W H Y
x N U T
B B N P
B Y P Y
B U H A
B N E P B P
As, Y x U= _ Y [ B Y P Y ]
Hence, possible values of Y and U are
Case 1 When Y={2, 4, 8} and U={6}
Case 2 When U={3, 7, 9} and Y={5} Detailed Explanation- Rule 3
Now proceed further...

1. C
2. A
3. B
W=2, H=3, Y=4, N=7, U=6, T=5, B=1, P=0, E=9
2 3 4
x 7 6 5
1 1 7 0
1 4 0 4
1 6 3 8
1 7 9 0 1 0

W H Y
x N U T
B B N P
B Y P Y
B U H A
B N E P B P
As, Y x U= _ Y [ B Y P Y ]
Hence, possible values of Y and U are
Case-1 When Y={2, 4, 8} and U={6}
Case-2 When Y={5} and U={3, 7, 9} Detailed Explanation- Rule 3
Firstly, take case-1
Take U=6 and rewrite the problem,
W H Y
x N 6 T
B B N P
B Y P Y
B 6 H A
B N E P B P
At this stage, collect some more clues,
W H Y
x N 6 T
B B N P
B Y P Y
B 6 H A
B N E P B P
B + 6 + (carry) = N [Carry may be either 0, 1 or 2]
Hence, Possible value of N= {7, 8, 9}
You have to start hit and trial with possible values of N={7, 8, 9} and Y={2, 4 ,8}
Firstly taking N=7 and Y=2. and check further,
when N=7 and Y=2 then B=9 [ As, N + Y = B] If we take take B=9 then,
W H Y
x N 6 T
B B N P
9 Y P Y
9 6 H A
0 N E P B P
Value of B=0 and B=9 will come in the same problem. i.e. you are getting the two values of B.Rejected.
Now, Check the with other possible values of Y and N
Let's take N=7 and Y=4
then B=1 as N + Y = B [last digit]
Taking B=1, N=7 and Y=4 rewrite the problem again,
W H 4
x 7 6 T
1 1 7 P
1 4 P 4
1 6 H A
1 7 E P 1 P
Now, you can easily predict the value of A=8 As, 7 x 4= _ A [last digit]
W H 4
x 7 6 T
1 1 7 P
1 4 P 4
1 6 H 8
1 7 E P 1 P
Now,
W H 4
x 7
1 6 H 8
You can easily predict the value of W=2
2 H 4
x 7 6 T
1 1 7 P
1 4 P 4
1 6 H 8
1 7 E P 1 P
Now you can easily predict other values.
Value of T=5 and P=0
2 3 4
x 7 6 5
1 1 7 0
1 4 0 4
1 6 3 8
1 7 9 0 1 0
[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.]

Can u plz explain how the possible values of N = {7,8,9}.

You have note discussed about B .

because B+U<=9 and B+U=N

and value of u=6

then B may be 1,2,3 if B=1 then N=6+1=7

b=2 then N=8 and B=3 then N=9

actually ,B + 6 = N , is not creating any carry otherwise ,at last B would not be B.

So,if (B + 6 + (0 or 1 or 2) = N)

then, N < 9 ,so possible values of, N={7, 8, 9}

now if carry =0 ; B ={ 1, 2 ,3}

else if carry = 1;B = { 0 , 1, 2}

else if carry = 2 ;B={ 0 ,1} and N = { 8, 9}

please explain why we assume u=6 instead of 5.

In the solution of the problem we taken two cases (Case1 and Case2) at the beginning of the solution.

Case-1 When Y={2, 4, 8} and U={6}

Case-2 When Y={5} and U={3, 7, 9}

We have to do hit and trial with the both the cases. You can also take Case1 and proceed. Later on, Y=5 will get rejected.

Please try once with Y=5.

(We haven’t made any assumption while providing solution to this Cryptarithmetic Problem.)

Please let us know if more information is required.

We have U*Y=Y

if we take u=5

then

5*1=5(not 1),5*2=0(not 2),5*3=5(not 3)……………means right digit never come

but in case of u=6

6*2=12(means 2)

6*4=24(means 4)

6*8=48(means 8)

these type of questions can only be solved by trial and error method .you have to take values from options to get a answer otherwise you will get stuck in those problems.

i.e. N+6=B and further, can any one explain this point please,

HOW N+6=B PLEASE EXPLAIN

how N+6=B?

n even if it is true then in answer how come N=7 and B=1?

Can anyone explain how W=2