**From New Scientist #2506, 2nd July 2005**

If you let A=1, B=2, C=3 etc then you can use these values to calculate the total score of any word. For example, ENIGMA has a score of 49. Interestingly SQUARE has a score which is square, PRIME has a score which is prime, ODD has a score which is odd, and EVEN has a score which is even.

I have now adjusted the values of some of the letters. For twenty-three letters their value is the same as before, but for the remaining three their values have been shuffled around. (For example I could have shuffled B, E and H to give B=8, E=2, H=5 with the other letters unchanged.) With these new values SQUARE has a different score from before but it is still square, PRIME has a lower score than before but it is still prime, ODD is still odd and EVEN is still even (but at least one of these last two words has a different score from before).

Which three letters have had their values changed, and what are their new values?

[enigma1347]

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Solution:The three changed letters are: D=21, R=4, U=18.I’m getting an additional answer:-

D = 9, I=21, U=4 giving 64, 73, 46 and 33 for square, prime, even and odd

Perhaps this is excluded because Enigma is changed but the setter does not seem to specify that it must remain unchanged.

The value of PRIME needs to be smaller using the new values than it was using the old values.

With the old values PRIME has a score of 61, but using the new values you give you get a score of 73. This is prime, but it cannot give a solution as it is larger than the original value.

There are two solutions where the new value of PRIME is larger than the old value: D=9, I=21, U=4 and E=17, Q=24, X=5.