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Thank you for your solutions to this tricky challenge. Sebastian and Josh from Swaffham Prior C of E Primary School Mega Maths Club found two solutions to the six-digit challenge and outlined their method for us:

$123654$ (Sebastian) and $321654$ (Josh)

We knew that the fifth number had to be the $5$.

We also knew that the second, fourth and sixth numbers had to be even because they had to be multiples of even numbers.

This meant that the first and third numbers could only be either $1$ or $3$.

Knowing this we used trial and error to find a number that followed the rules.

We used a calculator to find out if the numbers were multiples of $3$, $4$ and $6$.

We didn't need to use a calculator for the first, second and fifth numbers as these were obvious.

Zara from Oldfield Primary School gave us a bit more detail about how she worked on the solution.

First of all we decided to give each digit a position number eg: for the number $235614$ Position $1$ = $2$, Position $3$ = $5$

Next we worked out that position $5$ must be a $5$, because it has to be divisible by $5$ and there is no zero.

We then saw that position $2$, $4$ and $6$ must be even numbers, because they had to be divided by $2$, $4$ or $6$.

That left position $1$ and $3$, which had to be $1$ or $3$!

We knew that a number is divisible by $3$ if its digits when added together are divisible by $3$. So, the first three numbers must be $1$, $3$ and then an even number in position $2$, so when they are added together are divisible by $3$.

We tried $1 + 4 + 3 = 8$ this is not divisible by $3$

We tried $1 + 6 + 3 = 10$ this is not divisible by $3$

We tried $1 + 2 + 3 = 6$ this IS divisible by $3$

These are the first three numbers - and we found out that $1$ and $3$ can be swapped around $123$ or $321$.

We now had $123?5?$

We worked out the last two numbers by trying them and ended up with solutions:

$123654$ and $321654$.

Daniel also explained very clearly how he reached the solution. He started in the same way as Zara but then used a slightly different approach:

Since for the first operation (i.e. dividing the six-digit number by $6$) the number must be divisible by $6$, the last single digit number must be even.Since for the second operation (i.e. dividing the five-digit number by $5$) the number must be divisible by $5$, the second last single digit number can only be $5$.

Since for the third operation (i.e. dividing the four-digit number by $4$) the number must be divisible by $4$, the third last single digit number must be even.

Since for the fifth operation (i.e. dividing the two-digit number by $2$) the number must be divisible by $2$, the fifth last single digit number must be even.

By process of elimination, the other two numbers must be $1$ and $3$.

So for the format of the six-digit number, we have:

($1$ or $3$),($2$, $4$ or $6$),($1$ or $3$),($2$, $4$ or $6$),$5$,($2$, $4$ or $6$)

For the three-digit number to be divisible by $3$, its digits must add up to a multiple of $3$. Since this number contains $1$ and $3$, along with $2$, $4$ OR$ $6 we can work out:

$1+3+2=6$ (divisible by $3$)

$1+3+4=8$ (not divisible by $3$)

$1+3+6=10$ (not divisible by $3$)

Therefore the second number must be $2$.

So the format of the number is now:

($1$ or $3$),$2$,($1$ or $3$),($4$ or $6$),$5$,($4$ or $6$)

Since for the third operation (i.e. dividing the four-digit number by $4$) the number must be divisible by $4$, the number made by its last two digits must also be divisible by $4$. These digits are ($1$ or $3$) and ($4$ or $6$)

Therefore the possible numbers are $14$, $16$, $34$, and $36$

Since only $16$ and $36$ are divisible by $4$, the fourth digit of the number must be $6$.

By process of elimination the last number must therefore be $4$.

So we are left with ($1$ or $3$),$2$,($1$ or $3$),$6$,$5$,$4$

This gives the two numbers $123654$ and $321654$, both of which are correct, as shown above.

Hannah, Megan, Jess, Toby, Callum, Callum, Lydia, Nadia, Sam and Chris from Peterchurch Primary School also used very similar reasoning to start with, but used a different method again towards the end of their solution:

$123654$ and $321654$

We worked this out by the following process:

1. We knew that the fifth digit must be a $5$ because that's the only number divisible by $5$ (because we had no zero)

2. We knew that the second, fourth and sixth digits had to be even to be divisible by $2$, $4$ and $6$.

3. The first and third digits had to odd - $1$ or $3$

4. We worked out that the second digit must be $2$ because 143 or 163 would not be divisible by $3$

5. This meant that we only had four combinations left:

$123456$, $123654$, $321456$, $321654$

We tested these to find the only two possible solutions.

We also received fantastic solutions from Samuel at St Michael's Primary in Oxford; Orange Kingfishers at Wood Farm Primary; Alyssa, Caleb, Chelsea, Jake and Katie from Myland Primary and Krystof from Uhelny Trh, Prague.

Finally, Rachel from Greenacre School for Girls took the challenge even further. She used digits $0$ to $9$ and found one solution:

$3,816,547,290$

I used trial and improvement to solve this - it took me about two hours!

Very well done, Rachel. What perseverence! The challenge that Rachel set herself is in fact another NRICH problem in its own right: American Billions. Try it yourself!