If you put three beads onto a tens/ones abacus you could make the
numbers 3, 30, 12 or 21. What numbers can be made with six beads?
Lee was writing all the counting numbers from 1 to 20. She stopped
for a rest after writing seventeen digits. What was the last number
I am less than 25. My ones digit is twice my tens digit. My digits
add up to an even number.
It was interesting to see that in the solutions to this problem, some of you decided that zero could only be a units digit whereas others thought it could go in the tens column. Unfortunately, not many of you described exactly how you went about finding five numbers. However, someone who didn't give their name, wrote:
I will number the targets, $1$, $2$, $3$, $4$ and $5$:
$1$.Largest even number
$2$.Largest odd number
$3$.Smallest odd number
$4$.Largest multiple of $5$
$5$.Number closest to $50$
He or she then looked at prioritising each of these in turn:
Number $1$ in priority:
$98$, $73$, $01$, $65$, $42$
In other words, the above contains the largest even number that is possible, then the best you can for the other numbers after that.
Number $2$ in priority:
$86$, $97$, $01$, $45$, $32$
Number $3$ in priority:
$96$, $87$, $01$, $25$, $43$
Number $4$ in priority:
$86$, $73$, $01$, $95$, $42$
Number $5$ in priority:
$96$, $87$, $13$, $42$, $50$
(Although children from Balfour Junior School noticed that the fourth number isn't a multiple of five in this last set of numbers.)
How can we decide which is the 'best' solution, do you think? From the other solutions we had sent in, it looked to me as if many of you approached the problem by looking for the highest possible even number first, then the largest odd number like the first solution above.
However, Jenna, Max, Mitch and Davis from Highland Elementary School sent this solution:
largest even number: $96$
largest odd number: $73$
smallest odd number: $01$
largest multiple of $5$: $85$
closest number to $50$: $42$
Laura from Farnborough Hill gave these five numbers:
$72$, $83$, $01$, $95$, $46$
Max from BRES gave the following which only uses zero in the units column:
largest even number: $98$
largest odd number: $75$
smallest odd number: $13$
largest multiple of $5$: $60$
number closest to $50$: $42$
The difficulty still remains - how do we judge which solution is 'best'? Well, we could decide that 'best' means as near as possible to the ideal number which has each property. So, the ideals would be $98$, $97$, $01$, $95$ and $50$, assuming we can have zero in the tens column, or $98$, $97$, $13$, $95$ and $50$ if we decide that zero can only be in the units column. One
way to judge how close a solution is to the ideal might be to work out the difference between the ideal number and the one you have. So, for example, for this solution:
$98$, $73$, $01$, $65$, $42$
$98$ is the ideal therefore the difference is zero. $73$ is twenty four away from the ideal $97$. $01$ is the ideal. $65$ is thirty away from the ideal and $42$ is eight away. So, we could say that in total, the difference is sixty two.
Which of the above solutions is closest to all the ideal numbers i.e. has the lowest difference?
If you have another way of deciding which solution is best, then please let us know!