← Decompose into sums of unit fractions · Decompose a Number into Parts and Factors

Decompose into sums of unit fractions · 12 practice problems

4.NF.B.35.NF.A.1

Generated variants — 12

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

Variant 1 easy answer: 12+121+128\frac{1}{2} + \frac{1}{21} + \frac{1}{28}

Write 712\dfrac{7}{12} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 7 over 12 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 712\frac{7}{12}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 712\frac{7}{12} is 12\frac{1}{2}: one more 1th would overshoot.
712−12=112\frac{7}{12} - \frac{1}{2} = \frac{1}{12}
112\frac{1}{12} is still to be shared out.

2Split the leftover 112\frac{1}{12} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 112\frac{1}{12} is 121\frac{1}{21}, and what remains is already a unit fraction.
112−121=128\frac{1}{12} - \frac{1}{21} = \frac{1}{28}
The last piece is 128\frac{1}{28}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 28 to confirm they rebuild the target.
12+121+128=712\frac{1}{2} + \frac{1}{21} + \frac{1}{28} = \frac{7}{12}
Three different unit fractions, and they add back exactly.
Answer: 12+121+128\frac{1}{2} + \frac{1}{21} + \frac{1}{28}
4 · Reviewdoes it hold up?

All three denominators differ (2, 21, 28) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 2 easy answer: 13+121+128\frac{1}{3} + \frac{1}{21} + \frac{1}{28}

Write 512\dfrac{5}{12} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 5 over 12 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 512\frac{5}{12}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 512\frac{5}{12} is 13\frac{1}{3}: one more 2th would overshoot.
512−13=112\frac{5}{12} - \frac{1}{3} = \frac{1}{12}
112\frac{1}{12} is still to be shared out.

2Split the leftover 112\frac{1}{12} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 112\frac{1}{12} is 121\frac{1}{21}, and what remains is already a unit fraction.
112−121=128\frac{1}{12} - \frac{1}{21} = \frac{1}{28}
The last piece is 128\frac{1}{28}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 28 to confirm they rebuild the target.
13+121+128=512\frac{1}{3} + \frac{1}{21} + \frac{1}{28} = \frac{5}{12}
Three different unit fractions, and they add back exactly.
Answer: 13+121+128\frac{1}{3} + \frac{1}{21} + \frac{1}{28}
4 · Reviewdoes it hold up?

All three denominators differ (3, 21, 28) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 3 easy answer: 13+178+191\frac{1}{3} + \frac{1}{78} + \frac{1}{91}

Write 514\dfrac{5}{14} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 5 over 14 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 514\frac{5}{14}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 514\frac{5}{14} is 13\frac{1}{3}: one more 2th would overshoot.
514−13=142\frac{5}{14} - \frac{1}{3} = \frac{1}{42}
142\frac{1}{42} is still to be shared out.

2Split the leftover 142\frac{1}{42} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 142\frac{1}{42} is 178\frac{1}{78}, and what remains is already a unit fraction.
142−178=191\frac{1}{42} - \frac{1}{78} = \frac{1}{91}
The last piece is 191\frac{1}{91}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 91 to confirm they rebuild the target.
13+178+191=514\frac{1}{3} + \frac{1}{78} + \frac{1}{91} = \frac{5}{14}
Three different unit fractions, and they add back exactly.
Answer: 13+178+191\frac{1}{3} + \frac{1}{78} + \frac{1}{91}
4 · Reviewdoes it hold up?

All three denominators differ (3, 78, 91) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 4 easy answer: 13+112+120\frac{1}{3} + \frac{1}{12} + \frac{1}{20}

Write 715\dfrac{7}{15} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 7 over 15 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 715\frac{7}{15}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 715\frac{7}{15} is 13\frac{1}{3}: one more 2th would overshoot.
715−13=215\frac{7}{15} - \frac{1}{3} = \frac{2}{15}
215\frac{2}{15} is still to be shared out.

2Split the leftover 215\frac{2}{15} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 215\frac{2}{15} is 112\frac{1}{12}, and what remains is already a unit fraction.
215−112=120\frac{2}{15} - \frac{1}{12} = \frac{1}{20}
The last piece is 120\frac{1}{20}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 20 to confirm they rebuild the target.
13+112+120=715\frac{1}{3} + \frac{1}{12} + \frac{1}{20} = \frac{7}{15}
Three different unit fractions, and they add back exactly.
Answer: 13+112+120\frac{1}{3} + \frac{1}{12} + \frac{1}{20}
4 · Reviewdoes it hold up?

All three denominators differ (3, 12, 20) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 5 medium answer: 14+124+148\frac{1}{4} + \frac{1}{24} + \frac{1}{48}

Write 516\dfrac{5}{16} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 5 over 16 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 516\frac{5}{16}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 516\frac{5}{16} is 14\frac{1}{4}: one more 3th would overshoot.
516−14=116\frac{5}{16} - \frac{1}{4} = \frac{1}{16}
116\frac{1}{16} is still to be shared out.

