A P D
x A D
R P A D
D D C D
D P C E D
As, P + C= C
Hence value of P=9 Rule 1 - Case-II
Put P=9 and rewrite the problem,
A 9 D
x A D
R 9 A D
D D C D
D 9 C E D
further, you can see
A P D
x A D
R 9 A D
D D C D
D 9 C E D
Here D x D = _ D [ R 9 A D]
Hence possible values of D={5, 6} Detailed Explanation- Rule 2
Now proceed further...

1. C
2. D
3. D
A=7, P=9, D=5, R=3, C=6, E=2
7 9 5
x 7 5
3 9 7 5
5 5 6 5
5 9 6 2 5

A P D
x A D
R P A D
D D C D
D P C E D
As, P + C= _C
Hence value of P=9 Rule 1 - Case-II
Put P=9 and rewrite the problem,
A 9 D
x A D
R 9 A D
D D C D
D 9 C E D
further, you can see
A 9 D
x A D
R 9 A D
D D C D
D 9 C E D
Here D x D = _ D [ R 9 A D]
Hence possible values of D={5, 6} Detailed Explanation- Rule 2
Firstly take D=5 and rewrite the problem
A 9 5
x A 5
R 9 A 5
5 5 C 5
5 9 C E 5
A 9 5
x A 5
R 9 A 5
5 5 C 5
5 9 C E 5
Here, you can easily predict the value of R=3
So, the problem reduces to
A 9 5
x A 5
3 9 A 5
5 5 C 5
5 9 C E 5
As, A x 5 = _5 [ 5 5 C 5]
Hence possible values of A={3, 7, 9} Detailed Explanation
and as you have already taken R=3, Hence A cannot be equal to 3.
[In Cryptarithmetic, each variable should have unique and distinct value]
Hence possible value of A={7, 9}
Now, start hit and trial with the possible values of A
Firstly take A=7
Put A=7, and rewrite the problem again
7 9 5
x 7 5
3 9 7 5
5 5 C 5
5 9 C E 5
Now you can easily predict the value of C and E.
7 9 5
x 7 5
3 9 7 5
5 5 6 5
5 9 6 2 5
[Relax - it's going to take some time to understand the whole concept. If you are facing any difficulty. Please go through the Cryptarithmetic Tutorial.]

I googled so much to find a proper elitmus pattern and other information related to it,but no where i was able to find something so informative and proper like this.Commendable work!!

Thanks a lot

for P why 0 is not considered and for A why not 1 apart form 6 and 5?

`(For explanation only) Let's take P=0`

we don’t take p as zero bz if we do so, any no we multiply to p becomes zero ie a becomes zero (D*P=_A)

Because P+C=C, so this possible only when P ={0,9}. If we take P=0, multiplying anything with p i.e, 0 will give P. which is not coming as true here. so p=9. And for why A not equal to 1. because if A-1, when multiplied by A P D it should give same A P D which is not true here.

Because P+C=C, so this possible only when P ={0,9}. If we take P=0, what happens explanation is given. So, P=9. And for why A not equal to 1, because if A-1, when multiplied by A P D it should give same A P D which is not true here.

795*75

Here C could be 1,4

if c=1 solution in the option may be differ as expected.

can we solve this example by using unit digit method only without applying any other rules??

Hello Chatura,

Yes, you can solve this problem using the unit digit Method.

Unit Digit Method :

Really Great work 🙂

Start Analysing from D , as it is the only letter here Which should be either 5 or 6.

Because when D should be multiplied by D itself then the Units place of the new no. should be D only. And this is only possible if ,

D={5,6}

However the correct value of D is 5 .

Why cant we take D as 1?

Dear Karhik,

Yes D can be 1.

But assuming D=1 will not give you the solutions(Why ?)

Because if D=1 then Cryptarithmetic Problem will be so much easier to solve, which you should never expect from elitmus.

So, whenever these situation comes D*D=D(unit place) , then better to try first with D={5,6}

why c=6, we can take c=1,4,6,8…, i didnt get it make me understand.. thanks…

can c take 1?