5
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Each of the 9 letters of the word POSTERIAL has been assigned a different integer value between 1 and 9. With those same values, the product of the letters in each of the following words is as indicated:

P L O T = 144

S A I L O R = 6,048

A L E R T = 840

S A T I R E = 10,080

A P O S T L E = 5,760

a) Determine the value of each of the letters of POSTERIAL.

b) Altogether there are about 369 words (without repeated letters) that can be drawn from POSTERIAL ( A prolific nine-letter word). What is the least number of these that would have allowed to unequivocally deduce the values of the letters of POSTERIAL.

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  • 3
    $\begingroup$ Could you please double-check your numbers? I get the same contradiction as Deusovi. $\endgroup$ – Gareth McCaughan Oct 23 at 16:43
  • $\begingroup$ I got L * L * R * T * 6912 = 840... $\endgroup$ – ention everyone Oct 23 at 16:49
  • $\begingroup$ I keep getting a contradiction as well. All the values in PILOT are also in APOSTLE but one should be different because of the I $\endgroup$ – Anthony Ingram-Westover Oct 23 at 16:50
  • $\begingroup$ Sorry! I meant PLOT...not PILOT. Cofinement blues! $\endgroup$ – Bernardo Recamán Santos Oct 23 at 16:59
  • $\begingroup$ You should really go back and give the green checkmark to the answer to the Prolific Nine Letter Word answer, since you have now used his answer for making your puzzle. $\endgroup$ – Anthony Ingram-Westover Oct 23 at 17:58
9
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The solution:

From APOSTLE, we know that R and I are 7 and 9 in some order. ALERT is divisible by 7, so we must have R=7, I=9.
A factor of 5 is in only the last three words, so E=5.

Dividing out factors from words, TAL = 24 and STA = 32. This means either S=8 and L=6, or S=4 and L=3 (but we don't know which yet). APOSTLE/(E * STA) = 5760/(32*5); so POL = 36, and T=4.
This resolves the ALERT/SATIRE ambiguity (because we can't reuse 4). And finally, SAILOR gives us O=2, and anything else gives us P=3.

So, the final values are:
123456789
AOPTELRSI


You would need

at least 4 words to make it uniquely deducible. (If you have only 3 words, there are only 8 possible combinations of "which word do these letters appear in", and so two letters will appear in the exact same set of words, making them indistinguishable.)

This is pretty easy to do: one possible such set is POST=30, LAIR=2016, STIR=280, PART=945.

From prime factorizations we immediately get R=7, T=5. POS must then be {1,2,3} in some order, and SI must be {2,4}; this gives S=2, I=4. PA is 27, so it must be {3,9}; this gives P=3, O=1, A=9, L=8. And then the leftover letter, E, must be 6.

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  • $\begingroup$ For the least number of words proof, does that work for any set of values for the word? Or just the specific values you provided (since they are different than the actual values from the puzzle) $\endgroup$ – Anthony Ingram-Westover Oct 23 at 17:56
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    $\begingroup$ @AnthonyIngram-Westover I haven't checked that it works with any values, but since these letters are so common, finding a permutation of them that works (and that has each set anagram to a word) shouldn't be difficult. $\endgroup$ – Deusovi Oct 23 at 18:13
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    $\begingroup$ ... Just tried it, and it took about a minute: ALOE, STIR, ROTE, RILE works for the given set. $\endgroup$ – Deusovi Oct 23 at 18:15

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