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In this puzzle you have to find unique values for the variables to make all statements true. I have proceeded to here and now I seem to be stuck. Logic equation puzzle

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    $\begingroup$ F = C+H but C, H are at least 2 and 3, so F >= 5. H > B and B >= 2 so H > 2 $\endgroup$
    – Florian F
    Commented Sep 1, 2023 at 6:46
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    $\begingroup$ H > B so B < 9. 3I >= B so I < 3. i = 2. So C, H are at least 3 and 4, F >= 7 and C, H <= 6. $\endgroup$
    – Florian F
    Commented Sep 1, 2023 at 6:48
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    $\begingroup$ Where is this from? We have a policy of crediting the original creators of all puzzles. $\endgroup$
    – bobble
    Commented Sep 1, 2023 at 19:20

1 Answer 1

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Final answer:

A = 7
B = 4
C = 3
D = 9
E = 6
F = 8
G = 1
H = 5
I = 2

Below are all the steps that I took to find this answer:

Step 1

CG = C, ⟹ G has to be 1.
6C = CE, ⟹ E has to be 6.
step1

Step 2

3I ≤ 9, ⟹ I is 2 or 3.
H > B, and B ≥ 2, ⟹ H ≥ 3. And of course B < 9.
F = C + H, and with C ≥ 2 and H ≥ 3 makes F ≥ 5. And F can't be bigger than 9, ⟹ C ≤ 6 and H ≤ 7.
step2

Step 3

H > B, ⟹ B ≤ 6. I has to be 2 or 3:
- If I = 2, ⟹ B ≥ 3 ⟹ H ≥ 4.
- If I = 3, ⟹ H ≥ 4. In both cases: H ≥ 4.
step3

Step 4

Let's rule out H = 7.
Suppose H = 7. then C = 2 and F = 9 (F = C + H). Then I = 3.
Then: A + F + H + 1 = C + DI ⟹
(≥4) + 9 + 7 + 1 = 2 + 3D ⟹
2 + 3D ≥ 21 ⟹
D ≥ 6. That leaves D with 8 (other digits are taken).
⟹ A + F + H + 1 = 25 ⟹ A = 25 - 9 - 7 - 1 = 7. And that is not possible.
step4

Step 5

You can now see that 7, 8 and 9 can only be given to A, D and F. step5

Step 6

Let's substitute F = C + H in A + F + H + 1 = C + DI. ⟹
A + (C+H) + H + 1 = C + DI ⟹
A + 2H + 1 = DI ⟹
(7/8/9) + 2(4/5) + 1 = (2/3) * (7/8/9) ⟹
(16..20) = (2/3) * (7/8/9) ⟹
(16..20) = 2 * (8/9). So I = 2 and D = 8 or 9.

Lets try them both:
D = 8:
(7/9) + 2(4/5) + 1 = 16 ⟹ 7 + 2 * 4 + 1 = 16
gives: A = 7, H = 4, I = 2, D = 8.
This means F = 9, C = 5 (F=C+H), B = 3 (H>B)
but: E ≠ H + I ⟹ 6 ≠ 4 + 2, so this solution is not correct!

D = 9:
(7/8) + 2(4/5) + 1 = 18 ⟹ 7 + 2 * 5 + 1 = 18
gives: A = 7, H = 5, I = 2, D = 9. This means F = 8, C = 3 (F=C+H).
And that leaves B = 4
step6

Let's check our work:

CG = C ⟹ 5 * 1 = 5
6C = CE ⟹ 6 * 5 = 5 * 6
B + 3 ≠ 2A ⟹ 4 + 3 ≠ 2 * 7
3I ≥ B ⟹ 3 * 2 ≥ 4
F ≠ A + D ⟹ 8 ≠ 7 + 9
F = C + H ⟹ 8 = 3 + 5
3I ≤ 9 ⟹ 3 * 2 ≤ 9
C + E ≥ A ⟹ 3 + 6 ≥ 7
A + F + H + 1 = C + DI ⟹ 7 + 8 + 5 + 1 = 3 + 9 * 2
E ≠ H + I ⟹ 6 ≠ 5 + 2
H > B ⟹ 5 > 4

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