Cheeky Cryptarithm

A*D=D gives A=1 as any other number either gets a bigger 1 digit number than D or a multi-digit number

Note that If there is a carry in for AD, then AD<D (mod 10) which means that either A is less than 1 (can't be for a 4 digit number) or AD is multidigit (can't be since the result has 4 digits).

D*D=1 (remainder 10) gives D=1,9 and since A=1, D=9

B*D can't have a carry out or else, the result will have more digits as this carry will be added to a 9 on the left side of the number

So B=0,1 given A=1 then B=0

C D+8(carry from 9 * 9)= 0 (remainder 10) C D->2 Given D=9 then C =8

As given here, if D = 1, A = 1 if D = 2, A = 4 if D = 3, A = 9 ... if D = 9, A = 1.

The only alternative where the answer stays in the scope of 4 digits without violating the rules of a cryptogram is A = 1, D = 9.

so 1BC9 * 9 = 9CB1.

Now, we know that 9CB1 is a multiple of 9, so 9 + C + B + 1 is divisible by 9 Maximum value of C and B is 8 and 7, so maximum value of 9 + C + B + 1 is is 25, and minimum is 10 + 0 + 3 = 13. The only number satisfying all conditions for 9 + C + B + 1 is 18

It's obvious that A is 1 and D is 9. Since except 1and0 all others on multiplication with 9 gives a number greater than 10 so a remainder is taken to the next set of numbers... B has to be 0. From this we can easily understand that C has to be 8.

AD < 10, which means that either AD = D or AD = D + 1. The only solution which satisfies this is D = 9 and A = 1.

This makes the equation (1009 + 100B + 10C)(9) = 9001 + 100C + 100B

=> 9081 + 900B + 90C = 9001 + 100C + 10B

=> 80 + 890B = 10C

=> 8 + 89B = C

The only solution for C that satisfies 0 < C < 10 is B = 0 and therefore C = 8.

The equation is therefore (1089)(9) = 9081 and the solution is 18.

A has to equal 1 since otherwise A * D would either not be a single digit integer or A would be greater than D which is not possible since A * D equals D. Therefore D has to be 9, since no other square has the unit digit 1 except 1 itself which isn't possible since then A and D would not be distinct. However, knowing that the sum of digits of any number divisible by 9 has to be divisible by 9, we can assume that D + C + B + A has to be a multiple of 9 and therefore 10 + C + B has to be a multiple of nine, which is just given in the case C + B = 8. Thus D + B + C + A = 18

A and D are so that : $A<D \\ A\times D <10 \\ D^2 = A mod(10)$

Naturally comes that $A=1$ and $D=9$ .

Then B is so that $B\times 9<10$ and $B\neq 1$ . So $B=0$

At last $8+C\times 9 = 0 mod(10)$ . Then $C=8$

$\overline{ABCD}=1089$

We know that D*D ends with the digit A. For every value of D, A = 1 | 4 | 5 | 6 | 9.

D has to be higher than 1, as D = 0 would result in A = B = C = D = 0, and D = 1 in A = D & B = C.

Therefore, we cannot have A = 5, A = 6 or A = 9 as we would end up with a 5 digit number with the multiplication

We also know that D > A, because DCBA (which starts with the digit D) is greater than ABCD (which starts with the digit A), as D >=2. Therefore, A cannot be equal to 4, as we would at least get 4000 * 5, which is a 5 digit number.

We now know that A = 1.

D has to be equal to 9, so that 9 * 9 = 81 (the other choice being 1 * 1 = 1, which is illegal).
Then we can't have B >= 2, as 1200 * 9 is a 5 digit number, nor B = 1, as A = 1. We therefore end up with B = 0.

We now just need C, and can formulate that (only caring about the unit digit, not the tens):
D * C + 8 (from the 81) = B
=>
9 * C + 8 = 0
=>
9 * C = 2
=>
C = 8 (as 9*8 = 72).

We can verify that 1089 * 9 = 9081.

Easy to find 1089 x 9 = 9801. But a curiosity involves the double of 1089 (2178): 2178 x 4 = 8712. And this two 4-digit numbers (1089 and 2178) seems to be the unique which presents this curiosity.