← Common factor over 1 means not in lowest terms · Divisibility and Remainder Reasoning

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

4.OA.B.44.NF.A.1

Generated variants — 12

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

Variant 1 easy answer: 24

Among the proper fractions with denominator 45,
145, 245, 345, ⋯ , 4245, 4345, 4445,\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?

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=3×3×545 = 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=4445 - 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÷3−1=1445 \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÷5−1=845 \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÷15−1=245 \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=2014 + 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=2444 - 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 answer: 20

Among the proper fractions with denominator 50,
150, 250, 350, ⋯ , 4750, 4850, 4950,\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?

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=2×5×550 = 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=4950 - 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÷2−1=2450 \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÷5−1=950 \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÷10−1=450 \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=2924 + 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=2049 - 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 answer: 40

Among the proper fractions with denominator 55,
155, 255, 355, ⋯ , 5255, 5355, 5455,\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?

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=5×1155 = 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=5455 - 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÷5−1=1055 \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÷11−1=455 \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÷55−1=055 \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=1410 + 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=4054 - 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 answer: 36

Among the proper fractions with denominator 63,
163, 263, 363, ⋯ , 6063, 6163, 6263,\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?

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=3×3×763 = 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=6263 - 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÷3−1=2063 \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÷7−1=863 \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÷21−1=263 \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=2620 + 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=3662 - 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 answer: 40

Among the proper fractions with denominator 75,
175, 275, 375, ⋯ , 7275, 7375, 7475,\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?

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=3×5×575 = 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=7475 - 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÷3−1=2475 \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÷5−1=1475 \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÷15−1=475 \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=3424 + 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=4074 - 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 answer: 60

Among the proper fractions with denominator 77,
177, 277, 377, ⋯ , 7477, 7577, 7677,\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?

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=7×1177 = 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=7677 - 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÷7−1=1077 \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÷11−1=677 \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÷77−1=077 \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=1610 + 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=6076 - 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 answer: 40

Among the proper fractions with denominator 88,
188, 288, 388, ⋯ , 8588, 8688, 8788,\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?

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=2×2×2×1188 = 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=8788 - 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÷2−1=4388 \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÷11−1=788 \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÷22−1=388 \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=4743 + 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=4087 - 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 answer: 72

Among the proper fractions with denominator 91,
191, 291, 391, ⋯ , 8891, 8991, 9091,\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?

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=7×1391 = 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=9091 - 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÷7−1=1291 \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÷13−1=691 \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÷91−1=091 \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=1812 + 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=7290 - 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 answer: 42

Among the proper fractions with denominator 98,
198, 298, 398, ⋯ , 9598, 9698, 9798,\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?

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=2×7×798 = 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=9798 - 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÷2−1=4898 \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÷7−1=1398 \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÷14−1=698 \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=5548 + 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=4297 - 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 answer: 88

Among the proper fractions with denominator 115,
1115, 2115, 3115, ⋯ , 112115, 113115, 114115,\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?

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=5×23115 = 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=114115 - 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÷5−1=22115 \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÷23−1=4115 \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÷115−1=0115 \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=2622 + 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=88114 - 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 answer: 120

Among the proper fractions with denominator 143,
1143, 2143, 3143, ⋯ , 140143, 141143, 142143,\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?

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=11×13143 = 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=142143 - 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÷11−1=12143 \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÷13−1=10143 \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÷143−1=0143 \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=2212 + 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=120142 - 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 answer: 120

Among the proper fractions with denominator 175,
1175, 2175, 3175, ⋯ , 172175, 173175, 174175,\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?

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=5×5×7175 = 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=174175 - 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÷5−1=34175 \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÷7−1=24175 \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÷35−1=4175 \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=5434 + 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=120174 - 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.