Practice · Run the sum rule backwards on unit fractions · Sensim Math

Variant 1 easy

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 9 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 10.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 9 begins $\frac{1}{9}$ , $\frac{1}{A}$ ; row 10 begins $\frac{1}{10}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 8 is not drawn, but by the pattern it starts with $\frac{1}{8}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 8

#11 Work Backwards 6.NS.B.4

Row 8 starts with

\frac{1}{8}

, and the two numbers below it are

\frac{1}{9}

and

\frac{1}{A}

. So

\frac{1}{8} = \frac{1}{9} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

72

.

\frac{1}{A} = \frac{1}{8} - \frac{1}{9} = \frac{9-8}{72} = \frac{1}{72}

So

A = 72

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{9}

that starts row 9 are

\frac{1}{10}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{9} - \frac{1}{10} = \frac{10-9}{90} = \frac{1}{90}

So

B = 90

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{72}

are

\frac{1}{B} = \frac{1}{90}

and

\frac{1}{C}

. The least common multiple of

72

and

90

is

360

.

\frac{1}{C} = \frac{1}{72} - \frac{1}{90} = \frac{5-4}{360} = \frac{1}{360}

So

C = 360

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

72 + 90 + 360 = 522

A+B+C = 522

.

Answer: 522

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{90} + \frac{1}{360} = \frac{4+1}{360} = \frac{1}{72}$ and $\frac{1}{9} + \frac{1}{72} = \frac{8+1}{72} = \frac{1}{8}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 9 has $\frac{1}{9 \times 8}$ and row 10 has $\frac{1}{10 \times 9}$ , giving $A = 72$ and $B = 90$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $72$, $90$, $360$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 2 easy

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 6 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 7.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 6 begins $\frac{1}{6}$ , $\frac{1}{A}$ ; row 7 begins $\frac{1}{7}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 5 is not drawn, but by the pattern it starts with $\frac{1}{5}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 5

#11 Work Backwards 6.NS.B.4

Row 5 starts with

\frac{1}{5}

, and the two numbers below it are

\frac{1}{6}

and

\frac{1}{A}

. So

\frac{1}{5} = \frac{1}{6} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

30

.

\frac{1}{A} = \frac{1}{5} - \frac{1}{6} = \frac{6-5}{30} = \frac{1}{30}

So

A = 30

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{6}

that starts row 6 are

\frac{1}{7}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{6} - \frac{1}{7} = \frac{7-6}{42} = \frac{1}{42}

So

B = 42

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{30}

are

\frac{1}{B} = \frac{1}{42}

and

\frac{1}{C}

. The least common multiple of

30

and

42

is

210

.

\frac{1}{C} = \frac{1}{30} - \frac{1}{42} = \frac{7-5}{210} = \frac{2}{210} = \frac{1}{105}

So

C = 105

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

30 + 42 + 105 = 177

A+B+C = 177

.

Answer: 177

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{42} + \frac{1}{105} = \frac{5+2}{210} = \frac{1}{30}$ and $\frac{1}{6} + \frac{1}{30} = \frac{5+1}{30} = \frac{1}{5}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 6 has $\frac{1}{6 \times 5}$ and row 7 has $\frac{1}{7 \times 6}$ , giving $A = 30$ and $B = 42$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $30$, $42$, $210$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 3 easy

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 8 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 9.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 8 begins $\frac{1}{8}$ , $\frac{1}{A}$ ; row 9 begins $\frac{1}{9}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 7 is not drawn, but by the pattern it starts with $\frac{1}{7}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 7

#11 Work Backwards 6.NS.B.4

Row 7 starts with

\frac{1}{7}

, and the two numbers below it are

\frac{1}{8}

and

\frac{1}{A}

. So

\frac{1}{7} = \frac{1}{8} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

56

.

\frac{1}{A} = \frac{1}{7} - \frac{1}{8} = \frac{8-7}{56} = \frac{1}{56}

So

A = 56

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{8}

that starts row 8 are

\frac{1}{9}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{8} - \frac{1}{9} = \frac{9-8}{72} = \frac{1}{72}

So

B = 72

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{56}

are

\frac{1}{B} = \frac{1}{72}

and

\frac{1}{C}

. The least common multiple of

56

and

72

is

504

.

\frac{1}{C} = \frac{1}{56} - \frac{1}{72} = \frac{9-7}{504} = \frac{2}{504} = \frac{1}{252}

So

C = 252

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

56 + 72 + 252 = 380

A+B+C = 380

.

