If ABCDE x 4 = EDCBA what is the number ABCDE? As usual with this type of problem each letter is a digit and different letters represent different digits.
+2441 This problem just requires some logic to crack it.
Both numbers have 5 digits. A must be equal to or equal to or less than 2 because, otherwise, the result would exceed the 5-digit restriction. 4E is even, so A only has one value: 2. This leaves us with 2BCDE*4=EDCB2
4E must end in a 2. Only two values achieve this: 3 and 8. E cannot be 3 because 20000*4=80000. 8 is the only option then. This leaves us with 2BCD8*4=8DCB2
B2 must be a multiple of 4 because of the divisibility rules for 4. B cannot exceed 2 because 3000*4=12000. This would cause a carry for 2BCD8*4, and the result would be impossible. Only 1 works because that makes 8DCB2 end in a 12, which is divisible by 4. This leaves us with 21CD8*4=8DC12
4D+3 must end in a 1, so 7 and 1 are the only options. D cannot be 1 because all the digits must be unique, according to the original problem; therefore, d=7. This leaves us with 21C78*4=87C21
We have a carry of 3 from 4*7+3=31, and 4c+3 must result in a carry of 3 so that the result holds true. This leaves us with a basic algebra problem:
| \(4c+3=c+30\) | |
| \(3c=27\) | |
| \(c=9\) | |
c=9
Therefore, \(21978*4=87912\)
