1. | Value of B ? | |||

(a) 2 | (b) 4 | (c) 6 | (d) 8 |

2. | Value of W/2 ? | |||

(a) 1 | (b) 2 | (c) 3 | (d) 4 |

3. | Sum of E + S + D + D ? | |||

(a) 16 | (b) 17 | (c) 18 | (d) 19 |

B P W x M E N E S D D P D M A D E M B M A D A D D As, D + A = D Hence A=0 Detailed Explanation Put A=0 and rewrite the problem. B P W x M E N E S D D P D M 0 D E M B M 0 D 0 D D Now proceed further... [If you are sure about any particular variable, then replace it and write the problem again. It will help you in further analysis of the problem]

```
1. D
2. C
3. C
B=8, P=4, W=6, M=3, E=5, N=7, A=0, D=2
8 4 6
x 3 5 7
5 9 2 2
4 2 3 0
2 5 3 8
3 0 2 0 2 2
```

```
B P W
x M E N
E S D D
P D M A
D E M B
M A D A D D
As, D + A = D
Hence A=0 Detailed Explanation
Put A=0 and rewrite the problem.
B P W
x M E N
E S D D
P D M 0
D E M B
M 0 D 0 D D
Now,
B P W
x M E N
E S D D
P D M 0
D E M B
M 0 D 0 D D
Here, You can see, E x W = _0 [P D M 0]
which is only possible when E=5 and W={2, 4, 6, 8}
[To get O at unit digit you must have to multiply 5 with an even number {2, 4, 6, 8}]
[5 x even =_0 and 5 x odd=_5] Possible ways of getting 0 at Unit Digit
Put E=5 and rewrite the problem,
B P W
x M 5 N
5 S D D
P D M 0
D 5 M B
M 0 D 0 D D
Further,
B P W
x M 5 N
5 S D D
P D M 0
D 5 M B
M 0 D 0 D D
At this stage, You have two clues,
1. D + 1 (carry) = M
2. P + 5 = _0 [ P + 5 + 1 (carry) = 10]
Hence, Possible value of P=4
[Relax it's going to take some time to understand the concept.Please read... again!]
Let's take P=4 and rewrite the problem again,
B 4 W
x M 5 N
5 S D D
4 D M 0
D 5 M B
M 0 D 0 D D
At this stage, Divide the problem in three parts
(1) B 4 W
x N
5 S D D
(2) B 4 W
x 5
4 D M 0
(3) B 4 W
x M
D 5 M B
If you further analyse, (2)
(2) B 4 W
x 5
4 D M O
Here, You can see 4 is at thousand place.
which will be only possible when B={8, 9}
[5 x 8 = 40 , 5 x 9 = 45]
We have to check further with each possible values of B.
Let's take B=8,
(2) 8 4 W
x 5
4 2 M 0 [ 4 D M 0]
If you compare side by side,you have D=2
Hence M=3 [As, D + 1(carry) = M]
Now it reduces to,
8 4 W
x 5
4 2 3 0
You can easily predict the value of W=6.
8 4 6
x 5
4 2 3 0
Put these values in the main problem
B=8, P=4, W=6, M=3, E=5, A=0
8 4 6
x 3 5 7
5 9 2 2
4 2 3 0
2 5 3 8
3 0 2 0 2 2
[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.]
```

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Why did you assume that P + 5 = _0 [ P + 5 + 1 (carry) = 10],that the carry part will be for sure 1 and not 2?

if p*m=m,then p=6 or m=5 but we cant take m=5,so p must be 6..how it is not..plz do reply…

Take value of B and W = 2,4,6,8 as options given in question 1amd 2 by this solution will became easy.

proceed without options to learn better

just hint

AS D+A=D ,IT MEASN A=0

E * W=A MEANS

A=MUST BE A PART OF 10,20,30,0R 40

10=2*5

20=4*5

30=6*5

40=8*5

MEANS ONE OF THEM MUST BE 5 AND OTHER WILL BE AN EVEN NUMBER

IF W =5 THEN

FROM MULTIPLICATION B P W

* N

______

D ….D MUST BE 0 OR 5…D CANT BE ZERO AS A IS ZERO AND D CANT BE 5 ALSO

SO W IS NOT 5 ….E WILL BE 5…AND POSSIBLE VALUES OF W{2,4,6,8}