Almost one
Choose some fractions and add them together. Can you get close to 1?
Problem
Here is a set of six fractions: $$\frac{1}{6} \quad \frac{1}{25} \quad \frac{3}{5} \quad \frac{3}{20} \quad \frac{4}{15} \quad \frac{5}{8} $$
Choose some of the fractions and add them together. You can use as many fractions as you like, but you can only use each fraction once.
Can you get an answer that is close to 1?
What is the closest to 1 that you can get?
With thanks to Colin Foster who introduced us to this problem.
Getting Started
You could begin by choosing a fraction bigger than $\frac{1}{2}$ and adding on smaller fractions to get close to 1.
You could approximate each fraction to fractions that you are familiar with (with small denominators) and then use your approximations to estimate possible sums.
It is often easiest to add fractions when they have the same denominator...
Student Solutions
We received many solutions to this problem, and well done to everyone who found a sum close to 1.
Xaviar from Temora Public School in Australia, Demilade from Green Springs School in Nigeria, Ibrahim from Nigerian Tulip International College, Jonathan, Alex and Shafi from Greenacre Public School in Australia, Marissa, Jessica and Logan from Matamata Intermediate in New Zealand, Ethan from King Geroge V School in Hong Kong, Year 9 class at Wyedean School in England, John from South Hunsley Secondary School in the UK and Kenzo from ISL in Switzerland added fractions together to see what they could find.
Ethan found:
$$\tfrac1{25}+\tfrac35+\tfrac3{20}$$ I "simplified" them so it was easier to add them together.
$$\tfrac4{100}+\tfrac{60}{100}+\tfrac{15}{100}=\tfrac{79}{100}$$
It isn't very close but this is the closest one I managed to find.
Ibrahim found:
$$\begin{split}\tfrac35+\tfrac58+\tfrac3{20}&=\tfrac{24+25+6}{40}\\&=\tfrac{55}{40}\\&=1.375\end{split}$$ which, when given to 1 significant figure, is $1$
$$\begin{split}\tfrac16+\tfrac4{15}+\tfrac3{5}&=\tfrac{5+8+18}{30}\\&=\tfrac{31}{30}\\&=1.03\end{split}$$ which is very close to $1$
Logan took the idea of equivalent fractions a bit further:
First I analyzed the 6 fractions and looked at which ones could add together to get a number close to one. After a few tries not working I then chose to add $\frac58$ and $\frac4{15}$ which was equivalent to $\frac{107}{120}$ as $120$ is the least common multiple of the two denominators $15$ and $8$. I then found any other denominators that could multiply into $120$ and the fractions were $\frac35$, $\frac16$ and $\frac1{20}$. These fractions were equivalent to $\frac{72}{120}$, $\frac{20}{120}$ and $\frac{6}{120}$.
After adding all these numbers on in separate equations the closest answer was $127$ after adding the $\frac16$ or $\frac{20}{120}$. This is equivalent to $1.058$, just $0.058$ off the number $1$.
William from New End Primary School in the UK, James from The King's School Grantham in the UK, Will from WWSPS in Australia, Rishika from Nonsuch High School for Girls in the UK, Yesh from Manchester Grammar School in the UK, Aryaman from Bangkok Patana School in Thailand, Hondfa, Tony, Jacob, Nathan, Laurenc and Allan from Greenacre Public School, Isaac from Rugby School in the UK, Victor from South Hunsley in the UK, Daanyal from Caerleon Comprehensive School in Wales and Paul from Coventry University in the UK all used this method of expressing all of the fractions over the same denominator.
This is Rishika's working:
The easiest way to find an answer closest to 1 is to find the LCM (lowest common multiple) of all the denominators first, for which I used prime factorisation:
$6 = 2\times3\\
25= 5\times5 \hspace{3mm}(5^2)\\
5 = 5\\
20 = 2\times2\times5\hspace{3mm} (2^2\times5)\\
15 = 3\times5\\
8 = 2\times2\times2\hspace{3mm} (2^3)\\$
To find the LCM we need to multiply together the highest number of $2$s, $3$s and $5$s present in each set, overall. From above, we know that the highest number of $2$s is $3$ ($2\times2\times2$), the highest number of $3$s is $1$ ($2\times3$ and $3\times5$) and
the highest number of $5$s is $2$ ($5\times5$).
