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Fractions: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 Three fractions just above halfway

Difficulty: 1 of 3 stars, Stretch

Arrange $\frac{37}{73}$ , $\frac{38}{75}$ , and $\frac{39}{77}$ from smallest to largest without multiplying the three denominators together. Explain why your order is correct.

Then compare $\frac{100}{199}$ and $\frac{101}{201}$ using the same idea.
1. Hint 1
  Each numerator is just over half its denominator. How far above one-half is each fraction?
2. Hint 2
  Double the numerator, then subtract the denominator. The difference is 1 in every case.
Answer

$\frac{39}{77}<\frac{38}{75}<\frac{37}{73}$ , and $\frac{101}{201}<\frac{100}{199}$ .

Full solution

Comparing the numerators alone is misleading because the denominators also change.
Instead compare each fraction with the same simple reference value, one-half.
Doubling the numerator gives one more than the denominator in every case.
Therefore each fraction equals one-half plus one divided by twice its denominator.

Explicitly, $\frac{37}{73}=\frac{1}{2}+\frac{1}{146}$ , $\frac{38}{75}=\frac{1}{2}+\frac{1}{150}$ , and
$\frac{39}{77}=\frac{1}{2}+\frac{1}{154}.$
Of these small extra amounts, $\frac{1}{154}$ is smallest and $\frac{1}{146}$ is largest.
Adding the same one-half preserves their order, giving
$\frac{39}{77}<\frac{38}{75}<\frac{37}{73}.$

The same comparison works for the second pair: $\frac{100}{199}=\frac{1}{2}+\frac{1}{398}$ and
$\frac{101}{201}=\frac{1}{2}+\frac{1}{402}.$
Since $\frac{1}{402}<\frac{1}{398}$ , the fraction with numerator 101 is the smaller one.
The calculation stays small because we compare the distances from a benchmark instead of building a large common denominator.

Answer

$\frac{39}{77}<\frac{38}{75}<\frac{37}{73}$ , and $\frac{101}{201}<\frac{100}{199}$ .

Key idea

Fractions close to the same simple benchmark can be compared by their small differences from it.
Problem 1: Mark complete
Problem 2 A fair order for taking shares

Difficulty: 1 of 3 stars, Stretch

A bowl contains a positive amount of trail mix. Asha will take one-half of whatever remains when her turn begins, Ben will take one-third of what remains, and Cleo will take one-fourth of what remains. Each person takes exactly one turn.

(a) Prove that the fraction left in the bowl after all three turns is the same in every order.

(b) Find every order in which the three people receive equal amounts. Explain why no other order works.
1. Hint 1
  Instead of tracking the amount taken, track the fraction left after each turn.
2. Hint 2
  If the three people receive equal amounts, use the final leftover amount to find each share. That forces who must go first.
Answer

(a) One-fourth of the original amount remains. (b) The unique fair order is Cleo, then Ben, then Asha.

Full solution

Asha leaves one-half of the current amount, Ben leaves two-thirds, and Cleo leaves three-fourths.
The final amount is therefore the original amount multiplied by $\frac{1}{2}$ , $\frac{2}{3}$ , and $\frac{3}{4}$ , in whichever order the turns occur.
Multiplication gives the same product in any order, and $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}=\frac{1}{4}$ .
This proves part (a).

The three people together receive three-fourths of the original amount.
If their shares are equal, each must receive one-fourth.
The first person takes their stated fraction of the full original amount, so Cleo must go first: only her stated fraction is one-fourth.

After Cleo takes her share, three-fourths remains.
Ben then takes one-third of that, which is one-fourth of the original.
One-half remains, and Asha takes half of that, again one-fourth of the original.
Thus Cleo, Ben, Asha works.
If Asha went second, she would take three-eighths of the original, exceeding the required one-fourth.
So after the forced first turn, the other turns are forced too.

Answer

(a) One-fourth of the original amount remains. (b) The unique fair order is Cleo, then Ben, then Asha.

