Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

Marbles

I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?

More Marbles

I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?

How Much Can We Spend?

Stage: 3 Challenge Level:

Adam and Dylan had some great ideas, Sam explains his thoughts clearly:

The largest number that can't be made is 7. This is because the lowest run of three numbers you can make is 8,9 and 10. If you make this using 5 or 3 and then keep adding different quantities of 3 to them you should be able to make every number higher than them.

Using $3z$ and $4z$ the highest number that can't be made is 5. This is because you can make 6, 7 and 8.

Mrs Dillon's year 7 class had a different way of showing 7 is the largest amount that can't be made;

We think using a $3z$ & $5z$ coin you can make all numbers except: 1, 2, 4, 7
We know you can make all other numbers because: we made all of the numbers from 11 to 20, and we can then get the rest by adding multiples of 10 (two 5z coins) onto any of these numbers.
We have also noticed that if you could get change, the 1, 2, 4 & 7 could also be made. So if you had to pay $1z$ then you could pay $6z$ and be given $5z$ back.
You can make $2z$ by paying $5z$ and getting $3z$ in change.
For $4z$ you can pay $9z$ and get $5z$ back.
Finally $7z$ can be paid by you paying $10z$ and getting $3z$ change.

Fantastic!

Tze Liang Chee looked into other possible combinations for other countries:

For a pairing such as 2 and 7, then any number of the form $2X + 7Y$ can be made, where $X$ and $Y$ are whole numbers. Eg $10 = 2 \times 5 + 7 \times 0 = 2 + 2 + 2 + 2 + 2$ or $16 = 2 \times 1 + 7 \times 2 = 2 + 7 + 7$

The numbers that can't be made are any $C$ where for $C = 2X + 7Y$ means that $X$ and $Y$ are not whole numbers.

For this if you can make 2 consecutive numbers, then you can make everything higher by adding multiples of 2. So here the lowest consecutive numbers that can be made are $6 = 2 + 2 + 2 = 2 \times 3$ and $7 = 7 = 7 \times 1$ so everything higher than 6 can be made by adding multiples of two. So you can't make 1, 3, 5 and that's it.

Generally if your two coins are $a$ and $b$, where $a$ and $b$ are coprime (their only common factor is 1) and $a$ is smaller than $b$, then as soon as you can make $a$ consecutive numbers, then you can make everything higher by adding multiples of $a$.

Eg. If you have $4z$ and $5z$ then you need to make 4 consecutive numbers.

The lowest ones are

$12 = 4 + 4 + 4$,

$13 = 4 + 4 + 5$,

$14 = 4 + 5 + 5$,

$15 = 5 + 5 + 5$,

so every number bigger than 12 can be made.

If your numbers $c$ and $d$ are not coprime (they have a common factor greater than 1), then there will never be a highest number that they cannot make, as they will only be able to make multiples of their highest common factor, and never be able to make a set of consecutive numbers.

Eg. If you have $4z$ and $6z$, their highest common factor is 2, and so you can only make numbers in the 2 times table, and never make any odd numbers.

If you have $40z$ and $60z$, their highest common factor is 20, and so you can only make multiples of 20.

Making and testing hypotheses. Visualising. Generalising. Number theory. Mathematical reasoning & proof. Modulus arithmetic. smartphone. Working systematically. Creating expressions/formulae. Interactivities.

You may also like

Helen's Conjecture

Marbles

More Marbles

How Much Can We Spend?

Stage: 3 Challenge Level: