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## 'Reverse to Order' printed from http://nrich.maths.org/

In general, if $a$ and $b$ are single digits and $a > b$, $ba$ becomes $ab$.

Between the two numbers $ab$ and $ba$ there is a difference of

$$

\begin{align}

10 ( a - b ) + ( b - a ) &= 10 a - 10 b + b - a\\

&= 9 ( a - b )\; .

\end{align}

$$

So to turn $ab$ into $ba$ do the following calculation:

$$ba = ab - 9 ( a - b )\;.$$

Here is an alternative solution:

Add the digits, multiply by $11$ and then subtract the original number.

Can you justify it?

Correct solutions to this problem were received from: Archbishop Sancroft High School; Stephen and Adrian - South Greenhoe Middle School; Jack, Paul and Matthew - Smithdon High School.

We also received a great solution by Emily from Stanley Park Junior School. She thought about how to reverse the order of the digits of any three or four digit number:

If we reverse the order of the digits of the three digit number $htu$ we obtain $h+10t+100u$.

Reversing the digit order of the four digit number $Thtu$ yields $T + 10h + 100t + 1000u$

($u$ = units digit, $t$ = tens digit, $h$ = hundreds digit, $T$ = thousands digit).

For example, to reverse the digits in the number $1234$ you would do the following:

$1 + 10\times 2 + 100\times 3+1000\times 4$ .

The sum of all the numbers is $4321$, the reverse of $1234$.