Key idea

Track what remains to reveal an order-independent total, then use equality to force the order.
Problem 2: Mark complete
Problem 3 Recover the original fraction

Difficulty: 1 of 3 stars, Stretch

A positive proper fraction is written in simplest form. Its denominator is 7 greater than its numerator. After 4 is added to both numerator and denominator, the new fraction is equal to $\frac{3}{4}$ . What was the original fraction? Explain why there is exactly one answer.
1. Hint 1
  Adding the same amount to the numerator and denominator preserves their difference.
2. Hint 2
  Any fraction equal to three-fourths has numerator and denominator in the ratio 3 to 4. Their difference is one equal-sized part.
Answer

The original fraction is $\frac{17}{24}$ .

Full solution

Adding 4 to both numbers leaves the difference between the denominator and numerator equal to 7.
The resulting fraction is equal to three-fourths.
Its numerator therefore consists of three equal parts, while its denominator consists of four of the same parts.
Their difference is exactly one part.

That one part must be 7.
Hence the new numerator is $3\times7=21$ and the new denominator is
$4\times7=28.$
Subtract 4 from each to recover the original written fraction, $\frac{17}{24}$ .

Check all the conditions: it is positive and less than 1; the denominator exceeds the numerator by 7; and 17 and 24 have no common factor greater than 1, so the fraction is in simplest form.
Adding 4 to both gives $\frac{21}{28}=\frac{3}{4}$ , as required.
The fixed difference determines the size of a part uniquely, so neither the new pair nor the original pair has any other possibility.

Answer

The original fraction is $\frac{17}{24}$ .

Key idea

Equivalent fractions are scaled ratios; a known difference can determine the scale without equation-solving.
Problem 3: Mark complete
Problem 4 Split a sixth into two unit fractions

Difficulty: 2 of 3 stars, Challenge

A unit fraction is a fraction with numerator 1 and a positive integer denominator. Find every pair of distinct positive integers $a<b$ for which $\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$ . Prove that your list is complete.
1. Hint 1
  Each fraction is smaller than one-sixth. Also, because $a<b$ , the larger fraction is $\frac{1}{a}$ .
2. Hint 2
  Twice the larger fraction must exceed one-sixth. These two comparisons leave only five possible values of $a$ .
Answer

The pairs are $(7,42)$ , $(8,24)$ , $(9,18)$ , and $(10,15)$ .

Full solution

Because both terms are positive and their sum is one-sixth, each is smaller than one-sixth.
In particular $a>6$ .
Since $a<b$ , we have $\frac{1}{b}<\frac{1}{a}$ , so
$\frac{1}{6}<\frac{2}{a}.$
This forces $a<12$ .
Thus the only possibilities for the smaller denominator are 7, 8, 9, 10, and 11.

For each possible $a$ , subtract its unit fraction from one-sixth.
When $a=7$ , the remainder is
$\frac{1}{6}-\frac{1}{7}=\frac{1}{42}.$
When $a=8$ , it is $\frac{1}{24}$ .
When $a=9$ , it is $\frac{1}{18}$ .
When $a=10$ , it is $\frac{1}{15}$ .
These give the four listed pairs, and in each pair $b>a$ .

When $a=11$ , the remainder is $\frac{5}{66}$ .
This cannot equal a unit fraction with an integer denominator: its reciprocal is $\frac{66}{5}$ , which is not an integer.
No other value of $a$ is possible by the initial bounds.
The excluded boundary $a=12$ would give $b=12$ , violating the requirement that the denominators be distinct.
Hence the list is complete.

Answer

The pairs are $(7,42)$ , $(8,24)$ , $(9,18)$ , and $(10,15)$ .

Key idea

Bound the larger term first; an infinite-looking search may reduce to a handful of cases.
Problem 4: Mark complete
Problem 5 A sum that almost reaches one

Difficulty: 2 of 3 stars, Challenge

For a positive integer $n$ , define $S_n$ by adding the fractions $\frac{1}{1\times2},\frac{1}{2\times3},\frac{1}{3\times4}$ , and so on, ending at $\frac{1}{n(n+1)}$ .

(a) Find $S_{50}$ exactly without adding fifty fractions one at a time.

(b) Find the smallest positive integer $n$ for which $S_n>\frac{99}{100}$ .

(c) Can any such finite sum equal 1? Prove your answer.
1. Hint 1
  Find the difference between $\frac{1}{k}$ and $\frac{1}{k+1}$ .
2. Hint 2
  Write every term as a difference of two fractions. Most of the fractions then cancel in the whole sum.
Answer

(a) $S_{50}=\frac{50}{51}$ . (b) $n=100$ . (c) No finite sum equals 1.