2Split the leftover 116\frac{1}{16} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 116\frac{1}{16} is 124\frac{1}{24}, and what remains is already a unit fraction.
116−124=148\frac{1}{16} - \frac{1}{24} = \frac{1}{48}
The last piece is 148\frac{1}{48}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 48 to confirm they rebuild the target.
14+124+148=516\frac{1}{4} + \frac{1}{24} + \frac{1}{48} = \frac{5}{16}
Three different unit fractions, and they add back exactly.
Answer: 14+124+148\frac{1}{4} + \frac{1}{24} + \frac{1}{48}
4 · Reviewdoes it hold up?

All three denominators differ (4, 24, 48) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 6 medium answer: 13+130+145\frac{1}{3} + \frac{1}{30} + \frac{1}{45}

Write 718\dfrac{7}{18} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 7 over 18 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 718\frac{7}{18}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 718\frac{7}{18} is 13\frac{1}{3}: one more 2th would overshoot.
718−13=118\frac{7}{18} - \frac{1}{3} = \frac{1}{18}
118\frac{1}{18} is still to be shared out.

2Split the leftover 118\frac{1}{18} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 118\frac{1}{18} is 130\frac{1}{30}, and what remains is already a unit fraction.
118−130=145\frac{1}{18} - \frac{1}{30} = \frac{1}{45}
The last piece is 145\frac{1}{45}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 45 to confirm they rebuild the target.
13+130+145=718\frac{1}{3} + \frac{1}{30} + \frac{1}{45} = \frac{7}{18}
Three different unit fractions, and they add back exactly.
Answer: 13+130+145\frac{1}{3} + \frac{1}{30} + \frac{1}{45}
4 · Reviewdoes it hold up?

All three denominators differ (3, 30, 45) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 7 medium answer: 13+115+120\frac{1}{3} + \frac{1}{15} + \frac{1}{20}

Write 920\dfrac{9}{20} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 9 over 20 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 920\frac{9}{20}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 920\frac{9}{20} is 13\frac{1}{3}: one more 2th would overshoot.
920−13=760\frac{9}{20} - \frac{1}{3} = \frac{7}{60}
760\frac{7}{60} is still to be shared out.

2Split the leftover 760\frac{7}{60} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 760\frac{7}{60} is 115\frac{1}{15}, and what remains is already a unit fraction.
760−115=120\frac{7}{60} - \frac{1}{15} = \frac{1}{20}
The last piece is 120\frac{1}{20}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 20 to confirm they rebuild the target.
13+115+120=920\frac{1}{3} + \frac{1}{15} + \frac{1}{20} = \frac{9}{20}
Three different unit fractions, and they add back exactly.
Answer: 13+115+120\frac{1}{3} + \frac{1}{15} + \frac{1}{20}
4 · Reviewdoes it hold up?

All three denominators differ (3, 15, 20) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 8 medium answer: 13+122+133\frac{1}{3} + \frac{1}{22} + \frac{1}{33}

Write 922\dfrac{9}{22} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 9 over 22 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 922\frac{9}{22}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 922\frac{9}{22} is 13\frac{1}{3}: one more 2th would overshoot.
922−13=566\frac{9}{22} - \frac{1}{3} = \frac{5}{66}
566\frac{5}{66} is still to be shared out.

2Split the leftover 566\frac{5}{66} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 566\frac{5}{66} is 122\frac{1}{22}, and what remains is already a unit fraction.
566−122=133\frac{5}{66} - \frac{1}{22} = \frac{1}{33}
The last piece is 133\frac{1}{33}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 33 to confirm they rebuild the target.
13+122+133=922\frac{1}{3} + \frac{1}{22} + \frac{1}{33} = \frac{9}{22}
Three different unit fractions, and they add back exactly.
Answer: 13+122+133\frac{1}{3} + \frac{1}{22} + \frac{1}{33}
4 · Reviewdoes it hold up?

All three denominators differ (3, 22, 33) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 9 hard answer: 13+112+124\frac{1}{3} + \frac{1}{12} + \frac{1}{24}

Write 1124\dfrac{11}{24} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 11 over 24 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 1124\frac{11}{24}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 1124\frac{11}{24} is 13\frac{1}{3}: one more 2th would overshoot.
1124−13=18\frac{11}{24} - \frac{1}{3} = \frac{1}{8}
18\frac{1}{8} is still to be shared out.

2Split the leftover 18\frac{1}{8} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 18\frac{1}{8} is 112\frac{1}{12}, and what remains is already a unit fraction.
18−112=124\frac{1}{8} - \frac{1}{12} = \frac{1}{24}
The last piece is 124\frac{1}{24}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 24 to confirm they rebuild the target.
13+112+124=1124\frac{1}{3} + \frac{1}{12} + \frac{1}{24} = \frac{11}{24}
Three different unit fractions, and they add back exactly.
Answer: 13+112+124\frac{1}{3} + \frac{1}{12} + \frac{1}{24}
4 · Reviewdoes it hold up?