Answer: 380

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{72} + \frac{1}{252} = \frac{7+2}{504} = \frac{1}{56}$ and $\frac{1}{8} + \frac{1}{56} = \frac{7+1}{56} = \frac{1}{7}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 8 has $\frac{1}{8 \times 7}$ and row 9 has $\frac{1}{9 \times 8}$ , giving $A = 56$ and $B = 72$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $56$, $72$, $504$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 4 medium

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 10 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 11.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 10 begins $\frac{1}{10}$ , $\frac{1}{A}$ ; row 11 begins $\frac{1}{11}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 9 is not drawn, but by the pattern it starts with $\frac{1}{9}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 9

#11 Work Backwards 6.NS.B.4

Row 9 starts with

\frac{1}{9}

, and the two numbers below it are

\frac{1}{10}

and

\frac{1}{A}

. So

\frac{1}{9} = \frac{1}{10} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

90

.

\frac{1}{A} = \frac{1}{9} - \frac{1}{10} = \frac{10-9}{90} = \frac{1}{90}

So

A = 90

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{10}

that starts row 10 are

\frac{1}{11}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{10} - \frac{1}{11} = \frac{11-10}{110} = \frac{1}{110}

So

B = 110

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{90}

are

\frac{1}{B} = \frac{1}{110}

and

\frac{1}{C}

. The least common multiple of

90

and

110

is

990

.

\frac{1}{C} = \frac{1}{90} - \frac{1}{110} = \frac{11-9}{990} = \frac{2}{990} = \frac{1}{495}

So

C = 495

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

90 + 110 + 495 = 695

A+B+C = 695

.

Answer: 695

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{110} + \frac{1}{495} = \frac{9+2}{990} = \frac{1}{90}$ and $\frac{1}{10} + \frac{1}{90} = \frac{9+1}{90} = \frac{1}{9}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 10 has $\frac{1}{10 \times 9}$ and row 11 has $\frac{1}{11 \times 10}$ , giving $A = 90$ and $B = 110$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $90$, $110$, $990$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 5 easy

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 7 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 8.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 7 begins $\frac{1}{7}$ , $\frac{1}{A}$ ; row 8 begins $\frac{1}{8}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 6 is not drawn, but by the pattern it starts with $\frac{1}{6}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 6

#11 Work Backwards 6.NS.B.4

Row 6 starts with

\frac{1}{6}

, and the two numbers below it are

\frac{1}{7}

and

\frac{1}{A}

. So

\frac{1}{6} = \frac{1}{7} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

42

.

\frac{1}{A} = \frac{1}{6} - \frac{1}{7} = \frac{7-6}{42} = \frac{1}{42}

So

A = 42

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{7}

that starts row 7 are

\frac{1}{8}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{7} - \frac{1}{8} = \frac{8-7}{56} = \frac{1}{56}

So

B = 56

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{42}

are

\frac{1}{B} = \frac{1}{56}

and

\frac{1}{C}

. The least common multiple of

42

and

56

is

168

.

\frac{1}{C} = \frac{1}{42} - \frac{1}{56} = \frac{4-3}{168} = \frac{1}{168}

So

C = 168

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

42 + 56 + 168 = 266

A+B+C = 266

.

Answer: 266

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{56} + \frac{1}{168} = \frac{3+1}{168} = \frac{1}{42}$ and $\frac{1}{7} + \frac{1}{42} = \frac{6+1}{42} = \frac{1}{6}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 7 has $\frac{1}{7 \times 6}$ and row 8 has $\frac{1}{8 \times 7}$ , giving $A = 42$ and $B = 56$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $42$, $56$, $168$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 6 medium

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 11 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 12.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 11 begins $\frac{1}{11}$ , $\frac{1}{A}$ ; row 12 begins $\frac{1}{12}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 10 is not drawn, but by the pattern it starts with $\frac{1}{10}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 10

#11 Work Backwards 6.NS.B.4

Row 10 starts with

\frac{1}{10}

, and the two numbers below it are

\frac{1}{11}

and

\frac{1}{A}

. So

\frac{1}{10} = \frac{1}{11} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

110

.

