1. | Value of 2B + M ? | |||

(a) 5 | (b) 6 | (c) 7 | (d) 8 |

2. | Value of M + A + R + L + E + Y ? | |||

(a) 19 | (b) 20 | (c) 21 | (d) 22 |

3. | Value of Y ? | |||

(a) 6 | (b) 7 | (c) 8 | (d) 9 |

B O B x B O B M E O Y M I L O M E O Y M A R L E Y Here, B x O=_O [M I L O] Hence Possible Values of B and O are Case-I - When O={5} and B={3, 7, 9} Rule 3 Case-II - When B={6} and O={2, 4, 8} Here, you have one more clue for solving the problem, B O B x B O B M E O Y M I L O M E O Y M A R L E Y O + O = E i.e. 2O=E Now Proceed further..

```
1. C
2. D
3. D
B=3, O=5, M=1, Y=9, A=2, R=4, L=6, E=0
3 5 3
x 3 5 3
1 0 5 9
1 7 6 5
1 0 5 9
1 2 4 6 0 9
```

```
B O B
x B O B
M E O Y
M I L O
M E O Y
M A R L E Y
O + O = E i.e. 2O=E
and
B O B
x B O B
M E O Y
M I L O
M E O Y
M A R L E Y
Here, B x O=_O [M I L O]
Hence Possible Values of B and O are
Case-I- When O={5} and B={3, 7, 9} Rule 3
Case-II When B={6} and O={2, 4, 8}
Now, you have to start hit and trail with both the above cases.
Firstly, take O=5
When you will take O=5 then E=0 as [ O + O = _E(last digit)]
Put E=0 and O=5 and rewrite the problem
B 5 B
x B 5 B
M 0 5 Y
M I L 5
M O 5 Y
M A R L 0 Y
Now, divide the problem in three parts
(1) B 5 B
x B
M 0 5 Y
(2) B 5 B
x 5
M I L 5
(3) B 5 B
x B
M O 5 Y
Here, you can take Case (2).[In (2) you have maximum number of clues. You can select anyone based on your convenience.]
Now, you have to start hit and trial with the possible values i.e. B={3, 7, 9}
Firstly take B=3
Now Put B=3 in (2)
i.e. B 5 B
x 5
M I L 5
now it leads to
3 5 3
x 5
1 7 6 5
If you compare [ M I L 5] and [ 1 7 6 5]
M=1, I=7, L=6
At this stage, you have to check further whether these values violates any Basic Cryptaritmetic Rules
For checking this
put these values in
B O B
x B
M E O Y
i.e. 3 5 3
x 3
1 0 5 9 [M 0 5 Y]
From here you are also getting M=1, E=0, O=5 and Y=9.
Hence, It satisfies the Basic Cryptarithmetic Rules.
3 5 3
x 3 5 3
1 0 5 9
1 7 6 5
1 0 5 9
1 2 4 6 0 9
[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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heres the short answer:

After getting .

Case-I- When O={5} and B={3, 7, 9} Rule 3

Case-II When B={6} and O={2, 4, 8}.

Now observe that since O is the unit digit of O*B and B(at the unit place of first number)*O(at tens of second number)= O , i.e. the previous multiplication B*B (both at unit place) gives a digit and not a number , i.e Y is a digit ,i.e B^2 = Y , we get

Case-III {B,Y}={2,4},{3,9}.

B=3 satisfy case III as well as one of Case I/II . thus B=3 ,O=5 .rest you can get .

When you already know the combinations, the why do it in a complex way…!! Idiotic…

B*O = _ O

Possibilities :-

3*5= _,5

7*5=_,5

9*5=_,5

6*2=

6*4=

6*8=

So rather than trying solving you can get answer by checking :-

353

353

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Now try multiplying the patterns….

Funny thing is…u’ll be getting the sum done on the first go….

The guy above said the same thing….this is practical….

Notice the units tens digit n units digit in the 1st n 2nd line respectively…