All three denominators differ (3, 12, 24) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 10 hard answer: 13+112+121\frac{1}{3} + \frac{1}{12} + \frac{1}{21}

Write 1328\dfrac{13}{28} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 13 over 28 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 1328\frac{13}{28}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 1328\frac{13}{28} is 13\frac{1}{3}: one more 2th would overshoot.
1328−13=1184\frac{13}{28} - \frac{1}{3} = \frac{11}{84}
1184\frac{11}{84} is still to be shared out.

2Split the leftover 1184\frac{11}{84} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 1184\frac{11}{84} is 112\frac{1}{12}, and what remains is already a unit fraction.
1184−112=121\frac{11}{84} - \frac{1}{12} = \frac{1}{21}
The last piece is 121\frac{1}{21}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 21 to confirm they rebuild the target.
13+112+121=1328\frac{1}{3} + \frac{1}{12} + \frac{1}{21} = \frac{13}{28}
Three different unit fractions, and they add back exactly.
Answer: 13+112+121\frac{1}{3} + \frac{1}{12} + \frac{1}{21}
4 · Reviewdoes it hold up?

All three denominators differ (3, 12, 21) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 11 hard answer: 13+155+166\frac{1}{3} + \frac{1}{55} + \frac{1}{66}

Write 1130\dfrac{11}{30} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 11 over 30 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 1130\frac{11}{30}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 1130\frac{11}{30} is 13\frac{1}{3}: one more 2th would overshoot.
1130−13=130\frac{11}{30} - \frac{1}{3} = \frac{1}{30}
130\frac{1}{30} is still to be shared out.

2Split the leftover 130\frac{1}{30} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 130\frac{1}{30} is 155\frac{1}{55}, and what remains is already a unit fraction.
130−155=166\frac{1}{30} - \frac{1}{55} = \frac{1}{66}
The last piece is 166\frac{1}{66}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 66 to confirm they rebuild the target.
13+155+166=1130\frac{1}{3} + \frac{1}{55} + \frac{1}{66} = \frac{11}{30}
Three different unit fractions, and they add back exactly.
Answer: 13+155+166\frac{1}{3} + \frac{1}{55} + \frac{1}{66}
4 · Reviewdoes it hold up?

All three denominators differ (3, 55, 66) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.
Variant 12 hard answer: 13+163+184\frac{1}{3} + \frac{1}{63} + \frac{1}{84}

Write 1336\dfrac{13}{36} as a sum of three different unit fractions 1★+1♡+1▲\dfrac{1}{\bigstar}+\dfrac{1}{\heartsuit}+\dfrac{1}{\blacktriangle}.

Show solution
1 · Understandwhat's really being asked

We must write 13 over 36 as three unit fractions -- fractions with 1 on top -- added together, all with different bottoms.

Givens
  • The target is 1336\frac{13}{36}.
  • Exactly three unit fractions, all different.
Unknowns
  • The three denominators.
Constraints
  • Each piece has numerator 1, and no two may be the same.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #6 Guess and Check

Peel off the biggest unit fraction that still fits, then face the same question with a smaller leftover. Being greedy keeps the pieces distinct and shrinking, so it cannot loop.

3 · Execute3 carry out the plan

1Peel off the first unit fraction

#7 Identify Subproblems 4.NF.B.3
The largest unit fraction not bigger than 1336\frac{13}{36} is 13\frac{1}{3}: one more 2th would overshoot.
1336−13=136\frac{13}{36} - \frac{1}{3} = \frac{1}{36}
136\frac{1}{36} is still to be shared out.

2Split the leftover 136\frac{1}{36} into two unit fractions

#6 Guess and Check 5.NF.A.1
Repeat the move: the biggest unit fraction fitting inside 136\frac{1}{36} is 163\frac{1}{63}, and what remains is already a unit fraction.
136−163=184\frac{1}{36} - \frac{1}{63} = \frac{1}{84}
The last piece is 184\frac{1}{84}.

3Put the three pieces together and check

#6 Guess and Check 5.NF.A.1
Add all three over 84 to confirm they rebuild the target.
13+163+184=1336\frac{1}{3} + \frac{1}{63} + \frac{1}{84} = \frac{13}{36}
Three different unit fractions, and they add back exactly.
Answer: 13+163+184\frac{1}{3} + \frac{1}{63} + \frac{1}{84}
4 · Reviewdoes it hold up?

All three denominators differ (3, 63, 84) and each piece is smaller than the one before, which is what taking the biggest first guarantees.

Another way: Other splits exist -- this is not the only answer -- but the greedy route always finds one, and finds it without searching.

Standardsmin grade 5
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Reading the target as a sum of pieces with 1 on top.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators — Each subtraction, and the final addition that checks the split.
💡Takeaway. Take the biggest piece that fits, then ask the same question about what is left. The leftovers shrink until nothing remains.