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Numbers À Propos (Posted on 2005-09-20) Difficulty: 3 of 5
If letters are assigned the values A=1, B=2, ..., Z=26, then any word will have a total value equal to the total of the values of its letters.

Doing this will give square a value of 81, which is a perfect square. The word prime will have a value of 61, which is prime. Also, odd will evaluate as 23, which is odd, and even as 46, which is even.

If we were to change the values of three of the letters of the alphabet, so that the other 23 letters had the same values as above, but the chosen three had their values scrambled, words would have different values. An example of scrambling the values of three letters would be if I chose B, E and H, to give B=8 (ordinarily the value for H), E=2 (ordinarily the value for B) and H=5 (ordinarily the value for E).

With a particular scrambling of the values of some three letters, square has a different value, but is still a perfect square; prime has a lower value but is still a prime; and odd is still odd and even still even, though at least one of odd or even has a different value from before.

What are the three letters that had their values changed, and what are those values?

See The Solution Submitted by Charlie    
Rating: 4.2000 (5 votes)

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Solution Solution | Comment 1 of 3

The letter values that have to be "rotated" are 4, 21, and 18.

So D=21, U=18, R=4

square=64, prime=47, odd=57, and even=46

It didn't look like square could be increased to 100, so it seemed most likely that it should be 64.  Since prime had to be decreased as well, R seemed like a likely candidate.  Dropping R all the way to 1 would achieve this, but then A would have to change as well.  A prime value of 47 meant dropping R by 14 which was close to the 17 necessary.  R becomes D (4).   Nice too that odd will always be odd since D is doubled and O is odd.  So now we needed to lose 3 more.  U happened to be three away from R so that seemed a likely choice.  U becomes R(18) and D becomes U(21).


  Posted by Bob Smith on 2005-09-20 20:37:35
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