Full solution

The useful identity is
$\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}.$
To check it, use the common denominator $k(k+1)$ : the difference in the numerators is $(k+1)-k=1$ .

Now replace every term in the sum by this difference.
The beginning becomes $(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})$ , and the cancellation continues through the final pair $\frac{1}{n}-\frac{1}{n+1}$ .
All the middle fractions occur once with a plus sign and once with a minus sign.
What remains is
$S_n=1-\frac{1}{n+1}=\frac{n}{n+1}.$
This pattern of cancellation is called a telescoping sum.

For part (a),
$S_{50}=\frac{50}{51}.$
For part (b), exceeding $\frac{99}{100}=1-\frac{1}{100}$ means the missing fraction $\frac{1}{n+1}$ must be smaller than $\frac{1}{100}$ .
Thus $n+1>100$ , and the smallest allowed integer is $n=100$ .
At $n=99$ the sum is exactly the threshold, so it does not qualify.
Finally, for every finite positive integer $n$ , the fraction $\frac{1}{n+1}$ is positive.
The sum is therefore always strictly less than 1.

Answer

(a) $S_{50}=\frac{50}{51}$ . (b) $n=100$ . (c) No finite sum equals 1.

Key idea

A useful decomposition can turn a long sum into cancellation; distinguish reaching a threshold from exceeding it.
Problem 5: Mark complete
Problem 6 Equal amounts, different fractions

Difficulty: 2 of 3 stars, Challenge

Jug A begins with $\frac{3}{4}$ liter of pure juice. Jug B begins with $\frac{1}{2}$ liter of pure water. Pour $\frac{1}{4}$ liter from A into B and mix B thoroughly. Then pour $\frac{1}{4}$ liter of the mixture from B back into A. Assume nothing spills.

(a) After both transfers, how much water is in A and how much juice is in B?

(b) What fraction of the liquid in A is water, and what fraction of the liquid in B is juice?

(c) Explain why the two amounts in part (a) must be equal whenever equal volumes are transferred out and back, even before doing the detailed fraction calculations.
1. Hint 1
  After the first transfer, B contains one-fourth liter of juice in a total of three-fourths liter. Use that fraction for the return transfer.
2. Hint 2
  Jug A ends with its original total volume. Every bit of juice it has lost must have been replaced by something.
Answer

(a) Both amounts are $\frac{1}{6}$ liter. (b) A is $\frac{2}{9}$ water; B is $\frac{1}{3}$ juice.

Full solution

After the first transfer, A contains $\frac{1}{2}$ liter of juice.
B contains $\frac{1}{4}$ liter of juice and $\frac{1}{2}$ liter of water, totaling $\frac{3}{4}$ liter.
Thus B is one-third juice and two-thirds water.

The return transfer of $\frac{1}{4}$ liter contains $\frac{1}{4}\times\frac{1}{3}=\frac{1}{12}$ liter of juice and $\frac{1}{4}\times\frac{2}{3}=\frac{1}{6}$ liter of water.
So A finishes with $\frac{1}{6}$ liter of water.
B retains $\frac{1}{4}-\frac{1}{12}=\frac{1}{6}$ liter of juice.
The two foreign-liquid amounts are equal.

A ends with $\frac{3}{4}$ liter total, so its water fraction is
$\frac{1}{6}\div\frac{3}{4}=\frac{2}{9}.$
B ends with $\frac{1}{2}$ liter total, so its juice fraction is
$\frac{1}{6}\div\frac{1}{2}=\frac{1}{3}.$
Equal amounts need not be equal fractions when the total amounts differ.

For part (c), equal out-and-back volumes restore A to its starting volume.
Any original juice missing from A is now in B, since none was spilled.
To restore the total volume in A, precisely that missing amount must have been replaced by water.
Thus the water in A equals the juice left in B, independently of the mixing arithmetic.

Answer

(a) Both amounts are $\frac{1}{6}$ liter. (b) A is $\frac{2}{9}$ water; B is $\frac{1}{3}$ juice.