\frac{1}{A} = \frac{1}{10} - \frac{1}{11} = \frac{11-10}{110} = \frac{1}{110}

So

A = 110

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{11}

that starts row 11 are

\frac{1}{12}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{11} - \frac{1}{12} = \frac{12-11}{132} = \frac{1}{132}

So

B = 132

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{110}

are

\frac{1}{B} = \frac{1}{132}

and

\frac{1}{C}

. The least common multiple of

110

and

132

is

660

.

\frac{1}{C} = \frac{1}{110} - \frac{1}{132} = \frac{6-5}{660} = \frac{1}{660}

So

C = 660

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

110 + 132 + 660 = 902

A+B+C = 902

.

Answer: 902

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{132} + \frac{1}{660} = \frac{5+1}{660} = \frac{1}{110}$ and $\frac{1}{11} + \frac{1}{110} = \frac{10+1}{110} = \frac{1}{10}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 11 has $\frac{1}{11 \times 10}$ and row 12 has $\frac{1}{12 \times 11}$ , giving $A = 110$ and $B = 132$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $110$, $132$, $660$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 7 medium

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 12 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 13.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 12 begins $\frac{1}{12}$ , $\frac{1}{A}$ ; row 13 begins $\frac{1}{13}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 11 is not drawn, but by the pattern it starts with $\frac{1}{11}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 11

#11 Work Backwards 6.NS.B.4

Row 11 starts with

\frac{1}{11}

, and the two numbers below it are

\frac{1}{12}

and

\frac{1}{A}

. So

\frac{1}{11} = \frac{1}{12} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

132

.

\frac{1}{A} = \frac{1}{11} - \frac{1}{12} = \frac{12-11}{132} = \frac{1}{132}

So

A = 132

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{12}

that starts row 12 are

\frac{1}{13}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{12} - \frac{1}{13} = \frac{13-12}{156} = \frac{1}{156}

So

B = 156

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{132}

are

\frac{1}{B} = \frac{1}{156}

and

\frac{1}{C}

. The least common multiple of

132

and

156

is

1716

.

\frac{1}{C} = \frac{1}{132} - \frac{1}{156} = \frac{13-11}{1716} = \frac{2}{1716} = \frac{1}{858}

So

C = 858

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

132 + 156 + 858 = 1146

A+B+C = 1146

.

Answer: 1146

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{156} + \frac{1}{858} = \frac{11+2}{1716} = \frac{1}{132}$ and $\frac{1}{12} + \frac{1}{132} = \frac{11+1}{132} = \frac{1}{11}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 12 has $\frac{1}{12 \times 11}$ and row 13 has $\frac{1}{13 \times 12}$ , giving $A = 132$ and $B = 156$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $132$, $156$, $1716$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 8 medium

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 13 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 14.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 13 begins $\frac{1}{13}$ , $\frac{1}{A}$ ; row 14 begins $\frac{1}{14}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 12 is not drawn, but by the pattern it starts with $\frac{1}{12}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 12

#11 Work Backwards 6.NS.B.4

Row 12 starts with

\frac{1}{12}

, and the two numbers below it are

\frac{1}{13}

and

\frac{1}{A}

. So

\frac{1}{12} = \frac{1}{13} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

156

.

\frac{1}{A} = \frac{1}{12} - \frac{1}{13} = \frac{13-12}{156} = \frac{1}{156}

So

A = 156

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{13}

that starts row 13 are

\frac{1}{14}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{13} - \frac{1}{14} = \frac{14-13}{182} = \frac{1}{182}

So

B = 182

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{156}

are

\frac{1}{B} = \frac{1}{182}

and

\frac{1}{C}

. The least common multiple of

156

and

182

is

1092

.

\frac{1}{C} = \frac{1}{156} - \frac{1}{182} = \frac{7-6}{1092} = \frac{1}{1092}

So

C = 1092

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

156 + 182 + 1092 = 1430

A+B+C = 1430

.

Answer: 1430

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{182} + \frac{1}{1092} = \frac{6+1}{1092} = \frac{1}{156}$ and $\frac{1}{13} + \frac{1}{156} = \frac{12+1}{156} = \frac{1}{12}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 13 has $\frac{1}{13 \times 12}$ and row 14 has $\frac{1}{14 \times 13}$ , giving $A = 156$ and $B = 182$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $156$, $182$, $1092$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 9 hard

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 14 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 15.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 14 begins $\frac{1}{14}$ , $\frac{1}{A}$ ; row 15 begins $\frac{1}{15}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 13 is not drawn, but by the pattern it starts with $\frac{1}{13}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 13

#11 Work Backwards 6.NS.B.4

Row 13 starts with

\frac{1}{13}

, and the two numbers below it are

\frac{1}{14}

and

\frac{1}{A}

. So

\frac{1}{13} = \frac{1}{14} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

182

.

