, while 3 over 63 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 62 for a shared factor; it gives the same 36, at 62 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 63 and counting multiples of 3, 7 and 21.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 5 medium

Among the proper fractions with denominator 75,
$\frac{1}{75},\ \frac{2}{75},\ \frac{3}{75},\ \cdots,\ \frac{72}{75},\ \frac{73}{75},\ \frac{74}{75},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 75 has a numerator from 1 to 74. We must count how many of them cannot be reduced.

Givens

The denominator is always 75.
The numerators run 1 through 74.

Unknowns

How many of the 74 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 75's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 75 into primes to see which numerators can share a factor with it.

75 = 3 \times 5 \times 5

Only multiples of 3 or 5 can reduce.

2Count the total

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

The numerators run from 1 to 74.

75 - 1 = 74

74 fractions to sort through.

3Count numerators that are multiples of 3

#16 Count the Complement 4.OA.B.4

Every 3th numerator shares the factor 3. Since 3 divides 75 exactly, there are 75 / 3 multiples up to 75, and one of them is 75 itself.

75 \div 3 - 1 = 24

24 of them reduce because of 3.

4Count numerators that are multiples of 5

#16 Count the Complement 4.OA.B.4

Likewise every 5th numerator shares the factor 5.

75 \div 5 - 1 = 14

14 of them reduce because of 5.

5Fix the double-counting (multiples of 15)

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

A multiple of 15 was counted in both lists, so it has been counted twice.

75 \div 15 - 1 = 4

4 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

24 + 14 - 4 = 34

34 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

74 - 34 = 40

40 fractions are already in lowest terms.

Answer: 40

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 75 shares nothing with 75 and counts, while 3 over 75 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 74 for a shared factor; it gives the same 40, at 74 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 75 and counting multiples of 3, 5 and 15.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 6 medium

Among the proper fractions with denominator 77,
$\frac{1}{77},\ \frac{2}{77},\ \frac{3}{77},\ \cdots,\ \frac{74}{77},\ \frac{75}{77},\ \frac{76}{77},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 77 has a numerator from 1 to 76. We must count how many of them cannot be reduced.

Givens

The denominator is always 77.
The numerators run 1 through 76.

Unknowns

How many of the 76 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 77's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 77 into primes to see which numerators can share a factor with it.

77 = 7 \times 11

Only multiples of 7 or 11 can reduce.

2Count the total

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

The numerators run from 1 to 76.

77 - 1 = 76

76 fractions to sort through.

3Count numerators that are multiples of 7

#16 Count the Complement 4.OA.B.4

Every 7th numerator shares the factor 7. Since 7 divides 77 exactly, there are 77 / 7 multiples up to 77, and one of them is 77 itself.

77 \div 7 - 1 = 10

10 of them reduce because of 7.

4Count numerators that are multiples of 11

#16 Count the Complement 4.OA.B.4

Likewise every 11th numerator shares the factor 11.

77 \div 11 - 1 = 6

6 of them reduce because of 11.

5Fix the double-counting (multiples of 77)

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

A multiple of 77 was counted in both lists, so it has been counted twice.

77 \div 77 - 1 = 0

0 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

10 + 6 - 0 = 16

16 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

76 - 16 = 60

60 fractions are already in lowest terms.

Answer: 60

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 77 shares nothing with 77 and counts, while 7 over 77 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 76 for a shared factor; it gives the same 60, at 76 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 77 and counting multiples of 7, 11 and 77.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 7 medium

Among the proper fractions with denominator 88,
$\frac{1}{88},\ \frac{2}{88},\ \frac{3}{88},\ \cdots,\ \frac{85}{88},\ \frac{86}{88},\ \frac{87}{88},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 88 has a numerator from 1 to 87. We must count how many of them cannot be reduced.

Givens

The denominator is always 88.
The numerators run 1 through 87.

Unknowns

How many of the 87 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 88's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 88 into primes to see which numerators can share a factor with it.

88 = 2 \times 2 \times 2 \times 11

Only multiples of 2 or 11 can reduce.

2Count the total

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

The numerators run from 1 to 87.

88 - 1 = 87

87 fractions to sort through.

3Count numerators that are multiples of 2

#16 Count the Complement 4.OA.B.4

Every 2th numerator shares the factor 2. Since 2 divides 88 exactly, there are 88 / 2 multiples up to 88, and one of them is 88 itself.

88 \div 2 - 1 = 43

43 of them reduce because of 2.

4Count numerators that are multiples of 11

#16 Count the Complement 4.OA.B.4

Likewise every 11th numerator shares the factor 11.

88 \div 11 - 1 = 7

7 of them reduce because of 11.

5Fix the double-counting (multiples of 22)

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

A multiple of 22 was counted in both lists, so it has been counted twice.

88 \div 22 - 1 = 3

3 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

43 + 7 - 3 = 47

47 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

87 - 47 = 40

40 fractions are already in lowest terms.

