Re: Enigma 1531 - The year in question

On 30 Mar, 07:51, Richard Heathfield <r...@xxxxxxxxxxxxxxx> wrote:
[Puzzle text and initial prime table are spoiler space]

Chappy said:

Enigma 1531 - The year in question
New Scientist magazine, 7 February 2009.
By Susan Denham.

I have written down three 3-figure numbers
which, overall, use no digit more than once.
One of the numbers is a perfect square and
the two other numbers are primes.  Their
total is 2009.

With a little logic it is possible to
calculate the three numbers very quickly.
What (in increasing order) are they?

I assume no leading 0s.

The three-digit non-dup-digit primes are:

snip
and we can immediately see that 983+401=1384, so the answer is:

Perfect square: 625
Prime #1:       401
Prime #2:       983

I have not proved that this solution is unique.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

Good analysis Richard.
But just to justify the statement "it is possible to calculate the
three numbers very quickly":-)

3-digit primes must be odd, so the square left after substracting two
of them from an odd number must be odd.

By inspection of 11^2 upto 31^2 there are only 8 odd three digit
squares which have no dups.

On subtraction from 2009 these leave 4-digit even numbers, and the
last digit is p1 + p2 mod 10.
Odd primes end in 1 3 7 or 9 so you can work out the possible digits
for the primes,
and remove any dups from the digits in the squares.

Then you can list the five digits left, of which two go to complete p1
and two go to complete p2.

the resulting table is

square (2009-square) p1,p2 end digits remaining digits
169 1840 3,7 0 2 4 5 8
289 1720 3,7 0 1 4 5 6
361 1648 7,9 0 2 4 5 8
529 1480 3,7 0 1 4 6 8
625 1384 1,3 0 4 7 8 9

(other squares have no possible solutions)

The first row can be removed because 8xx + 5xx isnt big enough to be
1840
similarly for row 2 and 3

The rest is brute force and ignorance (apart from knowledge of primes)
applied to the final two rows.

I can confirm that your solution is unique.

The other row has a near miss, being 529 + 617 + 863 which dups 6 but
otherwise satisfies requirements.

Anyone see a quicker way?
HTH
JJ


.

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