11
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Each of the letters of the word EDUCATION is assigned a different integer value between 1 and 9, for a total value for the word of 45. With the same value for each of the letters, the total values of the letters in each of the seven words below are all different prime numbers. Moreover, the seven words have been organized from top to bottom in increasing order of value (thus UNITED has the greatest value).

CAT

NOD

DATE

CUTE

NICE

ACTION

UNITED

What is the value of each letter?

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  • $\begingroup$ What is the question? Do you want assignments of letter-value? Or something else? $\endgroup$
    – bobble
    May 28 '20 at 22:27
  • $\begingroup$ Value of each letter! $\endgroup$ May 28 '20 at 22:30
  • $\begingroup$ With the same value for each of the letters, the total values of the letters in each of the six words below are all different prime numbers. How will this be possible when all letters are assigned the "same" value? If they are assigned the same value, then the resulting number will be (length of the word) x value which is never a prime unless the value is 1 and length of every word is prime of which neither is true in this case $\endgroup$ May 28 '20 at 22:38
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    $\begingroup$ I believe that part means that each letter maintains its value in the different words; if E=4 in NICE then E=4 in UNITED, for example $\endgroup$
    – bobble
    May 28 '20 at 22:42
  • $\begingroup$ There are at least two solutions: the values for I and N are interchangeable. $\endgroup$ May 29 '20 at 0:09
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The solution is:

C = 1
D = 2
O = 3
A = 4
E = 5
T = 6
U = 7
N = 8
I = 9

This results in the following values for the words:
CAT 11
NOD 13
DATE 17
CUTE 19
NICE 23
ACTION 31
UNITED 37

Method:

The prime numbers in the range of interest for this problem, based on the minimum possible word score of a 3-letter word and the maximum possible word score for a 6-letter word are: 7, 11, 13, 17, 19, 23, 29, 31, 37.

CAT and NOD are the lowest-scoring words of the seven, so NOD can be at most 17. At a minimum, CAT and NOD have six different letters between them, so must add to at least 21. This rules out 7 and 11 for the two, and only allows {7, 17}, {11, 13}, {11, 17}, and {13, 17}.

So NOD is either 13 or 17, which means each of the higher remaining words can only have one of two possible values. For example, DATE can only be 17 or 19, and ACTION can only be 29 or 31. Furthermore, for the cases where NOD is 17 ({7, 17}, {11, 17}, and {13, 17}), we know what the exact values of all the higher words must be: DATE 19, CUTE 23, NICE 29, ACTION 31, UNITED 37.

For each of the above 4 possible sets of values of CAT and NOD, I created charts showing possible values of the 3 numbers comprising each word, as well as the leftover values, which must be assigned to the letters E, I, U. Having done that, I made the following observations: we know the sum of the letters in ACTDON, which is equivalent to CAT + NOD. It is 24 for {7, 17} and {11, 13}, 28 for {11, 17}, and 30 for {13, 17}. By knowing the value of ACTION (29 or 31), we know the difference between the values of D and I. Similarly, we know the sum of the letters in UNIOED (it is 45 - CAT), so by comparing it to UNITED, which must be 31 or 37, we know the difference between the values of T and O. In the aforementioned charts of possible values, we can determine the individual values of D, I, T, and O for each chart listing (sometimes down to one of several possibilities), and can eliminate sets of values that do not contain numbers which allow for the above D/I and T/O differences. If there are still multiple line items remaining, we can further narrow down and eliminate by next calculating the value of C, by subtracting the values of T and UE from the value of CUTE, noting that U+E is 45 - ACTDON - I, and we have already determined I (or at least know that it is one of two possible values). After this step, almost all possible line items in the charts have been eliminated. In a couple of cases, I needed to continue this method, using the word DATE, to finally eliminate all but one possibility. Once only one possibility remains, the values of all 9 letters can be determined with simple math.

I will not include all of my charts here, as it is lengthy, unless someone requests that information. I will only include the {11,13} chart, which looked like this:

 CAT NOD IUE
128 346 579
137 256 489
146 238 579
236 157 489
236 148 579
245 139 678

continued...