Answer: 40

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 88 shares nothing with 88 and counts, while 2 over 88 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 87 for a shared factor; it gives the same 40, at 87 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 88 and counting multiples of 2, 11 and 22.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 8 medium

Among the proper fractions with denominator 91,
$\frac{1}{91},\ \frac{2}{91},\ \frac{3}{91},\ \cdots,\ \frac{88}{91},\ \frac{89}{91},\ \frac{90}{91},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 91 has a numerator from 1 to 90. We must count how many of them cannot be reduced.

Givens

The denominator is always 91.
The numerators run 1 through 90.

Unknowns

How many of the 90 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 91's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 91 into primes to see which numerators can share a factor with it.

91 = 7 \times 13

Only multiples of 7 or 13 can reduce.

2Count the total

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

The numerators run from 1 to 90.

91 - 1 = 90

90 fractions to sort through.

3Count numerators that are multiples of 7

#16 Count the Complement 4.OA.B.4

Every 7th numerator shares the factor 7. Since 7 divides 91 exactly, there are 91 / 7 multiples up to 91, and one of them is 91 itself.

91 \div 7 - 1 = 12

12 of them reduce because of 7.

4Count numerators that are multiples of 13

#16 Count the Complement 4.OA.B.4

Likewise every 13th numerator shares the factor 13.

91 \div 13 - 1 = 6

6 of them reduce because of 13.

5Fix the double-counting (multiples of 91)

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

A multiple of 91 was counted in both lists, so it has been counted twice.

91 \div 91 - 1 = 0

0 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

12 + 6 - 0 = 18

18 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

90 - 18 = 72

72 fractions are already in lowest terms.

Answer: 72

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 91 shares nothing with 91 and counts, while 7 over 91 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 90 for a shared factor; it gives the same 72, at 90 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 91 and counting multiples of 7, 13 and 91.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 9 hard

Among the proper fractions with denominator 98,
$\frac{1}{98},\ \frac{2}{98},\ \frac{3}{98},\ \cdots,\ \frac{95}{98},\ \frac{96}{98},\ \frac{97}{98},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 98 has a numerator from 1 to 97. We must count how many of them cannot be reduced.

Givens

The denominator is always 98.
The numerators run 1 through 97.

Unknowns

How many of the 97 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 98's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 98 into primes to see which numerators can share a factor with it.

98 = 2 \times 7 \times 7

Only multiples of 2 or 7 can reduce.

2Count the total

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

The numerators run from 1 to 97.

98 - 1 = 97

97 fractions to sort through.

3Count numerators that are multiples of 2

#16 Count the Complement 4.OA.B.4

Every 2th numerator shares the factor 2. Since 2 divides 98 exactly, there are 98 / 2 multiples up to 98, and one of them is 98 itself.

98 \div 2 - 1 = 48

48 of them reduce because of 2.

4Count numerators that are multiples of 7

#16 Count the Complement 4.OA.B.4

Likewise every 7th numerator shares the factor 7.

98 \div 7 - 1 = 13

13 of them reduce because of 7.

5Fix the double-counting (multiples of 14)

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

A multiple of 14 was counted in both lists, so it has been counted twice.

98 \div 14 - 1 = 6

6 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

48 + 13 - 6 = 55

55 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

97 - 55 = 42

42 fractions are already in lowest terms.

Answer: 42

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 98 shares nothing with 98 and counts, while 2 over 98 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 97 for a shared factor; it gives the same 42, at 97 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 98 and counting multiples of 2, 7 and 14.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 10 hard

Among the proper fractions with denominator 115,
$\frac{1}{115},\ \frac{2}{115},\ \frac{3}{115},\ \cdots,\ \frac{112}{115},\ \frac{113}{115},\ \frac{114}{115},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 115 has a numerator from 1 to 114. We must count how many of them cannot be reduced.

Givens

The denominator is always 115.
The numerators run 1 through 114.

Unknowns

How many of the 114 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 115's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 115 into primes to see which numerators can share a factor with it.

115 = 5 \times 23

Only multiples of 5 or 23 can reduce.

2Count the total

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

The numerators run from 1 to 114.

115 - 1 = 114

114 fractions to sort through.

3Count numerators that are multiples of 5

#16 Count the Complement 4.OA.B.4

Every 5th numerator shares the factor 5. Since 5 divides 115 exactly, there are 115 / 5 multiples up to 115, and one of them is 115 itself.

115 \div 5 - 1 = 22

22 of them reduce because of 5.

4Count numerators that are multiples of 23

#16 Count the Complement 4.OA.B.4

Likewise every 23th numerator shares the factor 23.

115 \div 23 - 1 = 4

4 of them reduce because of 23.

5Fix the double-counting (multiples of 115)

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

A multiple of 115 was counted in both lists, so it has been counted twice.

