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Sunday Times Teaser 2756 – Terrible Teens

by Victor Bryant

Published: 19 July 2015 (link)

I have allocated a numerical value (possibly negative) to each letter of the alphabet, where some different letters may have the same value. I can now work out the value of any word by adding up the values of its individual letters. In this way NONE has value 0, ONE has value 1, TWO has value 2, and so on up to ELEVEN having value 11. Unfortunately, looking at the words for the numbers TWELVE to NINETEEN, I find that only two have values equal to the number itself.

Which two?

4 Comments Leave one →
  1. Brian Gladman permalink

    Here is a solution using SymPy:

    Here is a manual solution (this analysis has been updated to take account of Tony’s comment below).

    First we have: \[\begin{align}\small FOURTEEN&=\small 4+\small TEEN \\ \small SIXTEEN&=\small 6+\small TEEN \\ \small SEVENTEEN&=\small 7+\small TEEN \\ \small NINETEEN&=\small 9+\small TEEN\end{align}\]

    So, whatever value \(\small TEEN\) has, these are all correct or all incorrect. And, since only two can be correct, this means that they are all incorrect.

    Second, we can show that \(\small TWELVE\) is correct as follows:

    \[\begin{align}\small TWELVE&=\small TW+\small LV+\small 2\small E \\&=(\small TWO-\small O)+(\small ELEVEN-\small N-\small 3\small E)+\small 2\small E\\ \small &=\small TWO+\small ELEVEN-\small ONE \\ &=\small 12\end{align}\]

    Third, we have \[\begin{align}\small THIRTEEN&=\small THREE+\small TEN+\small I-\small E \\ \small &=\small THREE+\small TEN+\small NINE-\small 2(\small N+\small E)\\ \small &=\small 22 -\small 2 \small E – \small 2(\small NONE – \small ONE) \\ &= \small 24-\small 2\small E\\ \small EIGHTEEN&=\small EIGHT+\small N+\small 2\small E\\ \small &=\small EIGHT+\small NONE-\small ONE+\small 2E\\ \small&=\small 7+\small 2E\end{align}\]

    To make \(\small THIRTEEN\) correct, \(\small E\) has to be 5.5 but this value also makes \(\small EIGHTEEN\) correct. However, if both are correct we would have three correct values when there are only two. So they must both be incorrect.

    Hence only \(\small TWELVE\) and \(\small FIFTEEN\) can be correct.

  2. Tony Smith permalink

    (1) The above does not show that FIFTEEN is actually correct (although your algebra on the other site does).
    (2) On the other site John Crabtree correctly points out that letter values are not restricted to integers.
    Allowing non-integers still gives the solution TWELVE, FIFTEEN but for a reason different from the above.
    Your algebra leads to TWELVE=12, FIFTEEN=15, THIRTEEN=24-2E and EIGHTEEN=2E+7.
    If E=5.5 THIRTEEN and EIGHTEEN would both be correct (not allowed)
    Also if E=0 FOURTEEN, SIXTEEN, SEVENTEEN and NINETEEN would also be correct (not allowed).
    Any value of E other than 5.5 or 0 and any value of G (real or complex in each case) would lead to a valid set of letter values.

  3. Brian Gladman permalink

    Hi Tony,

    Yes I assumed integer values in my proof above, which, as you say, is not necessary to obtain a solution. I’ll update the analysis as you suggest. But, given the integer values assumption, \(\small FIFTEEN\) is shown to be correct indirectly by showing that other words in \(\small THIRTEEN\) to \(\small NINETEEN\) are incorrect (unless, of course, the teaser is wrong).

    • geoffrounce permalink

      Here is a solution in MiniZinc:

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