Key idea

Separate an amount from its fraction of a whole, and use conservation to explain equalities.
Problem 6: Mark complete
Problem 7 The jars with equal blue counts

Difficulty: 2 of 3 stars, Challenge

Three jars contain only red and blue beads. In the first jar, one-half of the beads are red. In the second, one-third are red. In the third, one-fourth are red. Each jar contains the same positive number of blue beads.

(a) When all the beads are combined, what fraction are red?

(b) What is the smallest possible total number of beads in the three jars? Explain why smaller totals are impossible.
1. Hint 1
  The jars do not contain equal total numbers of beads. Match their blue counts instead.
2. Hint 2
  The red-to-blue ratios are 1 to 1, 1 to 2, and 1 to 3. Choose a common blue count that makes all three red counts whole numbers.
Answer

(a) $\frac{11}{29}$ of the combined beads are red. (b) The smallest total is 29 beads.

Full solution

In the first jar, equal fractions are red and blue, so the red and blue counts are equal.
In the second jar, the red-to-blue ratio is 1 to 2; its red count is half its blue count.
In the third jar, the red-to-blue ratio is 1 to 3; its red count is one-third its blue count.

Choose a common blue count of 6 for a model of the proportions.
The three jars then have 6, 3, and 2 red beads, respectively.
Each has 6 blue beads.
Combined, there are $6+3+2=11$ red beads and 18 blue beads, making the red fraction $\frac{11}{29}$ .
Any other allowed configuration scales all these counts by the same factor, so the fraction is unchanged.

To justify the minimum, the common blue count must be divisible by 2 so that the second red count is whole, and by 3 so that the third red count is whole.
Its smallest positive value is therefore 6.
The corresponding jars have 12, 9, and 8 beads, totaling 29.
Every other possible common blue count is a positive multiple of 6, so every other total is a positive multiple of 29.
The displayed configuration proves the minimum is attainable.

Answer

(a) $\frac{11}{29}$ of the combined beads are red. (b) The smallest total is 29 beads.

Key idea

When combining fractions, identify what is equal across the groups before choosing a common scale.
Problem 7: Mark complete
Problem 8 A fraction in a narrow gap

Difficulty: 3 of 3 stars, Deep challenge

Find the positive fraction in simplest form with the smallest possible denominator that lies strictly between $\frac{2}{5}$ and $\frac{3}{7}$ . Prove both that it lies in the gap and that no smaller denominator can work.
1. Hint 1
  Try adding the two numerators and adding the two denominators to obtain a candidate, but do not assume that proves it is optimal.
2. Hint 2
  If the denominator were at most 11, the numerator would be at most 4. Examine those four possible numerators rather than every denominator.
Answer

The unique fraction with the smallest possible denominator is $\frac{5}{12}$ .

Full solution

The candidate obtained by adding numerators and denominators is
$\frac{2+3}{5+7}=\frac{5}{12}.$
It lies in the required interval: $\frac{5}{12}-\frac{2}{5}=\frac{1}{60}>0$ and
$\frac{3}{7}-\frac{5}{12}=\frac{1}{84}>0.$
It is also in simplest form.

To prove minimality, suppose a fraction in the gap had denominator at most 11.
Its numerator could not be 5 or more, since even $\frac{5}{11}>\frac{3}{7}$ , and a smaller denominator only makes that fraction larger.
Thus the numerator is 1, 2, 3, or 4.

For numerator 1, a denominator putting the fraction in the gap would have to be strictly between $\frac{7}{3}$ and $\frac{5}{2}$ , which contains no integer.
For numerator 2, the denominator would have to lie strictly between $\frac{14}{3}$ and 5.
For numerator 3, it would have to lie strictly between 7 and $\frac{15}{2}$ .
For numerator 4, it would have to lie strictly between $\frac{28}{3}$ and 10.
None of these open intervals contains an integer.

No denominator below 12 works.
With denominator 12, numerators at most 4 are too small, while numerators at least 6 are too large.
Therefore $\frac{5}{12}$ is the unique fraction attaining the smallest denominator.

Answer

The unique fraction with the smallest possible denominator is $\frac{5}{12}$ .

Key idea

A good candidate is only half an optimization proof; eliminate all smaller possibilities with a short, organized search.
Problem 8: Mark complete
Problem 9 The fraction-writing machine

Difficulty: 3 of 3 stars, Deep challenge

A machine starts with the written fraction $\frac{1}{2}$ . It can apply either move any number of times: move A adds 1 to the numerator and 2 to the denominator; move B adds 2 to the numerator and 3 to the denominator. The machine never simplifies its written fraction between moves.

