14
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Every letter is a different value from 1-26

  1. A x A = L + G
  2. C x C = U
  3. J x H = P + R
  4. W - K = N
  5. G + L = E + Z
  6. S + J = P - Y
  7. E x C = Z x B
  8. D + U = Z + E
  9. G x F = D + U
  10. O + X = V
  11. S x K = Z + E
  12. K x R = M
  13. D + U = R + P
  14. Q is higher than T
  15. O is higher than I

Can you work out their values?

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5
  • $\begingroup$ Hi! Welcome to Puzzling. Is this your own puzzle? If not, we need to have proper attribution for puzzles that you didn't create; this helps prevent plagiarism on our site. Thanks, and again, welcome! $\endgroup$
    – Stevo
    Commented Jan 21 at 10:05
  • 1
    $\begingroup$ Hi, I wrote it, sorry i didnt know i have to mention that $\endgroup$
    – LizardofOz
    Commented Jan 21 at 10:15
  • $\begingroup$ @LizardofOz you don't have to mention that it is your own puzzle; if it is not yours you have to provide proper attribution. $\endgroup$
    – Sny
    Commented Jan 21 at 13:56
  • $\begingroup$ It is still safer to mention it is one's own puzzle if it looks like it could have been taken from somewhere else, or it might get closed. $\endgroup$
    – Florian F
    Commented Jan 21 at 18:19
  • $\begingroup$ A and C can only be 3, 4 or 5. $\endgroup$ Commented Jan 21 at 18:22

2 Answers 2

7
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The key to this puzzle is that

Lines 1, 3, 5, 8, 9, 11 and 13 all have the same value: a square with at least 7 divisors greater than 1. These are A,F,G,H,J,K and S. The only possible value for this is square is 36, so we have A = 6 and the other six are 2, 3, 4, 9, 12, and 18.

We can proceed with:

C x C = U must be 5 x 5 = 25, as C is not a factor of 36.
U + D = 36 gives us D = 11.
F x G = G + L = 36, and this can only be 3 x 12 = 12 + 24.
R + P = 36 gives R > 9 and K x R = M gives us K = 2
S x K = 36 gives us S = 18
E x C = Z x B with Z + E = 36 gives us 36 / E = (5 + B) / B.
Trial and error finds E = 21 and B = 7

Further trial and error and a process of elimination complete the solution:

 A =  6  |  N = 17
 B =  7  |  O = 14
 C =  5  |  P = 23
 D = 11  |  Q = 20
 E = 21  |  R = 13
 F =  3  |  S = 18
 G = 12  |  T = 16
 H =  9  |  U = 25
 I = 10  |  V = 22
 J =  4  |  W = 19
 K =  2  |  X =  8
 L = 24  |  Y =  1
 M = 26  |  Z = 15

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  • $\begingroup$ perfectly done, nice job $\endgroup$
    – LizardofOz
    Commented Jan 21 at 20:28
7
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Looks like a correct answer got posted while I was working on this, but the following identifies all the values without trial and error:

First observe that there are lots of common elements in (1), (3), (5), (8), (9), (11) and (13).
Combining them we have A*A = G*F = S*K = J*H = L+G = E+Z = D+U = R+P.
The sums must all be between 3 and 51, so we need a square between 4 and 49 that can be factored four different ways.
The only possibility is 36 = 6*6 = 4*9 = 3*12 = 2*18.
So A = 6.

All the letters in the sums must be at least 10, but cannot be 18, and as G appears in both a sum and a product we know G = 12.
And that means F = 3 and L = 24.

Now look at (12): K x R = M
We know R >= 10 from the sums to 36, so K <= 2. But K=1 gives R=M.
So K = 2.
From the sums to 36, we know R >= 10, so 10 <= R <= 13, and 20 <= M <= 26.
And from the products to 36, we have S = 18, leaving J and H as 4 and 9 (but we don't know which way round yet).

Now look at (6): S + J = P - Y
We know S = 18, and J = 4 or 9. But the difference can be at most 25, so J = 4 .

So we have 22 = P - Y.
So Y <= 4, and the only free option is Y = 1.
Leaving P = 23.

And from the sums and products to 36, that means R = 13 and H = 9.
And from (12), M = 26.

Now look at (2): C x C = U.
The square gives us 2 <= C <= 5, and the only remaining option is C = 5.
So U = 25. And from the sums to 36, that means D = 11.

Now look at (7) - E x C = Z x B
We know C = 5, so Z*B must be a multiple of 5.
We also know E and Z sum to 36, and with several values taken, both must be between 14 and 22.
Rearranging we have E/Z = B/5.
With all lower values taken, we know B >= 7, and that also means E > Z.

If B >= 10 we would have E/Z >= 2, which is impossible given the possible values.
So 7 <= B <= 9

And that means Z must be a multiple of 5, and the only possibility is Z = 15.
So E = 21. And B = 7.

Now look at (10): O + X = V
With most values taken, there is only one possibility here: 8 + 14 = 22.
So V = 22.
And O,X are 8,14 in some order.

Looking at (15): O is higher than I
O cannot be 8, as there are no lower values free for I.
So O = 14 and X = 8.
And I = 10 (The only lower value available.)

Now look at (4): W - 2 = N
There is only one option among the remaining values: 19 - 2 = 17.
So W = 19 and N = 17

And finally (14): Q is higher than T makes Q = 20 and T = 16

1   Y
2   K
3   F
4   J
5   C
6   A
7   B
8   X
9   H
10  I
11  D
12  G
13  R
14  O
15  Z
16  T
17  N
18  S
19  W
20  Q
21  E
22  V
23  P
24  L
25  U
26  M

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