Pebbles

Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

GOT IT Now

For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?

Ben's Game

Stage: 3 Challenge Level:

We received several responses mentioning that a strategy of trial and error had been used to arrive at the result. This is a valuable strategy but it may be difficult to tell if there is more than one solution.

Alice used a spreadsheet to help her consider the many possibilities :

First of all I considered what number of counters each of them could have.

Ben's have to be a multiple of $3$, but not $3$, Emma's have to go by $5$, but not $5$, and Jacks had to go by $4$ but not be $4$.

I decided to do a spread-sheet:
Ben has $2/3$ of his left and $1/5$ of Emma's, ($2B/3 + E/5$),
Jack has $3/4$ of his left and $1/3$ of Ben's ($3J/4 + B/3$),
and Emma has $4/5$ of hers left and $1/4$ of Jack's ($4E/5 + J/4$).

After a few goes at putting numbers into the spreadsheet, the answer turned out to be $12$ for Ben, $8$ for Jack and $10$ for Emily.

Zak and Sam from Norwich School for boys showed that this solution works:

These are the number of counters they started with:
Ben $12$
Jack $8$
Emma $10$
Ben gives four to Jack and receives two from Emma.
Jack gives two to Emma and recieves four from Ben.
Emma gives two to Ben and receives two from Jack.
They all end up with ten.
There are no other solutions to this problem - the figures doubled would give too much and the figures halved would mean Emma gives only one counter.

Ian from Myton School reasoned as follows:

Since they give $1/3, 1/4$ and $1/5$ ($B, J$ and $E$) and more than $1$ the minimum they could give is $6, 8$ and $10$ totalling $24$.
Ben must start with the most since he gives the most and gets the least so the minimums are $12, 8$ and $10$, with other options of $B, J$ and $E$:
$12, 8$ and $10$
$15, 8$ and $10$
$18, 8$ and $10$
(since the total number of counters must be a multiple of 3).
So I attempted the top answer using the three formulas
$2/3B+1/5E=X$
$8+2=X$
$3/4J+1/3B=X$
$6+4=X$
$1/4J+4/5E=X$
$2+8=X$
all of which give $X=10$ showing a correct solution.

Ben's approach confirmed that there is a single solution:

$E$ is a multiple of $5$, $B$ of $3$ and $J$ of $4$.

$E$ can start with $10, 15, 20\ldots$ etc (starts at $10$ because cannot pass just $1$ counter)
If we use $15$ then passing $1/5$ (ie.$3$) leaves $12$. We cannot receive $1$ counter on its own so the finishing total for $E$ would be greater than $13$ which is not possible as this would require more than $40$ counters altogether. Therefore $E$ must have started with $10$.

If $E$ starts with $10$ then $B + J < 30$.
So $B$ can start with $6, 9, 12, 15, 18, 21\ldots$
and $J$ with $8, 12, 16, 20\ldots$

Only way to get $E, B$ and $J$ to finish with $13$ (the maximum possible),
is if $J = 20$ and $B = 9$,
or if $J = 8$ and $B = 21$
However these do not work, so $E, B$ and $J$ must finish with less than $13$.

Therefore we can eliminate other combinations leaving the options of
$B$ starting with either $6, 9, 12$ or $15$,
and $J$ starting with either $8, 12$ or $16$.

If $E = 10$,
$B + J = 14, 17, 20, 23, 26$ (since the total must be a multiple of $3$)
so try $6 + 8.$
This doesn't work so elimate 6 since it doesn't contribute to any other total.

Try $9 + 8$.
This doesn't work so elimate 9.

Therefore $B$ starts with either $12$ or $15$,
and $J$ starts with either $8, 12 or 16$.

$12 + 12 = 24$ and $12 + 15 = 27$, so elimate $J$ starting with $12$.

Therefore $B$ starts with either $12$ or $15$,
and $J$ starts with either $8$ or $16$.

$12 + 16 = 28$ and $15 + 16 = 31$,so elimate $J$ starting with $16$.

Therefore $B$ starts with either $12$ or $15$,
and $J$ starts with $8$.

But $B = 15$ and $J = 8$ does not work

so $B = 12$ and $J = 8$
$E= 10, B = 12, J = 8$

All finish with $10$ therefore use $30$ counters.

A student from Carres Grammar School used simultaneous equations to arrive at the solution:

Ben starts with $x$ counters
Jack starts with $y$ counters
Emma starts with $z$ counters
We know that:
$x + y + z < 40$
$x/3 + 3y/4 = y/4 + 4z/5 = z/5 + 2x/3$
Therefore:
$20x/60 + 45y/60 = 15y/60 + 48z/60 = 12z/60 + 40x/60$
$20x + 45y = 15y + 48z = 12z + 40x$
(A) $20x + 30y = 48z$ (from the first two equations)
(B) $15y + 36z = 40x$ (from the last two equations)
(C) $12z + 20x = 45y$ (from the first and last equations)
$40x + 60y = 96z$ (from A)
(D)$40x = 96z - 60y$
$15y + 36z = 96z - 60y$ (from B and D)
$75y + 36z = 96z$
$75y = 60z$
$15y = 12z$
(E) $45y = 36z$
$12z + 20x = 36z$(from C and E)
$20x = 24z$
$20x = 30y = 24z$
$2x/3 = y = 4z/5$

Knowing that $x$ is a multiple of $3$, $y$ is a multiple of $4$ and $z$ is a multiple of $5$ then leads to the solution.

Well done to you all.