Largest product

Which set of numbers that add to 100 have the largest product?

Age

11 to 14

Challenge level

Secondary

NUMBER

Number Operations and Calculation Methods

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Largest Product printable worksheet

Here are some different ways in which we can split 100:

$30 + 70 = 100$
$20 + 80 = 100$
$21 + 56 + 23 = 100$
$10 + 10 + 10 + 10 + 10 + 10 + 20 + 20 = 100$

The products of these sets are all different:

$30 \times70 = 2100$
$20 \times 80 = 1600$
$21 \times 56 \times23 = 27048$
$10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 20 \times 20 = 400000000$

What is the largest product that can be made from whole numbers that add up to 100?

Choose another starting number and split it in a variety of ways.

What is the largest product this time?

Can you find a strategy for splitting any number so that you always get the largest product?

Click here for a poster of this problem.

The solution below is to an earlier question, which asked what is the largest product that can be made from whole numbers that add up to 10?

Thank you to all those who tried this problem, there were a large number of solutions received. Luke and Alex from Aqueduct Primary sent in their solution:

The largest product that we could find was $2.5 \times 2.5 \times 2.5 \times2.5 = 39.06$.

We spent hours looking at other numbers but realised that larger numbers often made smaller products.

This was a really fun challenge!

Matthew from Bydales realised that we have to exclude negatives, otherwise solutions like $99\times - 44\times - 45 = 196 020 $ are possible. He wrote:

Using negative numbers you can make the number as big as you like as long as the other two numbers are negative and will leave you with 10 when they are added. You can therefore get close to inifinity if negatives are allowed!

Thank you to all of the children at St George's CE Primary School who had a go at this problem. Nathan, Otis, Hannah and Leon all sent in the correct answer. Saif fom Durston School, Sion and Daniel from TES and Cameron from Tokoroa Intermediate in New Zealand all sent in correct solutions as well.

Andy from Garden International School sent in a very clearly explained solution showing his working:

First we divided ten by $2$, then ten by $3$, then ten by $4$ and so on until we reach $9$.

Then we multiplied the divided number. In the case of $2$ is:

$5\times 5$

In the case of $3$ is: $3\frac{ 1}{3}\times3\frac{ 1}{3}\times3\frac{ 1}{3}$.

We divided ten because the same number multiplied together would give the largest product.

We started with two because there's no point using one and we end with $9$ because there's no point multiplying $1$ by $1$ nine times. In the end $2 \frac{1}{2} \times 2 \frac{1}{2} \times 2 \frac{1}{2} \times 2 \frac{1}{2}$ gives the largest product and the product is $39.0625$.

Kang Yun Seok, also from Garden International School sent in a complete solution and discussed how similar numbers give a larger product due to the properties of shapes like squares and circles. This was also discussed in other solutions including the one sent in by Mikey of Tadcaster School.

The largest product of any two numbers is from numbers that are as similar as possible. That is why a $20 \times20$ square has bigger area than a rectangle of the same perimeter - say, $39 \times1$ ...(that is because the biggest shape you can make with a length of string is a circle or square instead of a long thin rectangle).

So the largest product from ten is... with

$2$ numbers that add up to ten: $5 \times5$

$3$ numbers that add up to ten: $3.333 \times3.333 \times3.33$

4...

5...

and the largest of these products is: $2 \frac{1}{2} \times 2 \frac{1}{2} \times 2 \frac{1}{2} \times 2 \frac{1}{2}= 39.0625$.

It soon becomes apparent that when $10$ is split into $n$ parts (where $n$ is a whole number), the largest product is given by $(\frac{10}{n})^n$ . So the problem is in fact to find the largest value of $(\frac{10}{n})^ n$ . The largest value of $(\frac{10}{n})^ n$ is when $n=4$ that is $39.0625$. This is the largest product.

Thomas Hu from A Y Jackson school used his knowledge of the euler number ($e=2.71828...$) to find the maximal solution for all values of $x$

Given the product $(\frac{x}{n})^n$ (where $x$ is the number in question and $n$ in the number of parts it is being divided into). It is already clear that repeated multiplication of the same number $(2.5^4)$ is greater than that of two different numbers $(4*6)$ due to maximization and difference of squares $(x-a)(x+a)= x^2-a^2$.

Although the optimal $n$ for $10$ was stated to be $4$, that is only true if one assumes that $n$ must be an integer. Otherwise, $n = 3.7, 3.68, 3.679, 3.6788$ each provides increasingly larger products. But these cannot work as the number must still add to the value $x$.

But to maximise $(\frac{x}{n})^n$, where $n$ is an integer, then $n$ must be chosen to find $\frac{x}{n}$ as close as possible to e. Or in technical notation, to minimise the absolute value of $\frac{x}{n}-e$ subject to $n$ an integer for given $x$.

Well done Thomas, you've really got the hang of this problem. For those who don't understand his notation, you have to find an integer $n$ to make $\frac{x}{n}$ as close as possible to 2.7.You then use this integer to find the answer. Abover $x=10$, $n=4$ and so $\frac{x}{n}$ is 2.5. So the sum is $10$, and the product is $(\frac{x}{n})^n$, which is $39.0625$ in this case.

Well done everyone on a very tough problem!

Why do this problem?

This problem can be tackled at very different intellectual levels. It is an excellent context to practise adding and multiplying while trying to solve a problem.

Part of the problem is to realise that there is, in fact, a question to consider.

Possible approach

This printable worksheet may be useful: Largest Product.

Ask the students for various sets of numbers that add to 100 (exclude negatives!). Ask students to work out the products of some of these sets - students might demonstrate the methods on the board to help the group remember the algorithms.

Present the problem for the lesson - to find the biggest product, for numbers that add to $100$, to keep a sensible record of things that they have tried and what did and didn't work out.

After some time working, students could feed back on successful strategies they are using, and the class 'best' could be put up on the board for others to beat. Make sure that students know that there are ways of adapting the problem if they feel that they are making no further progress.

Key questions

How shall we start to make some progress with this challenge ?
Can you generalise your results somehow?
What is the same and what is different about the solution if the $100$ is changed to another number? (smaller numbers encourage the use of decimals)

Possible support

With students who aren't ready for working with these decimals, perhaps consider integer solutions only for various sums.

Possible extension

What about the sum being $0.5$ or $1.1$ or $-8$ etc., how would the strategy/solution change?

Students often work by reasoning from a few 'spot' values and showing that improvement occurs as they adjust towards a particular value, which they therefore declare to be the optimum. It is certainly good to draw attention to the logical possibility that an even better 'local maximum' might exist somewhere else. And also, if they are up to it, that adjustments are always incremental so how can we be sure that we haven't jumped right over an interval which contains something important.

Can we reason that this function is continuous ?

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