Therefore we multiply $2\times2\times2\times3\times5\times5 = 600$, which is the LCM.
Then we can put all the fractions over the LCM:
$\frac{100}{600}\\
\frac{24}{600}\\
\frac{360}{600}\\
\frac{90}{600}\\
\frac{160}{600}\\ \frac{75}{600}$
To find the fractions that add to give an answer closest to $1$, I first added all the above fractions together, giving $\frac{809}{600}$ (remember $1 =\frac{600}{600}$).
I needed $\frac{209}{600}$ less to make $1$.
The fractions that add to make the closest to $\frac{209}{600}$ were: $\frac{100}{600}, \frac{24}{600}$ and $\frac{75}{600}$ ($\frac16,\frac1{25}$ and $\frac{5}{8}$), summing to $\frac{199}{600}$.
Therefore, I needed to add other fractions to give me the answer closest to $1$:
$$\begin{split}\tfrac35+\tfrac3{20}+\tfrac4{15}&=\tfrac{360}{600}+\tfrac{90}{600}+\tfrac{160}{600}\\&=\tfrac{610}{600}\\&=1\tfrac1{60}\end{split}$$ which is the closest answer to $1$.
Daanyal used a denominator of $1800$ instead of $600$, and wanted to prove that $\tfrac35+\tfrac3{20}+\tfrac4{15}$ was the best possible sum. This is Daanyal's working:
I used trial and error to find a combination close to $1800$. Quite quickly I got to: $$\tfrac{1080}{1800}+\tfrac{270}{1800}+\tfrac{480}{1800}=\tfrac{1830}{1800}=\tfrac{61}{60}$$ I decided that this was the closest I could get by trial and error. Next, I needed to prove/disprove that this was the closest to $1$ you can get.
To improve this equation, you need to:
a) Remove one or more terms
b) Replace it with one or more terms
To keep it brief, I am going to refrain from using the denominators as they are not very significant.
Three numerators used: Three numerators not used:
$$1080\hspace{70mm}300\\
270\hspace{72mm}72\\
480\hspace{69mm}1125$$
The three used numerators can be used to create a list of seven options for removing terms from the equation. Likewise, the three unused numerators can be used to create a similar list of replacement terms.
Removal Replacement
$$480\hspace{70mm}300\\
270\hspace{72mm}72\\
1080\hspace{69mm}1125\\
480+270=750\hspace{45mm}300+72=372\\
270+1080=1350\hspace{40mm}72+1125=1197\\
480+1080=1560\hspace{39mm}300+1125=1425\\
180+270+1080=1830\hspace{27mm}300+72+1125=1497$$ If you remove any term on the left and replace it with any term on the right, it results in a total which is further from $1800$ than $1830$ is. This proves that $\frac{1830}{1800}$ or $\frac{61}{60}$ is the closest to $1$ you can get.
Paul said that we could probably solve this by programming some software to try every iteration.
Flynn from Parkside Primary School in Australia and Aidan from Sheldon School in England expressed all of the fractions as fractions with denominators of 100, which led to some very strange fractions (which mathematicians don't usually allow). Flynn got
$\frac{62.5}{100}+\frac{16.6666666}{100}+\frac{15}{100}+\frac{4}{100}$ or $1.02$ (Rounded)
and Aidan got
This is similar to what Saroja from India did using percentages:
$\frac35,\frac3{20}$ and $\frac4{15}$.
The first fraction is 60%.
The third one is a little more than 25%.
The total of these two is about 85%
Hence we have to choose a fraction which is about 15%.
