Enigma 1743 - Order, order!

[Chappy]

Enigma 1743 - Order, order!
New Scientist magazine, 06 April 2013.
By Susan Denham.

I have written down a list of four positive
whole numbers in increasing order. They are
all less than 100, and no two of them have a
common factor greater than 1. If I write the
numbers in words, then each begins with a
different letter of the alphabet and my list
is also in alphabetical order. If I tripled
the numbers, then the answers would be in
reverse alphabetical order when written in
words. If I then doubled those answers, the
resulting numbers, when written in words,
would be in alphabetical order once again.

What is my list of numbers?

Ciao,
Chappy.

[John Jones]

On Wed, 15 May 2013 23:52:08 -0700 (PDT), "Chappy"
<petergreg...@hotmail.com> wrote in article <ed4735e2-d8fa-4fde-
8a20-38e...@googlegroups.com>...

Start by writing a table of 100 numbers as a 10 x 10 array.
write in the top of each cell the letter of the corresponding wordy
number, so cell (1,1)=eleven->e.
Now in the bottom write the letter of the triple. So 11->33->t.

Now you have a complete list of first letters and what they map to as
triples.
tedious, but here is the result in alpha order:
e -> t, f
f -> t, f, o
n -> t, f
o -> t
s -> e, t, f, o
t -> s, n, t, e, o

There are four numbers with different letters, and in alpha order.
The mapping must be in reverse alpha, ie t s o n f e

It doesnt look good for o, since if used it maps to t. Which means
everything else would have to, and if that was the answer it would be a
bit of a travesty and our Susan wouldnt do that, so that's out.

so the starting numbers are either efns, efnt, efst, enst, or fnst.
Juggling these with mappings shows that there is no solution unless the
triples are allowed to have at least one occurrence of two letters the
same.

The allowable mappings are then
e -> t , f
f -> f
n -> f
s -> o,f,t,e
t -> e

To keep the spirit of only two letters the same a candidate is
e->t f->f n->f t->e.

(I tried s->e and it didnt work)
From the table,
e->t can be 8->24 or 11->33. But when doubled 33->66(s) and that is
penultimate in the desired order e f n o s t so cant be right.
so e->t is 8->24 -> 48 = f.

f->f can be 5->15 14->42 15->45. 14->42 is out because (8,14)<>1.
doubling, 15->t and 45->n. It has to be n, so
f->f is 15->45.

n->f can only be 19->57.

t->e can be 27->81 28->84 29->87. 28 and 27 are out becuase they share
divisors with prior numbers. So 29->87 it is.

Final check
8(e) 24(t) 48(f)
15(f) 45(f) 90(n)
19(n) 57(f) 114(o)
29(t) 87(e) 174(o)

ie 8, 15, 19, 29.

HTH
JJ