\frac{1}{A} = \frac{1}{13} - \frac{1}{14} = \frac{14-13}{182} = \frac{1}{182}

So

A = 182

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{14}

that starts row 14 are

\frac{1}{15}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{14} - \frac{1}{15} = \frac{15-14}{210} = \frac{1}{210}

So

B = 210

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{182}

are

\frac{1}{B} = \frac{1}{210}

and

\frac{1}{C}

. The least common multiple of

182

and

210

is

2730

.

\frac{1}{C} = \frac{1}{182} - \frac{1}{210} = \frac{15-13}{2730} = \frac{2}{2730} = \frac{1}{1365}

So

C = 1365

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

182 + 210 + 1365 = 1757

A+B+C = 1757

.

Answer: 1757

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{210} + \frac{1}{1365} = \frac{13+2}{2730} = \frac{1}{182}$ and $\frac{1}{14} + \frac{1}{182} = \frac{13+1}{182} = \frac{1}{13}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 14 has $\frac{1}{14 \times 13}$ and row 15 has $\frac{1}{15 \times 14}$ , giving $A = 182$ and $B = 210$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $182$, $210$, $2730$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 10 hard

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 15 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 16.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 15 begins $\frac{1}{15}$ , $\frac{1}{A}$ ; row 16 begins $\frac{1}{16}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 14 is not drawn, but by the pattern it starts with $\frac{1}{14}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 14

#11 Work Backwards 6.NS.B.4

Row 14 starts with

\frac{1}{14}

, and the two numbers below it are

\frac{1}{15}

and

\frac{1}{A}

. So

\frac{1}{14} = \frac{1}{15} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

210

.

\frac{1}{A} = \frac{1}{14} - \frac{1}{15} = \frac{15-14}{210} = \frac{1}{210}

So

A = 210

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{15}

that starts row 15 are

\frac{1}{16}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{15} - \frac{1}{16} = \frac{16-15}{240} = \frac{1}{240}

So

B = 240

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{210}

are

\frac{1}{B} = \frac{1}{240}

and

\frac{1}{C}

. The least common multiple of

210

and

240

is

1680

.

\frac{1}{C} = \frac{1}{210} - \frac{1}{240} = \frac{8-7}{1680} = \frac{1}{1680}

So

C = 1680

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

210 + 240 + 1680 = 2130

A+B+C = 2130

.

Answer: 2130

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{240} + \frac{1}{1680} = \frac{7+1}{1680} = \frac{1}{210}$ and $\frac{1}{15} + \frac{1}{210} = \frac{14+1}{210} = \frac{1}{14}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 15 has $\frac{1}{15 \times 14}$ and row 16 has $\frac{1}{16 \times 15}$ , giving $A = 210$ and $B = 240$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $210$, $240$, $1680$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 11 hard

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 16 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 17.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 16 begins $\frac{1}{16}$ , $\frac{1}{A}$ ; row 17 begins $\frac{1}{17}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 15 is not drawn, but by the pattern it starts with $\frac{1}{15}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 15

#11 Work Backwards 6.NS.B.4

Row 15 starts with

\frac{1}{15}

, and the two numbers below it are

\frac{1}{16}

and

\frac{1}{A}

. So

\frac{1}{15} = \frac{1}{16} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

240

.

\frac{1}{A} = \frac{1}{15} - \frac{1}{16} = \frac{16-15}{240} = \frac{1}{240}

So

A = 240

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{16}

that starts row 16 are

\frac{1}{17}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{16} - \frac{1}{17} = \frac{17-16}{272} = \frac{1}{272}

So

B = 272

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{240}

are

\frac{1}{B} = \frac{1}{272}

and

\frac{1}{C}

. The least common multiple of

240

and

272

is

4080

.

\frac{1}{C} = \frac{1}{240} - \frac{1}{272} = \frac{17-15}{4080} = \frac{2}{4080} = \frac{1}{2040}

So

C = 2040

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

240 + 272 + 2040 = 2552

A+B+C = 2552

.

