Friday 17 June 2011

Cambridge Cryptogram

This is the IQ puzzle that I referred to in my post "The last supper". It is copied verbatim from the book of IQ puzzles that Ray gave me, including the somewhat curious double,value of "N" in the example. I think it is just a proofing error, but who knows?

Cambridge Cryptogram

Each letter has a numerical value attached to it, and the total of all the letters equals the professor's value. For example, if the letters N,E, W, T, O and N had values of 12, 7, 9, 14, 21 and 5, respectively, then Isaac Newton would have a numermical value of 68.

Your objective is to figure our Hawking's numerical value.
BARROW 71                TURTON 80
NEWTON 70                AIRY 46
WHISTON 104              BABBAGE 84
SAUNDERSON 129           KING 45
COLSON 51                LARMOR 58
WARING 92                DIRAC 52
MILNER 58                LIGHTHILL 130
WOODHOUSE 108            HAWKING ?

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3 comments:

  1. David Barnes17 June 2011 at 18:34

    After the earlier teaser, I am glad you decided to post it! Do you think there is a constraint on the letter values being distinct?

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  2. Patrick O'Hara7 August 2011 at 18:21

    I used the internet to help! The solution is 106 with these values:

    A B C D E G H I K L M N O R S T U W Y
    17 5 9 1 13 22 26 11 2 6 4 10 3 14 20 16 21 18 4


    My method:
    1. Lookup simultaneous equations on Wikipedia: http://en.wikipedia.org/wiki/Simultaneous_equations
    2. Follow the link to Gaussian Elimination: http://en.wikipedia.org/wiki/Gaussian_elimination
    3. In the References is this interesting item: A program that performs Gaussian elimination similarly to a human working on paper Exact solutions to systems with rational coefficients, the link is: http://marekrychlik.com/cgi-bin/gauss.cgi. I pasted the problem matrix and it reduced it to the reduced row echelon form.
    4. Wrote a program to try combinations of the four free variables left after the elimination. I chose to use Python, which has very nice rich data types, including Fractions and arrays.
    5. Iterated fixing finger-trouble for some hours.

    The moral of the story is that most problems can benefit from the internet. Though I could not find a page with the solution. After this post there will be one!

    Cheers
    Patrick

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  3. David Barnes8 August 2011 at 18:31

    Nice to see someone persevered! I followed the same first three steps and gave up at that point!

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