Copyright © University of Cambridge. All rights reserved.

## 'Numbered Cars' printed from http://nrich.maths.org/

Nicky from Westende Junior School approached
this problem very logically. He wrote:

Subtract the value of the numbers from each of the five number
plates.

So, for the illustrated car,

$2+0+8=10$ $65-10=55$ so S+V+B+J=$55$.

And for the 4 other cars

$2+5+3=10$ $65-10=55$ so V+H+D+S=$55$

$5+1+6=12$ $65-12=53$ so R+J+S+H=$53$

$2+0+2=4$ $65-4=61$ so V+B+D+S=$61$

$9+6+8=23$ $65-23=42$ so T+H+T+R=$42$.

Look at the letters on the two number plates beginning with V
(VHDS & VBDS). There is only one letter different. The plate
with H has a value 6 less than the one with a B, so that implies
the alphabet has been written backwards as the numbers increase
e.g.

$1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
26$

Z Y X W V U T S R Q P O N M L K J I H G F E D C B A

Check this solution works for the $5$ number plates which add up
to $65$. It does.

Use the code to find the value of the two new number plates:

T$584$YME and P$214$DOR both add up to $62$.

Excellent reasoning Nicky, although you were
lucky you picked Z as $1$ to start with! Bronya from Tattingstone
School used a similar method and noticed that it seemed to be that
the closer the letter is to the beginning of the alphabet, the
bigger the amount it is worth. She then tried A = $26$, B = $25$
etc as Nicky did.

Many of you read the clue and worked from
there. Very sensible! Leila and Sheya from St Andrew's C of E
Primary School in Totteridge told us:

First we tried A=$1$ B=$2$ C=$3$ and so on but
that did not work.

Then we tried A=$2$ B=$3$ C=$4$ and so on but that did not work
either.

The we tried the alphabet in reverse order A=$26$ B=$25$ C=$24$ and
so on and that worked!

Then using the code we worked out the total of the bottom number
plates:

T$584$YME=$7+5+8+4+2+14+22$=$62$

P$214$DOR=$11+2+1+4+23+12+9$=$62$

I wonder if you read the clue, girls?!