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14 Divisors

What is the smallest number with exactly 14 divisors?

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Summing Consecutive Numbers

Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?

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Dozens

Do you know a quick way to check if a number is a multiple of two? How about three, four or six?

Largest Product

Stage: 3 Challenge Level: Challenge Level:1

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!