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How about 2178?

I started off the problem as follows:

First off the number is in the form of 1000*a + 100*b + 10*c + d. If a is any higher than 2, then when it's multiplied by 4 it won't be 4 digits any more, so a must be either 1 or 2. 10*b + a must be divisible by 4, because the reversed number of 1000*d + 100*c + 10*b + a must be divisible by four, and any number divisible by four has its last two digits divisible by four. That means that a can't be 1 because no multiply of four has a 1 in the ones digit. As a matter of fact, the only multiples of four that are less than 100 with a 2 in the ones place are 12, 32, 52, 72, and 92. 2500 doesn't work out, and when it is multiplied by 4, it becomes 5 digits. Therefore, b must be 1 or 3. The maximum, 2399, when multiplied out is 8796. That means that there will be no carry over from the multiplication of the previous digits, and that since 2*4 = 8, d must be 8. So far, we have a = 2, b = 1 or 3, and d = 8. That leaves 20 different numbers to check. I brute force checked them to find this one.

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Q: Write a 4 digit no. which when reversed is 4 times of the no.?
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