115 \div 115 - 1 = 0

0 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

22 + 4 - 0 = 26

26 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

114 - 26 = 88

88 fractions are already in lowest terms.

Answer: 88

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 115 shares nothing with 115 and counts, while 5 over 115 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 114 for a shared factor; it gives the same 88, at 114 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 115 and counting multiples of 5, 23 and 115.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 11 hard

Among the proper fractions with denominator 143,
$\frac{1}{143},\ \frac{2}{143},\ \frac{3}{143},\ \cdots,\ \frac{140}{143},\ \frac{141}{143},\ \frac{142}{143},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 143 has a numerator from 1 to 142. We must count how many of them cannot be reduced.

Givens

The denominator is always 143.
The numerators run 1 through 142.

Unknowns

How many of the 142 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 143's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 143 into primes to see which numerators can share a factor with it.

143 = 11 \times 13

Only multiples of 11 or 13 can reduce.

2Count the total

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

The numerators run from 1 to 142.

143 - 1 = 142

142 fractions to sort through.

3Count numerators that are multiples of 11

#16 Count the Complement 4.OA.B.4

Every 11th numerator shares the factor 11. Since 11 divides 143 exactly, there are 143 / 11 multiples up to 143, and one of them is 143 itself.

143 \div 11 - 1 = 12

12 of them reduce because of 11.

4Count numerators that are multiples of 13

#16 Count the Complement 4.OA.B.4

Likewise every 13th numerator shares the factor 13.

143 \div 13 - 1 = 10

10 of them reduce because of 13.

5Fix the double-counting (multiples of 143)

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

A multiple of 143 was counted in both lists, so it has been counted twice.

143 \div 143 - 1 = 0

0 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

12 + 10 - 0 = 22

22 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

142 - 22 = 120

120 fractions are already in lowest terms.

Answer: 120

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 143 shares nothing with 143 and counts, while 11 over 143 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 142 for a shared factor; it gives the same 120, at 142 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 143 and counting multiples of 11, 13 and 143.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Variant 12 hard

Among the proper fractions with denominator 175,
$\frac{1}{175},\ \frac{2}{175},\ \frac{3}{175},\ \cdots,\ \frac{172}{175},\ \frac{173}{175},\ \frac{174}{175},$
how many are in lowest terms?

Your answer

Show solution

1 · Understandwhat's really being asked

Every proper fraction with denominator 175 has a numerator from 1 to 174. We must count how many of them cannot be reduced.

Givens

The denominator is always 175.
The numerators run 1 through 174.

Unknowns

How many of the 174 fractions are already in lowest terms.

Constraints

A fraction reduces exactly when its numerator and denominator share a factor bigger than 1.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Counting the reducible ones is easier than counting the rest: they are just the multiples of 175's prime factors. Count those, subtract from the total, and watch for numerators counted twice.

3 · Execute7 carry out the plan

1Factor the denominator

#9 Solve an Easier Related Problem 4.OA.B.4

Break 175 into primes to see which numerators can share a factor with it.

175 = 5 \times 5 \times 7

Only multiples of 5 or 7 can reduce.

2Count the total

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

The numerators run from 1 to 174.

175 - 1 = 174

174 fractions to sort through.

3Count numerators that are multiples of 5

#16 Count the Complement 4.OA.B.4

Every 5th numerator shares the factor 5. Since 5 divides 175 exactly, there are 175 / 5 multiples up to 175, and one of them is 175 itself.

175 \div 5 - 1 = 34

34 of them reduce because of 5.

4Count numerators that are multiples of 7

#16 Count the Complement 4.OA.B.4

Likewise every 7th numerator shares the factor 7.

175 \div 7 - 1 = 24

24 of them reduce because of 7.

5Fix the double-counting (multiples of 35)

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

A multiple of 35 was counted in both lists, so it has been counted twice.

175 \div 35 - 1 = 4

4 numerators counted twice.

6Combine to find the 'not lowest terms' count

#16 Count the Complement 4.NF.A.1

Add the two lists, then take off the overlap once.

34 + 24 - 4 = 54

54 of the fractions can be reduced.

7Subtract from the total

#16 Count the Complement 4.NF.A.1

Whatever is left cannot be reduced.

174 - 54 = 120

120 fractions are already in lowest terms.

Answer: 120

4 · Reviewdoes it hold up?

Spot-check the smallest: 1 over 175 shares nothing with 175 and counts, while 5 over 175 reduces -- both land on the right side of the tally.

Another way: Test every numerator from 1 to 174 for a shared factor; it gives the same 120, at 174 checks instead of four divisions.

Standardsmin grade 4

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Factoring 175 and counting multiples of 5, 7 and 35.
4.NF.A.1 Explain why a fraction is equivalent to another fraction — Reading 'shares a factor' as 'can be reduced'.

💡Takeaway. Counting what you do NOT want is often easier -- just remember to subtract the ones you counted twice.

Common factor over 1 means not in lowest terms · 12 practice problems

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