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Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

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Have You Got It?

Can you explain the strategy for winning this game with any target?

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Pair Sums

Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?

Almost One

Age 11 to 14 Challenge Level:

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
$\frac{3}{5}\hspace{3mm}\left(\frac{60}{100}\right)\hspace{3mm}+\frac3{20}\hspace{3mm}\left(\frac{15}{100}\right)\hspace{3mm}+\frac4{15}\hspace{3mm}\left(\frac{26.7}{100}\right)\hspace{3mm}\text{(rounded up)}\hspace{3mm}=\frac{101.7}{100}$


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.