(a) Can it produce the exact written fraction $\frac{17}{27}$ ? If so, describe every possible choice of the number of A moves and B moves.

(b) Can it produce the exact written fraction $\frac{18}{27}$ ? Justify your answer.

(c) Prove that the value of its fraction is always less than $\frac{2}{3}$ , regardless of which moves it makes.
1. Hint 1
  Both moves increase the denominator-minus-numerator difference by exactly 1.
2. Hint 2
  For the upper bound, compare three times the numerator with twice the denominator. Track how each move changes those two quantities.
Answer

(a) Yes: exactly two A moves and seven B moves, in any order. (b) No. (c) Every reachable fraction is strictly less than $\frac{2}{3}$ .

Full solution

The initial denominator exceeds the numerator by 1.
Either move increases that difference by 1.
The target $\frac{17}{27}$ has difference 10, so it requires exactly nine moves.
If all nine were A moves, the numerator would become $1+9=10$ .
Replacing an A move by a B move adds one extra to the final numerator.
Reaching 17 therefore requires seven B moves and two A moves.
These moves produce numerator $1+2+14=17$ and denominator $2+4+21=27$ .
Their order does not matter because the same additions are made in total.

For $\frac{18}{27}$ , the difference is 9, so exactly eight moves would be required.
Each move adds at most 2 to the numerator, so the largest possible numerator after eight moves is $1+16=17$ .
Thus this written fraction cannot occur.

For the general bound, initially three times the numerator is 3, less than twice the denominator, which is 4.
Move A increases those two quantities by 3 and 4, respectively, preserving the strict inequality.
Move B increases both by 6, also preserving it.
Hence three times the numerator is always less than twice the denominator.
Since the denominator is positive, the fraction is always less than two-thirds.

Answer

(a) Yes: exactly two A moves and seven B moves, in any order. (b) No. (c) Every reachable fraction is strictly less than $\frac{2}{3}$ .

Key idea

Track simple differences that behave predictably under every allowed operation.
Problem 9: Mark complete
Problem 10 Why the reciprocal sum cannot be whole

Difficulty: 3 of 3 stars, Deep challenge

Consider the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{20}$ , including every integer denominator from 2 through 20.

(a) Prove that this sum is not a whole number without calculating its exact value.

(b) More generally, prove that for every integer $n\geq2$ , the sum of the unit fractions with denominators 2, 3, 4, ..., $n$ is not a whole number.
1. Hint 1
  Use the least common multiple of the denominators as a common denominator. Pay attention only to factors of 2.
2. Hint 2
  For denominators through 20, 16 contains more factors of 2 than any other denominator. What does that do to the parity of the converted numerators?
Answer

Neither the sum through 20 nor any such sum ending at an integer $n\geq2$ is a whole number.

Full solution

For part (a), let $L$ be the least common multiple of 2 through 20.
Exactly four copies of 2 occur in $L$ , because 16 is the largest power of 2 among these denominators.
After writing the fractions over $L$ , the numerator belonging to $\frac{1}{16}$ is $L\div16$ , an odd integer.
Every other denominator contains fewer than four copies of 2, so every other converted numerator is even.

The numerator of the whole sum is therefore one odd integer plus several even integers, which is odd.
Its denominator $L$ is even.
An odd number divided by an even number cannot be a whole number: any whole-number multiple of an even number is even.

For part (b), let $q$ be the largest power of 2 not exceeding $n$ .
Then
$q\leq n<2q.$
Among the denominators 2 through $n$ , only $q$ is divisible by $q$ , since the next positive multiple would be $2q>n$ .
Consequently $q$ contains strictly more copies of 2 than any other denominator.
Their LCM has exactly that many copies.
Once again, the term with denominator $q$ produces the unique odd numerator over this LCM, while all others produce even numerators.
The total numerator is odd and the denominator is even, proving the result for every allowed $n$ , including $n=2$ .

Answer

Neither the sum through 20 nor any such sum ending at an integer $n\geq2$ is a whole number.

Key idea

A common denominator can expose parity or divisibility even when its numerical value is unnecessary.
Problem 10: Mark complete