This chart shows all the various possible number assignments for CAT which add to 11, and then the remaining possible assignments for NOD which add to 13, plus the 3 remaining leftover letters. Here is an example of how a line item can be eliminated: we know that UNIOED (45 - CAT) adds to 34. We know that UNITED is 31 or 37. So we know that either T = O + 3 or T = O - 3. In the second to last row, there is no possible value of T among 2,3,6 in the CAT column which is three greater or three less than a value of O in the NOD column. So this row can be completely eliminated.

Now look at row 1. We know that ACTDON adds to 24 (11 + 13 for CAT and NOD in this chart), and ACTION is either 29 or 31. So we know that either I = D + 5, or I = D + 7. Look at the second and third columns of the first row, to find possible values of D and I which meet this criterion. The only possibility is with D as 4 and I as 9 (and, therefore, ACTION as 29). But now, on this same row, look for values of T and O which meet the above-mentioned criterion (T = O + 3 or T = O - 3). Only values of T = 1 and O = 4 satisfy this, but we already determined that D must be 4, and it is not possible to assign the same value to two different numbers, so this line item can also be eliminated. I will not include all of my step-by-step reasoning here, as it is lengthy, but using the above-described method, I was able to gradually eliminate all but one row among all of the four charts, leaving only one possible solution.

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Partial answer

The 7 words given have prime sums in ascending order. This limits their possible sums to the following:

enter image description here

Here are the possible combinations of how 3 letter words could have sums of 7, 11, 13 and 17:

enter image description here

Let's assume that CAT has the sum 7, which only has 1 combination. The word NOD must then have different numbers (as it has no common letters with CAT) which sum to 11, 13 or 17. The only possible combinations for NOD are given in yellow:

enter image description here

Both of the combinations for NOD sum to 17.

But if NOD is 17, then DATE must be 19 (see first figure), CUTE must be 23, NICE must be 29 and ACTION must be 31. But ACTION = CAT + NOI. And we know that CAT + NOD = 7 + 17 = 24. So "I" must be a number which is 7 greater than "D". But none of the two possible combinations of numbers for NOD has a number that can be increased by 7 (the smallest number is 3). So our assumption that CAT = 7 fails.

So therefore CAT must be 11 or 13. But if CAT is at least 11, this means NOD must be at least 13, DATE at least 17, etc. We can thus update the first figure to the following:

enter image description here

A work in progress...

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  • $\begingroup$ Consider CAT+NOD+EIU=45...? $\endgroup$ May 30 '20 at 0:34
-2
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This is a very fun puzzle, thanks!

Answer: THIS IS NOT POSSIBLE!

Posting as I figure things out so everyone can work off me :)

Part 0: Impossible word values (Credits to bobble)

2,3,5, 41, 43

Part 1: Comparisons that come from comparing similar words:

1) U > C (United vs Action)
2) U + E > A (Cute vs Cat)
3) N + I > T + U (Nice vs Cute)
4) D + E > A (Date vs Cat)
5) C + U > D + A (Cute vs Cat)

Part 2: Size of above comparisons:

Each aforementioned comparison has to be greater/less by at least 2. This is because there are no Prime pairs with a difference of 1 (except 2 and 3, neither of which can be reached).They also must be even, for a similar reason.

Part 3: Value of UNITED

UNITED = 37. It cannot be higher than 39 according to part 0. Also it cannot be lower than 36, as one more letter and it has to add to 45. 37 is the only prime between 36 and 39.

Part 4: Answer: This is impossible

If United = 37, C=8. This of course is impossible as U-C>=2 and U can't be > 9

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    $\begingroup$ Whoever downvoted this... why? It's quite common to post like this in order to help the community. We should cooperate not compete. $\endgroup$
    – Ankit
    May 28 '20 at 23:37
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    $\begingroup$ If UNITED = 37, then C+A+O = 8. $\endgroup$ May 29 '20 at 0:57

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