 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Wonderful problem (Posted on 2009-06-07) “The Square Root of Wonderful” was the name of play on Broadway.

If each letter in WONDERFUL stands for a different digit (zero excluded) and if OODDF, using the same code, represents the square root, then what is the square root of WONDERFUL?

 See The Solution Submitted by pcbouhid Rating: 2.0000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Not Quite Pencil and Paper | Comment 4 of 7 | O O D D F   W O N D E R F U L     X     X^2

It's construction requires just values for the bolded O, D and F with the O and D values copied to adjacent cells.  The 5 respective values are then multiplied by the appropriate power and summed to form X (the value of OODDF).  This is then squared.

Individual values of WONDERFUL are "stripped" from X^2.

That leaves 3 different values; a choice of 9 for O, 8 for D and 7 for F!

O <= 2 otherwise the square is 10 digits.
If O = 1 W will always be 1!  O = 2.

There are now 8 values for D and 7 for F.

The idea now is to get a value for D such that the value for O in WONDERFUL is either 1 (expecting a carry-over value) or 2.

Let F = 1 and proceed incrementing D from 1.  At D=7  O=1.  Testing F by increment from 1 does nothing for the O value.

Incrementing D to 8 suddenly gives an O value of 2.  It takes little then to find the 7 to give the desired square and pandigital.

I arrived at 22887 and 523814769.

When I originally wrote the above I forgot that I had given D the value of 9 but gave values of 8 for N and D, and 0 for F and U.

Edited on June 8, 2009, 11:06 pm
 Posted by brianjn on 2009-06-08 04:09:22 Please log in:

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