Answer: 2552

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{272} + \frac{1}{2040} = \frac{15+2}{4080} = \frac{1}{240}$ and $\frac{1}{16} + \frac{1}{240} = \frac{15+1}{240} = \frac{1}{15}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 16 has $\frac{1}{16 \times 15}$ and row 17 has $\frac{1}{17 \times 16}$ , giving $A = 240$ and $B = 272$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $240$, $272$, $4080$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Variant 12 hard

The array of numbers shown below is called Leibniz's harmonic triangle. In this triangle, the second row starts and ends with $\frac{1}{2}$ , the third row starts and ends with $\frac{1}{3}$ , the fourth row starts and ends with $\frac{1}{4}$ , and so on. Also, every number equals the sum of the two numbers directly below it; for example, $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ . Find the value of $A+B+C$ for the three numbers $A$ , $B$ , $C$ .

Your answer

Show solution

1 · Understandwhat's really being asked

In this triangle each row starts and ends with $\frac{1}{n}$ for row $n$ , and every number is the sum of the two numbers just below it. We must find the hidden entries $\frac{1}{A}$ in row 17 and $\frac{1}{B}$ , $\frac{1}{C}$ in row 18.

Givens

Row $n$ starts and ends with $\frac{1}{n}$ .
Every number equals the sum of the two numbers directly below it, as in $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$ .
Row 17 begins $\frac{1}{17}$ , $\frac{1}{A}$ ; row 18 begins $\frac{1}{18}$ , $\frac{1}{B}$ , $\frac{1}{C}$ .

Unknowns

$A+B+C$

Constraints

Row 16 is not drawn, but by the pattern it starts with $\frac{1}{16}$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #5 Look for a Pattern

The rule builds a number from the two below it, so turn it around: a number below equals the number above minus its neighbor. Start where everything is known, at the left edge, and work inward one entry at a time.

3 · Execute4 carry out the plan

1Find $\frac{1}{A}$ from row 16

#11 Work Backwards 6.NS.B.4

Row 16 starts with

\frac{1}{16}

, and the two numbers below it are

\frac{1}{17}

and

\frac{1}{A}

. So

\frac{1}{16} = \frac{1}{17} + \frac{1}{A}

; subtract to undo the sum, using the common denominator

272

.

\frac{1}{A} = \frac{1}{16} - \frac{1}{17} = \frac{17-16}{272} = \frac{1}{272}

So

A = 272

.

2Find $\frac{1}{B}$ the same way

#5 Look for a Pattern 7.NS.A.1

Below the

\frac{1}{17}

that starts row 17 are

\frac{1}{18}

and

\frac{1}{B}

.

\frac{1}{B} = \frac{1}{17} - \frac{1}{18} = \frac{18-17}{306} = \frac{1}{306}

So

B = 306

. The pattern: next to

\frac{1}{n}

sits

\frac{1}{n(n-1)}

.

3Find $\frac{1}{C}$ under $\frac{1}{A}$

#11 Work Backwards 6.NS.B.4

Below

\frac{1}{A} = \frac{1}{272}

are

\frac{1}{B} = \frac{1}{306}

and

\frac{1}{C}

. The least common multiple of

272

and

306

is

2448

.

\frac{1}{C} = \frac{1}{272} - \frac{1}{306} = \frac{9-8}{2448} = \frac{1}{2448}

So

C = 2448

.

4Add

#11 Work Backwards 7.NS.A.1

Add the three denominators.

272 + 306 + 2448 = 3026

A+B+C = 3026

.

Answer: 3026

4 · Reviewdoes it hold up?

Check the sums go back up: $\frac{1}{306} + \frac{1}{2448} = \frac{8+1}{2448} = \frac{1}{272}$ and $\frac{1}{17} + \frac{1}{272} = \frac{16+1}{272} = \frac{1}{16}$ . Both match the rule.

Another way: Use the visible pattern of the second entries: row 3 has $\frac{1}{6} = \frac{1}{3 \times 2}$ and row 4 has $\frac{1}{12} = \frac{1}{4 \times 3}$ , so row 17 has $\frac{1}{17 \times 16}$ and row 18 has $\frac{1}{18 \times 17}$ , giving $A = 272$ and $B = 306$ before computing $C$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiples $272$, $306$, $2448$ as common denominators.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting fractions to undo a sum and adding the results.

💡Takeaway. If a rule adds two numbers to make a third, subtracting runs it backwards.

Run the sum rule backwards on unit fractions · 12 practice problems

Generated variants — 12

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