cryptarithmetic problems - elitmus cryptarithmetic problems PART 3

                  B C H                * G E I          ---------------                B A D B              A A I J           ...


                 
B C H
               * G E I
         ---------------
               B A D B
             A A I J
           A F F F
        ----------------
           A H J F D B
Let's start this way: 

                  B C H
               * G E I
         ---------------
               B A D B
             A A I J
           A F F F
        ----------------
           A H J F D B

J will be 0.

                  B C H
                * G E I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B

Now focus here:

                  B C H
                * G E I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B

E * H gives a result whose unit place digit is 0. Which means either one of them is 5 and other is an even digit or vice versa.
So we have two possible cases:
H = 5 and E is even
             OR
E = 5 and H is even

Let's proceed with the first case:
Case I: H = 5 and E is even
First of all check whether H = 5 is possible or not
We will try to use the following symmetry in some way. Let's see how can we use this:
                  B C H
                * G E I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B

Look carefully:
                  B C H
                * G E I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B

A + F + (Some Carry from last step) = H
A + F + (Some Carry from last step) = 5(as we are proceeding for H = 5) -----------(1)
What do you think will be the carry from last step ? Let's see this:

                  B C H
                * G E I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B
B + A + F + (Some carry from last step) = X0 (X is carry, just ignore it) -------------(2)
And what you think will be carry from the last step ?
Let's see that too:

                  B C H
                * G E I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B

A + I + F + (Some carry from last step)= XF (X is carry, just ignore it)
The last step D + 0 = D will not produce any carry. So In this case carry will be 0.
which implies:
A + I + F = XF => A + I will be 10
So, A + I + F will produce carry 1
Putting this carry in the equation (2), we get:
B + A + F + 1 = X0                                                                       ---------------(3)
Also from equation (1) we had: A + F + (Some Carry from last step) = 5, which means that A + F cannot be more than 5.
So, B + A + F + 1 cannot be other than 10, which means this step produces a carry 1.
Putting this carry in eq (1) we get:
A + F = 4
Substituting this value of (A + F) in eq (3):
B + 4 + 1 = X0 => B will be 5
WAIT !!  What we got ?? We assumed H = 5 and following that we got B = 5. Not acceptable. Reject this solution and pick the other option.

Case II: E = 5 and H is even
                  B C H
                * G 5 I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A H 0 F D B
Possible values of H = {2, 4, 6, 8}
Let's start one by one:
When H = 2:
                  B C 2
                * G 5 I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A 2 0 F D B
First check whether H = 2 is possible or not.
                  B C 2
                * G 5 I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A 2 0 F D B
A + F + (some carry from last step) = 2
Clearly the last step is producing some carry as the result is having unit place digit 0.
A + F + (some non zero carry from last step) = 2
Sum of three positive integers can't be 2(no one can be 0 because 0 is already there)
So H = 2 is not possible.
When H = 4:
                  B C 4
                * G 5 I
         ---------------
                B A D B
              A A I 0
            A F F F
        ----------------
            A 4 0 F D B
Follow the same step we followed in case I when we rejected H = 2. You can easily find B = 6.
                  6 C 4
                * G 5 I
         ---------------
                6 A D 6
              A A I 0
            A F F F
        ----------------
            A 4 0 F D 6
Notice:
                  6 C 4
                * G 5 I
         ---------------
                6 A D 6
              A A I 0
            A F F F
        ----------------
            A 4 0 F D 6
I can be either 4 or 9. But 4 is already there so I can be 9 only
                  6 C 4
                * G 5 9
         ---------------
                6 A D 6
              A A 9 0
            A F F F
        ----------------
            A 4 0 F D 6
Look carefully:
                  6 C 4
                * G 5 9
         ---------------
                6 A D 6
              A A 9 0
            A F F F
        ----------------
            A 4 0 F D 6

5 * 4 = 20, so carry is 2. Now in the next multiplication 5 * C is giving unit place digit 9.
Is, 5 * C + 2 = X9 possible ? (X is just a carry, ignore it)
In this case also we are getting wrong result. So we will move to the next case.

When H = 6:
                  B C 6
                * G 5 I
         ---------------
                B A D
              A A I 0
            A F F F
        ----------------
            A 6 0 F D B
Follow the same step we followed in case I when we rejected H = 2. You can easily find B = 4.
                  4 C 6
                * G 5 I
         ---------------
                4 A D 4
              A A I 0
            A F F F
        ----------------
            A 6 0 F D 4
Focus here:
                  4 C 6
                * G 5 I
         ---------------
                4 A D 4
              A A I 0
            A F F F
        ----------------
            A 6 0 F D 4
I can be none other than 9
                  4 C 6
                * G 5 9
         ---------------
                4 A D 4
              A A 9 0
            A F F F
        ----------------
            A 6 0 F D 4
Again because of the same reason as in previous case, 
                  4 C 6
                * G 5 9
         ---------------
                4 A D 4
              A A 9 0
            A F F F
        ----------------
            A 6 0 F D 4
5 * C + 3 = X9 is not possible, so we can reject this case also.
Now we are left with just one case H = 8
Let's proceed with that too.

When H = 8:
                  B C 8
                * G 5 I
         ---------------
                B A D
              A A I 0
            A F F F
        ----------------
            A 8 0 F D B
Follow the same step we followed in case I when we rejected H = 2. You can easily find B = 2.
                  2 C 8
                * G 5 I
         ---------------
                2 A D 2
              A A I 0
            A F F F
        ----------------
            A 8 0 F D 2
Look at the highlighted characters:
                  2 C 8
                * G 5 I
         ---------------
                2 A D 2
              A A I 0
            A F F F
        ----------------
            A 8 0 F D 2
Clearly I can be either 4 or 9
When I = 4:
                  2 C 8
                * G 5 4
         ---------------
                2 A D 2
              A A 4 0
            A F F F
        ----------------
            A 8 0 F D 2
Again check if this value of 4 is acceptable or not
                  2 C 8
                * G 5 4
         ---------------
                2 A D 2
              A A 4 0
            A F F F
        ----------------
            A 8 0 F D 2
So you think 4 * 2 + (some carry from last step) can produce a result like 2A. Its never possible. So we reject this case also.
Now the last case when I is 9.
When I = 9:
                  2 C 8
                * G 5 9
         ---------------
                2 A D 2
              A A 9 0
            A F F F
        ----------------
            A 8 0 F D 2

See the highlighted characters: 

                  2 C 8
                * G 5 9
         ---------------
                2 A D 2
              A A 9 0
            A F F F
        ----------------
            A 8 0 F D 2
A + 9 + F = F (No carry from the last step)
=> A + 9 = 10
=> A = 1
                  2 C 8
                * G 5 9
         ---------------
                2 1 D 2
              1 1 9 0
            1 F F F
        ----------------
            1 8 0 F D 2
Now you have everything in your hand. You can easily reach to the solution:
                  2 3 8
                * 7 5 9
         ---------------
                2 1 4 2
              1 1 9 0
            1 6 6 6
        ----------------
            1 8 0 6 4 2 

You must be thinking its too long. I have explained all the steps thats why it went so long. When you will be practicing a lot, you will be in a position to solve these problems in not more than 15 minutes. Solving 3 question in eLitmus in 15 minutes is worth. Don't leave this problem. It just needs some practice and nothing else.





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MRA Job Seekers | Bangalore Job Seekers Hub: cryptarithmetic problems - elitmus cryptarithmetic problems PART 3
cryptarithmetic problems - elitmus cryptarithmetic problems PART 3
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