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This is a letter to number conversion and pattern puzzle.

What number goes in the blank circle and why?

enter image description here

Each letter is a distinctly seperate digit from 0 to 9. ( There are 10 letters representating 10 seperate digits from 0 to 9).

All the vowel digits add up to a Prime number.

N^M = A

I^S = D

N! = C(T-O)^I

N = I^2

Please explain the logic

Attribution: Part of the puzzle inspired by a famous puzzle from the Orient which I will reveal after the answers are in
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    $\begingroup$ Should [no-computers] be added? $\endgroup$ Jul 12 '20 at 13:04
  • $\begingroup$ Finding the values for each letter is easy enough, but something in the diagram isn't adding up...? $\endgroup$ Jul 12 '20 at 13:36
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    $\begingroup$ Very well put @DanielMathias $\endgroup$
    – DrD
    Jul 12 '20 at 13:37
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    $\begingroup$ Are you sure the very bottom one shouldn't contain the letters $AM$ rather than $AA$? $\endgroup$
    – Earlien
    Jul 12 '20 at 14:08
  • $\begingroup$ Yes I am sure @Earlien $\endgroup$
    – DrD
    Jul 12 '20 at 14:56
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The final answer is

9

To figure out the digits (building off some of the logic in Earlien's solution):

$N=I^2$ means $I$=2 or 3 and $N$=4 or 9. 0 or 1 are not possible if $N$ and $I$ are to be unique.

Since $N$ = 4 or 9, $N^M=A$ means $M$=0 and $A$=1 ($M$=1 will result in non-uniqueness, and larger $M$ result in $A$>9).

$I^S=D$ means ${I,S,D}$ = {3,2,9} or {2,3,8} since 0 and 1 are taken. However, $I$=3 and $D$=9 is impossible because $N$=9 when $I$=3. So now we can confirm $I$=2, $N$=4, $D$=8, and $S$=3.

$N! = C(T-O)^I$ simplifies to $24/C = (T-O)^2$. $C$=6 because that is the only possibility that gives a perfect square. Then $T$=9 or 7 and $O$=7 or 5.

Since the vowels must add up to a prime number, $E$=3, 5, or 9 if $O$=5, or $E$=1, 3, 7, or 9 if $O$=7. Since all those digits are taken except for $E$=9, then $O$=5, $E$=9, and $T$=7. We now have all the digits we need.

To figure out the missing number:

My first instinct was also to subtract the inputs from each other. However, this fails in the last step, where 20-10 != 11.

Instead, it seems you add all the digits of the inputs, so 1+3+2+3 = 9. Incredible how most of the numbers satisfy both patterns, until the last step befuddles you! Honestly, if I wasn't thinking about digits already, this would have been even tougher.

Upon some searching it seems this puzzle is inspired by:

Nob's number puzzle, a famous and similar number tree puzzle designed by Japanese puzzle guru Nobuyuki Yoshigahara.

Very clever, thanks for the puzzle!

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  • $\begingroup$ Very nice logical thinking @BrainEaser. Glad you liked it $\endgroup$
    – DrD
    Jul 12 '20 at 21:49
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Starting with $N=I^2$,

$N$ has to be either 4 or 9 (with $I$ being 2 or 3 respectively) otherwise $N$ and $I$ won't be unique (both 1) or too large. Given that there is a also an $N!$, it seems unlikely $N$ would be 9, so $N=4$ is a fairly safe assumption. And $I=2$.

Then

$N^M=A$ implies that $M$ must be zero and $A$ = 1 for the same reasons above - $A$ will be too large, or if $M=1$, then $A=N$ and are thus not unique.

From here,

it is pretty clear from $I^S=D$ must mean that $S=3$ and $D=8$ as there are no other small enough numbers for $S$ that work. $N! = C(T-O)^I$ simplifies to $24/C = (T-O)^2$. The only way that 24 will divide evenly with the remaining choices is if $C=6$, which means $T-O=2$. Thus $T=7$ and $O=5$. (As pointed out in the comments, at this stage, another possibility is $T=9$ and $O=7$, however, this will not satisify the last condition, discussed in the next step).

Which just leaves

$E=9$. And the vowels add up to 17 which is a prime number as advised.


For the second part, I do not know what the numbers (or equivalently, letters) should be in the missing circle.

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  • $\begingroup$ You need to eliminate the possibility of $T-O=9-7$ (easy to do) $\endgroup$ Jul 12 '20 at 13:35
  • $\begingroup$ In the first step above, there was another possibility as well, but making a few assumptions along the way is helpful and they are soon confirmed as correct. $\endgroup$
    – Earlien
    Jul 12 '20 at 13:36
  • $\begingroup$ I'm intrigued what the famous puzzle from the Orient is and how it relates to this one. $\endgroup$
    – Earlien
    Jul 12 '20 at 13:59

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