A P R
x O C T
P U R A
R O J R
R E C U
R A A J A A
As, R x C= _C [ R O J R]
Hence Possible values of R and C are
Case 1 When C={6} then R={2, 4, 8}
Case 2 When R={5} then C={3, 7, 9} Detailed Explanation -Rule 3
At this stage you can also collect one more clue
A P R
x O C T
P U R A
R O J R
R E C U
R A A J A A
i.e. R + R = A
Now proceed further...

1. A
2. B
3. D
A=4, P=3, R=2, O=5, C=6, T=7, J=9, U=0, E=1
4 3 2
X 5 6 7
3 0 2 4
2 5 9 2
2 1 6 0
2 4 4 9 4 4

A P R
x O C T
P U R A
R O J R
R E C U
R A A J A A
As, R x C= _C [ R O J R]
Hence Possible values of R and C are
Case 1 When C={6} then R={2, 4, 8}
Case 2 When R={5} then C={3, 7, 9} Detailed Explanation -Rule 3
At this stage you can also collect one more clue
A P R
x O C T
P U R A
R O J R
R E C U
R A A J A A
i.e. R + R = A
Now firstly take Case-1
Case 1- C={6} then R={2, 4, 8}
take C=6 and R=2
Put C=6 and R=2 and rewrite the problem again.
A P 2
x O 6 T
P U 2 A
2 O J 2
2 E 6 U
2 A A J A A
A P R
x O C T
P U R A
R O J R
R E C U
R A A J A A
At this stage you can easily predict the value of A=4
Put A=4 and rewrite the problem again
4 P 2
x O 6 T
P U 2 4
2 O J 2
2 E 6 U
2 4 4 J 4 4
At this you can easily predict the value of T=7
(As, 2 x T= _ 4) (T cannot be equal to 2 as you have already taken R=2)
Put T=7 and rewrite the problem.
4 P 2
x O 6 7
P U 2 4
2 O J 2
2 E 6 U
2 4 4 J 4 4
At this stage, you can see 2 + E + 1(carry) = 4
Hence E=1
Now,
4 P 2
x O
2 E 6 U
you can see, 2 is present at thousand place, which is only possible when value of O=5
(Value of O cannot be equal to 6,7 as you have already taken these values)
Hence put O=5 and E=1 and rewrite the problem.
4 P 2
x 5 6 7
P U 2 4
2 5 J 2
2 1 6 U
2 4 4 J 4 4
Now you can easily solve the problem.
4 3 2
x 5 6 7
3 0 2 4
2 5 9 2
2 1 6 0
2 4 4 9 4 4
[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.]

how to solve those ques in which there is a ques mark in place of alphabet..plz sir help

Here at 10th place R+R=A, which means that A is an even number. now check ques.3, even values of A in option are 2 and 4, which means R= 1 or 2 (A=2R)

now lets R=1, but in first multiplication T*R !=T, so R cannot be 1. hence R=2 and A=4

now in first multiplication T*2=4 which is possible for only 2 values of T(2,7) but T cannot be 2 so T=7

NOw solve it by yourself

You have to find the set which is containing the odd numbers only.