In east Asia 2017 is the year of a rooster.
$$ABCD=\left(A \times CA^A + \frac{C}{D} \right) \times D$$
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2$\begingroup$ So what's the puzzle? $\endgroup$– blehCommented Jan 6, 2017 at 2:16
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1$\begingroup$ Did you find that image and puzzle elsewhere? If so, please cite your source. (I just wondered because of the text at the bottom of the image.) $\endgroup$– Rand al'ThorCommented Jan 6, 2017 at 2:21
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1$\begingroup$ @bleh This is an alphametic puzzle. Find each letter (or rooster). $\endgroup$– P.-S. ParkCommented Jan 6, 2017 at 2:27
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$\begingroup$ @P.-S.Park But like put that in the puzzle? $\endgroup$– blehCommented Jan 6, 2017 at 2:29
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1$\begingroup$ @randal'thor I made the image. But the weathercock icon was made by Tintins. I inserted the text by the copyright policy of the site. $\endgroup$– P.-S. ParkCommented Jan 6, 2017 at 2:29
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3 Answers
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The solution is, of course,
A = 2, B = 0, C = 1, D = 7.
We can prove that this is a unique solution like so:
Looking at A,
A is the first digit of a 4-digit number, and so cannot be 0. A cannot be greater than 3 either, since 144 is already 5 digits. A cannot be 3, since the only possible value for C keeping the result under 4 digits would be 1, but then there would be no solution for D. A cannot be 1, since then we would have (CD+C) equaling a 4-digit number, which is impossible. Therefore A must be 2.
Looking at C,
C is the first digit of a 2-digit number, and so cannot be 0. C cannot be greater than 3 either, since 2*422 is 3528, whose first digit is greater than 2. C cannot be 2, since A is 2. C cannot be 3, since 2*322 is 2048, and to keep the first digit 2, D is forced to be 1. However, subbing these values for A, C, and D in the right-hand side, the result is 2051, which doesn't fit the left-hand side. Therefore C must be 1.
Looking at D,
2*122 is 288. To get a 4-digit number starting with 2, D must be greater than or equal to 7. However, D cannot be 8 or 9 since the right-hand side would be 2305 or 2592 respectively, neither of which matches the left-hand side. D = 7 results in 2017, which fits and forces B to be 0.
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With the question, as it is now.
This could be accepted
A,B,C = 0
D = 1
answered Jan 6, 2017 at 2:38
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$\begingroup$ Yes it does. ABCD without the concatenation sign on top is A*B*C*D $\endgroup$– blehCommented Jan 6, 2017 at 3:10
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I don't know if there are several answers, but here is one possibility:
C is equal to 0, A and B are equal to any existing number, and D is equal to any number different of 0.
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