# Tag Info

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First of all, let's see why your brute-forcing fails. (This is the puzzle part, the rest is plain old math.) No matter which you chose, the number at the bottom right would have to be both odd and even at the same time, so there's no integer solution. However, there are four equations and four unknowns, so we should have at least one solution (unless the ...

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Transcription of the grid: 00000000 D8 FF E0 FF 10 00 46 42 11 21 01 00 48 00 27 6E 00000010 2E 5F 2E 5F 6F 73 65 6D 61 52 64 6E 6D 6F 68 43 00000020 72 61 63 61 65 74 73 72 5F 2E 5F 2E 0A 0A 6E 4F 00000030 79 6C 61 20 20 6E 6C 69 75 6C 69 73 65 76 6D 20 00000040 6E 69 20 64 69 77 68 74 69 20 6C 6C 73 75 76 69 00000050 20 65 68 74 75 6F 68 67 73 74 63 20 6E ...

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The answer is The individual answers, starting from top left and moving right, are Then use A-Z = 0-25.

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The numbers If we then and And I think the secret word is Confession:

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The answer could require an intuitive understanding of long multiplication (as opposed to long division). $11^2$ $= 11(10 + 1)$ $= 110 + 11$ $= 121$ Similarly $111^2$ $= 111(100 + 10 + 1)$ $= 11100 + 1110 + 111$ $= 12321$ From the above two examples, you can notice a pattern: the middle number is equal to the length of the number, i.e., the number of ...

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Overall solution The nine symbols are, in the arrangement given: Step-by-step deductions From the factorial relationship, But from the perfect-power relationship, So The top left (division) relationship From the square root relationship, Going back to the perfect-power relationship, Now for the big product relationship: From the division ...

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We know that: It then follows that So the question boils down to finding The solutions are therefore:

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Second attempt The two-digit number is: I calculated it as follows: Previous attempt I will take a stab at it, and guess that the two-digit number is: I calculated it as follows:

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If you know what happens when you multiply numbers with small digits, then you might know $111111 \times 111111 = 12345654321$, so $11.1111 \times 11.1111 = 123.45654321$.

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Here's how to think about this, if you don't already know the neat solution and don't have enough computational power to divide into prime factors. $n^2= 123.45654321$ is an (even) power of $10$ times an integer, so it's enough to find the square root of $12345654321$. That's going to be a six-digit number. Since the number we have is so beautiful and ...

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2648: 1942: 2899: 2869: 1278:

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The missing numbers are Reasoning So

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Well, judging from the hint, each $[m\ast n]$ means In our case, we seem to have only $[n\ast n]$, in which case So rewriting all the expressions we've been given: Word 1 (2 letters): $[3*3]+[1*1]+[2*2]+[2*2]$ Word 2 (5 letters): $((-[1*1] + [3*3] + [4*4] + [8*8] + [27*27]) \cdot [2*2] + [1*1] + [1*1] + [5*5] \cdot [27*27]) \cdot [2*2]$ Word 3 (2 ...

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Solving the number puzzles: See Jens's answer - I didn't manage to get this one. See Jens's answer - I didn't manage to get this one. Turning the numbers into letters: The final step, motivated by the appearance of is to yielding the solution.

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My first small idea here is to notice that our number this is actually $12345654321 \cdot 10^{-8}$ so actually to find its square root we can divide it into parts $\sqrt{12345654321} \cdot \sqrt{10^{-8}}$ which makes us work with the integer number $12345654321$ instead of float. Then we can proceed with any factorization technique which can be used to ...

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The same 56 solutions but with logical reasoning instead of bruteforcing:

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The answer is most likely Because

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Each of the 4 weights has 3 possibilities. It can be in the left pan, right pan or off the scale. That gives 3 to the fourth = 81 possibilities. Ignoring the case where all the weights are off the scale, the same weight is represented twice, once with the left side tipping and once with the right side tipping, giving 80/2 = 40 possibilities. We need at least ...

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There are 56 solutions: I wrote an algorithm to find these solutions.

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Sum: is Method:

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Methodology:

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I know this is answered and ticked, but:

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