Tricky number
Find the five digit number such that = . All digits are different.
Just go through your basics!!
Edit: Problem has been corrected.
The answer is 21978.
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From the beginning, a had to be an even number because edcba is a multiple of 4 thus the last digit is even. Also by considering the first digit of abcde, a cannot be 0 because that would make it a 4 digit number. It also cannot be greater than 2 because the resulting number will be a six digit number. Hence a = 2.
Using modular arithmetic, 4e = 2 mod 10 so e = 3 or e = 8. If e was 3 then abcde would be a four digit number as 40000 > edcba implying 10000 > 1/4 ebcda = abcde. Hence by force, e = 8.
Using the properties of multiples of 4, ba is a multiple of 4 so b is odd. Also, since abcde starts with 2 and, when multiplied, becomes 8, there is no carry from the thousands digit so b isn't greater than 2. Hence b = 1
From the last 2 digits of the numbers, 4*(10d + 8) = 12 mod 100. leading to 40d + 32 = 10 mod 1000 and 40d = 80 mod 100. This means d is either 2 or 7 and since 2 is already used for a, d = 7.
Lastly, the thousands digits is 1 and when multiplied by 4 becomes 7. This means the must be a carry of 4 from the previous place so c is either 8 or 9. As 8 is already taken, c = 9. Hence abcde = 21978 and a quick check shows that 21978 * 4 = 87912.