Hence $\frac3{20}$ (equivalent to 15%) fits in well. The sum of the three fractions is $\frac{61}{60}$ - nearly 1.
Zane from Shireland Collegiate Academy, Soham from Sutton Grammar School and Sheila, all in the UK, converted the fractions into decimals to solve the problem. Click here here to see Soham's complete solution, with explanation.
Teachers' Resources
The suggestions in these notes are adapted from Colin Foster's article, Sum Fractions.
Why do this problem?
Adding and subtracting fractions is a procedure which students often find very difficult to master. It is important to address the area without it feeling like an exact repetition of what they have done many times before.
One way to avoid the tedium of lots of repetitive practice is to embed practice in a bigger problem which students are trying to solve. This idea is explored in Colin Foster's article, Mathematical Etudes, and this problem is an example of a mathematical etude.
Possible approach
"What can you say about these six fractions?" $$\frac{1}{6} \quad \frac{1}{25} \quad \frac{3}{5} \quad \frac{3}{20} \quad \frac{4}{15} \quad \frac{5}{8} $$
Students might note that they are all different, that they are all less than 1, that they are all positive, that they are all expressed in their simplest terms, that four are less than a half and two are greater than a half, that they are not in order of size, and so on.
Encourage students to say as many things as they can think of. Questions like this are a good way to encourage students to be mathematically observant.
"Which fraction do you think is the largest? Which is the smallest? Why?"
Since all of the fractions are expressed in their simplest terms, it is easy to see that none of them are equal. Students may compare fractions by making their denominators equal or converting them to decimals.
Encourage students to use 'informal' methods of comparing fractions, and only calculate when it becomes absolutely necessary. For example, $\frac{1}{20}$ is bigger than $\frac{1}{25}$, so $\frac{3}{20}$ will certainly be bigger than $\frac{1}{25}$.
"Write a fraction that is equal to $\frac35$. And another, and another..."
Students could write their fractions on mini-whiteboards.
They will probably list equivalent fractions such as $\frac{6}{10}, \frac{30}{50}$ etc.
You could encourage a wider range of answers by introducing some constraints, for example... "Write down one with an odd denominator" or "Write down one where the numerator is a five-digit number that does not end in 0".
Then ask them to do the same with $\frac58$.
"How would you add $\frac35$ and $\frac58$ without a calculator?"
"The answer to $\frac35 +\frac58$ is a little bit more than 1. Is there any way that you could have predicted that the answer was going to be more than 1 without working it out exactly?"
Both fractions are more than $\frac12$, so their total must be more than 1.
This sort of reasoning can be very useful for estimating the size of an answer so that mistakes can be spotted. Estimation will be important in the main activity that follows.
Return to the original set of six fractions:
$$\frac{1}{6} \quad \frac{1}{25} \quad \frac{3}{5} \quad \frac{3}{20} \quad \frac{4}{15} \quad \frac{5}{8} $$
"Choose some of the fractions and add them together. You can use as many fractions as you like, but you can only use each fraction once."
"Can you get an answer that is close to 1?"
"What is the closest to 1 that you can get?"
Make it clear that calculators are not to be used!
If some students are unsure how to start, encourage them to talk to their partner.
Give students some time to work on the problem. This will be a good opportunity to circulate and see how students are getting on.
If students obtain an answer like $\frac{11}{12}$ (from $\frac{1}{6} + \frac{3}{5} + \frac{3}{20}$), they may think that they are as close as possible, as their answer is "only 1 away", but because the "one" is "one twelfth" they are not really that close ($\frac{1}{12}$ is more than 8%), so they should aim to get even closer!
Allow plenty of time at the end of the lesson for students to share their approaches and reasoning.
Possible support
If students are not secure with equivalent fractions they could do some work with Fractional Wall.
Possible extension
Students could be asked to find the set of fractions which add up to as near to $\frac{1}{2}$ as possible.
For other rich contexts that offer students an opportunity to practise manipulating fractions see Peaches Today, Peaches Tomorrow and Keep it Simple.