## Solution to Puzzle #167: 4 Rats in a Star

This was a relatively simple puzzle, and I received correct answers from Mahi Saraf, Pratik Poddar, “Mashex” and Prakhar. Well done all!

Here is the explanation:

Since the product is four digits long, RATS can be no greater than 9999/4. Thus the largest RATS can be 2499. This tells us that R is either 1 or 2.

Can R be 1? We also know that S x 4 ends in R. Since 4 is even and an even number times anything results in an even number, R must be even. Thus R can’t be 1, so must be 2.

2ATS x 4 = STA2

Since R = 2, we know 2ATS is somewhere between 2000 and 2499. Thus the product is somewhere between 8000 and 9996, and S must be 8 or 9.

Can S be 9? We know that S x 4 results in a number that ends in 2. 9 x 4 = 36, which ends in 6. 8 x 4 = 32, which does end in 2. Thus S = 8.

2AT8 x 4 = 8TA2

When we multiply 8 by 4, we get 32. We write down the 2 and carry the 3. Thus we know that T x 4 + 3 ends in A. Since T x 4 is even, T x 4 + 3 must be odd, so A is odd. Since the product is between 8002 and 8992, 2AT8 must be between 2001 and 2248.  So A is between 0 and 2.  The only odd number in that range is 1, so A = 1.

21T8 x 4 = 8T12

For T x 4 + 3 to end in 1, T x 4 must end in 8. The only values that will work are 2 and 7. R is already 2, so T can’t be 2. Thus T must be 7. Checking our work, we indeed get that

2178 x 4 = 8712

So we see that 2178 times 4 equals its reverse. Furthermore, 2178 x 2 = 4356 and 2178 x 3 = 6534, and those values are the reverse of each other as well.

Hope you all enjoyed the puzzle!

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