I will try to solve this problem in least number of complex mathematical steps possible. I will not use any maths greater than grade 8. ABCD*4 = DCBA implies A must be less than 3 since A is at 1000s place and 3*4 = 12, which makes the given number a five digit number. Thus A can be 1 or 2. Since DCBA is a multiple of 4 it implies that A must be even and hence A is 2. A is 2 again implies D must be 8 or 9. (D cannot be less than 8 since it is at the 1000s place of DCBA, which

The basic concept employed in this solution is that 100,000 divided by 271 leaves remainder as 1. Hence all the terms having powers greater than 5 or multiples of 5 are giving only remainder as 1 multiplied by corresponding coefficients. For example remainder when 10^11 is divided by 271 = 1*10 = 10, since 10^10 will give remainder as 1 only. This simplifies the problem. Refer attached image for full solution. #Maths #Remainder #basicarithematic