Look, 25 points y'all
this is a logic-math combo question that was from my 5th grade math book (singapore math). just for fun. i never solved it on my own. must show all work
there are 4 single digit whole numbers represented by a, b, c, d. they are all different numbers.
when you multiply abcd with d, you get dcba
solve for the values of a, b,c, and d
(note: if a=2,b=3,c=4,d=5 then abcd=2,345, but that is not the solution)
show all work and do not copy

To start, ABCD(D)=DCBA means that D×A=D because they're all single digit numbers and since nothing carries over to make DCBA a 5 digit number A must be 1. If A is 1 for ABCD(D)=DCBA that would mean D^2 is a number ending in 1 so the only possibility for D is 9. Since D is 9, D×B has to be less than 10 because we know nothing carries over to the D×A (9×1) so B has to be 0. So far we have 10*9(9)=9*01 and since D×D (9×9) is 81 we need a number divisible by 9 that ends with 2 to add to the 8 digit which the only possibility is 72 from 8×9. SO A=1, B=0, C=8, D=9
1089(9)=9801