An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Find a great variety of ways of asking questions which make 8.
Choose any three by three square of dates on a calendar page...
We enjoyed the solutions here in different languages and hope
this will prompt more contributions to the Further Inspirations
section giving solutions using the number words in other languages.
NRICH has members in many countries. How many different solutions
to 'As easy as' can we collect?
One difficulty in writing mathematics is that the reader may
interpret your words differently than you intended. Paradoxically,
this can be an advantage - each interpretation can lead to a
different problem and each problem may have interesting
Some students assumed they had to use integers [whole numbers]
in As easy as 1, 2, 3
Amongst solvers of the "whole number" problem were James Dotti,
Shabbir Tejani and Mark Wang of Jack Hunt School, Peterborough who
submitted similar but not identical solutions.
Here is Shabbir's argument.
To make ONE=1, we must have O=1, N=1 and E=1.
How do we make TWO=2 ? Looking ahead, T is in THREE but W is not
in any number name up to 10, so we must put T=1 and W=2.
Similar arguments give E=1, F=1, H=3, I=1, N=1, O=1, R=1, S=1,
T=1, U=4, V=5, W=2, X=6 and give ONE=1, TWO=2, THREE=3, FOUR=4,
FIVE=5 and SIX=6
But how about SEVEN=7. We already have SEVEN=1 x 1 x 5 x 1 x 1=5
so we cannot reach 7.
James and Mark reached the same conclusion, but James went
further, saying "I thought maybe fractions would get the total
Using the same approach as before, James managed to reach 8
using fractional values.
So well done James, Shabbir and Mark.
Three students from Madras College in St Andrews, Jonathan Hyne,
Elizabeth Whitmore and James Wylde, came nearest to a full solution
to the problem. They begin with a correct statement of the solution
"This problem can be solved as far as eleven in an innumerable
number of ways. This can be proved and it can also be proved that
this is the maximum value".
However they don't give a solution up to eleven [instead they
indicate how it could be done] and their argument as to why TWELVE
cannot be reached is flawed.
But Jonathan, Elizabeth and James don't stop there. They go on
to attack the problem in German and here their argument is
They begin by finding values for individual letters to correctly
give EIN=1, ZWEI=2, DREI=3, VIER=4, FUNF=5, SECHS=6, SIEBEN=7,
ACHT=8, NEUN=9, ZEHN=10, ELF=11, ZWOLF=12 thus proving that 12 can
be reached in German [I leave finding the values to the reader] and
then point out that since DREI=3 and ZEHN=10 one is forced to admit
that DREIZEHN = 3 x 10 = 30 and not 13 as we would like. This
contradiction shows that we cannot reach 13 in German.
This "proof by contradiction" is very powerful. Well done
Jonathan, Elizabeth and James.
Not content with a good attempt at the English problem and a
full solution to the German problem, they than attacked the Latin
problem and use an even more sophisticated method of proof.
They find values for individual letters to correctly give all
Latin numbers up to TREDECIM=13.
Then they say
"The word for 14 is QUATTUORDECIM and has no solution.
This can be shown as DUODECIM=12 means that DECIM=12/DUO=12/2=6
and QUATTUORDECIM=14 means that DECIM=
This is a contradiction so 14 cannot be reached". Well done
The arguments by contradiction used by Jonathan, Elizabeth and
James in solving the German and Latin problems can be used to prove
their correct assertion that, in English, we can reach no more than
So the most we can hope for is a maximum reach of 11, and it
turns out this is possible in more than 1 way.
The number names are ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN,
EIGHT, NINE, TEN, ELEVEN.
First note that W occurs only in TWO, U occurs only in FOUR, X
occurs only in SIX and G occurs only in EIGHT. So we can leave
these number-names to the end and adjust the W, U, X and G as we
please. So we work on ONE, THREE, FIVE, SEVEN, NINE, TEN,
One solution is as follows;
Set E = 1 So we still require;
ON = 1
THR = 3
FIV = 5
SVN = 7
NIN = 9
TN = 10
LVN = 11
Set N = 3 and I =1 So we still require;
3O = 1
FV = 5
3SV = 7
NINE = 9
3T = 10
3LV = 11
We must now put O = 1/3 and T = 10/3 leaving;
ONE = 1
HR = 9/10
TEN = 10
We have some choice now and putting V=1 and F= 5 gives;
HR = 3/5
FIVE = 5
3S = 7
3L = 11
Finally, S = 7/3, L= 11/3 fixes SEVEN and ELEVEN and R = 1, H =
1/5 fixes THREE
The equation TWO = 2 is now 10/3*W x 1/3 = 2 ie 10W/9 = 2 ie W =
The equation FOUR = 4 becomes 5 x 1/3 x U x 1 = 4 and so U =
The equation SIX = 6 gives 7/3 x 1 x X = 6 and X = 18/7
The equation EIGHT = 8 becomes 1 x 1 x G x 1/5 x 10/3 = 8 and G
E = 1 R = 1
F= 5 S = 7/3
G = 12 T = 10/3
H = 1/5 U = 12/5
I =1 V = 1
L= 11/3 W = 9/5
N = 3 X = 18/7
O = 1/3