As Easy as 1,2,3
Problem
My calculator has 26 memories - one per letter of the alphabet. When I type a sequence of letters the calculator gives the product of all the numbers in the corresponding memories. I want to put numbers in the various stores so that when I type the word ONE it returns 1, and when I type the word TWO it returns 2, and when I type the word THREE it returns 3 and so on. How far can you get ? Is there an integer above which it is impossible to get ?
Student Solutions
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 features.
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 higher"
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 by saying:
"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 correct.
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= QUATTUORDECIM/QUATTUOR=14/4=3 ½ .
This is a contradiction so 14 cannot be reached". Well done again.
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 11.
- I can use the "German argument" to show we cannot reach 21, because to do so we would need ONE=1 and TWENTY=20 but multiplying these two gives TWENTYONE=20. We have a contradiction.
- I can use the "Latin argument" to show we cannot reach 16. We would need FOUR=4, TEN=10 and SIXTEEN=16. It follows that 6 x 10 x E=16 =>E=16/60=4/15. But we also need FOURTEEN=14 so 4 x 10 x E=14 =>E=14/40=7/20. Once again, we have a contradiction.
- Finally, an extension of the "Latin argument" shows we cannot reach 12. We would need ELEVEN=11 and TWELVE=12. Dividing these two equations gives ELEVEN/TWELVE=11/12 and cancelling produces 2 x EN=11 x TW . But we also need ONE=1 and TWO=2 and division here gives 2 x EN = TW. Another contradiction.
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, ELEVEN.
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 |
THR = 3 |
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 |
FV = 5 |
3SV = 7 |
|
NINE = 9 |
TEN = 10 |
3LV = 11 |
We have some choice now and putting V=1 and F= 5 gives;
|
ONE = 1 |
HR = 3/5 |
FIVE = 5 |
3S = 7 |
|
NINE = 9 |
TEN = 10 |
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 = 9/5
The equation FOUR = 4 becomes 5 x 1/3 x U x 1 = 4 and so U = 12/5
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 = 12
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