cryptarithmetic problems - elitmus cryptarithmetic problems PART 1

                   W H Y                  *N U T          ---------------                 O O N P              O Y P Y      ...


                   W H Y
                 *N U T
         ---------------
                O O N P
             O Y P Y
          O U H A
        ----------------
          O N E P O P

Step 1: Figure out the characters which can not take value 0
Now you will ask the question "how do I figure out which character can be or can not be 0 ?" Follow these rules and you will be able to find them very easily:
  • No leading digit of a number can be 0. What does that mean ? That simply means the left most digit in a number can't be 0(ever saw a number like 023 or 007 ?). Following this rule in our case, none of these can be 0: W, N O
     
  • None of the characters in multiplier(N, U and T in our case) will be 0. You can argue "Hey are you crazy ? You are wrong !". OK wait a moment.

    Can you tell me what is the value of U in the question below ?
                      W H Y
                     *N U T
             ---------------
                   O O N P
                   U U U
             O U H A
           ----------------
             O N E P O P


    Yes, you are thinking right. U will be 0. I think you know the reason. After all you are an engineer.

    Do you really think eLitmus is gonna ask you such an easy question ? Don't even think of that.

    So simple rule of thumb is that just proceed assuming no digit in multiplier will be 0 and you can land on the solution faster

    So where did we reach ? None of these will be 0: W, N, O, U, T
  • No trailing digit in multiplicand will be 0

    OK let me ask you a question again

    What will be the value of E in the following question ?
                      W H Y
                     *N U T
             ---------------
                   O O N Y
                O Y P Y
             O U H Y
           ----------------
             O N E P O Y

    No wonder again. Yes, Y will be 0.

    Do you again think eLitmus will ask you so simple question ? No, I don't think so.

    So like in the previous step, just proceed assuming that the first(from the first bullet point) and the last digit in multiplicand will not be 0

    Now we have the following result for non zeroes: None of these will be 0: W, N, O, U, T, Y
Step 2: Figure out the characters which can not take value 1
  • No digit in multiplier will be 1

     
    OK, answer my this question and you will get to know why so ?
                      W H Y
                     *N U T
             ---------------
                      W H Y
                 O Y P Y
              O U H A
            ----------------
              O N E P O Y

    OR you can even see this one:
                      W H Y
                     *N U T
             ---------------
                    O O N P
                   W H Y
              O U H A
             ----------------
              O N E P O P 


    Again eLitmus won't ask you so simple question.

    So stick to the rule that no digit in multiplier will be 1. Following this rule in our case none of these will be 1: N, U, T
  • The trailing digit in multiplicand will not be 1. An example to prove the above point:
                      W H Y
                     *N U T
             ---------------
                    O O N T
                 O Y P U
              O U H N
            ----------------
              O N E P O T

    So following this rule we find that Y will not be 1

    By far digits which can't be 1 are: N, U, T, Y
Following the above two steps we find:

Non 0 digits: W, N, O, U, T, Y and non 1 digits: N, U, T, Y

So digits which may be 0 are: H, P, A, E and digits which may be 1 are: W, H, O, P, A, E
Let's proceed further.

Step 3: Identify 6 or 5

Let's see some mathematical property:
  • 6*0 = 0
  • 6*2 = 12
  • 6*4 = 24
  • 6*6 = 36
  • 6*8 = 48  
  • 5*1 = 5
  • 5*3 = 15
  • 5*5 = 25
  • 5*7 = 35
  • 5*9 = 45
Notice when 6 is multiplied with an even digit, the unit place digit of the result has the same even digit.
And similarly when 5 is multiplied with an odd digit, the unit place digit of the result is 5

So what do we get from these rules ?

Let's apply these rules to solve our problem.

 
                  W H Y
                 *N U T
         ---------------
                O O N P
             O Y P Y
          O U H A
        ----------------
          O N E P O P


Focus on the digits highlighted in blue. Let's write it in this format for better understanding. C is the carry, so just ignore it.

             Y
          * U 
          ----
          C Y

From the above rule we discussed we can say these:
Y is 5 and U is odd
         OR

Y is even and U is 6

So we have two ways starting from here. Follow one path(any one) and see if you are able to solve the problem. If not, take another path and proceed further.

Case 1:
 Y is 5 and U is odd

                  W H 5
                 *N U T
         ---------------
                O O N P
             O 5 P 5
          O U H A
        ----------------
          O N E P O P


Now ?? What next ?? Let me admit one thing here. I can't teach you from here how to solve this. No one can tell you what will be the next step from here. It is just you who can solve this by hit and trial and applying your common sense.
Anyhow I will proceed with my way but remember one thing you have to solve it by yourself from here keeping in mind the results of step 1 and 2 and applying own logic.
Follow my steps and you can learn many tricks from my experience.
What are the unit place digit when 5 is multiplied with any number ?? You will say, "Hey I know, they can be either 0 or 5". OK, you the answer. But did you notice when 5 is multiplied by N, U and T they are giving three unit place digits P, 5 and A ?
                  W H 5
                 *U T
         ---------------
                O O N P
             O 5 P 5
          O U H A
        ----------------
          O N E P O P
Do you think it is possible ? So we assumed something wrong and so landed upon a wrong output. So we will now start with the second case.
Case 2: Y is even and U is 6
                  W H Y
                 *N 6 T
         ---------------
                O O N P
             O Y P Y
          O 6 H A
        ----------------
          O N E P O P
Now since Y is even, the unit place digit of all the result where Y is involved will be even. So P and A will also be even. C is just a carry for demonstration. Please ignore it. Note we didn't say anything about Y and N yet(even or odd).
                   Y
                * T
                ----
                C P
             
                  Y
               * N
               ----
               C A

So we have the following till now:
                  W H Y
                 *N 6 T
         ---------------
                O O N P
             O Y P Y
          O 6 H A
        ----------------
          O N E P O P
So we have found four even digits: Y, U(6), P and A. Notice the second column from right in addition(C is carry, ignore it):
                N
             + Y
             ----
             C O
N and Y both will be odd because if any one of them is even the other will automatically be even so we will get another two even numbers. But we already have four even number in our hand so we cannot afford two more even numbers.
                  W H Y
                 *N 6 T
         ---------------
                O O N P
             O Y P Y
          O 6 H A
        ----------------
          O N E P O P
Now notice the second column in addition from left, Since no carry is being passed to 1st column from left, the sum be less than or equal to 9. So the possible value of O can be either 1 or 3(since O is odd).
Let's proceed with any one case
When O is 1
                  W H Y
                 *N 6 T
         ---------------
                1 1 N P
             1 Y P Y
          1 6 H A
        ----------------
          1 N E P 1 P
Y is even and it can take following values: (2, 4, 8). U is already 6 and Y cannot be 0 as we discussed in step 1. Let's put Y = 2 and proceed.
                  W H 2
                 *N 6 T
         ---------------
                1 1 N P
             1 2 P 2
          1 6 H A
        ----------------
          1 N E P 1 P
Notice second column from right
                    N
                 + 2
                 ----
                 C 1
From here it is clear that N will be 9.
                  W H 2
                 *9 6 T
         ---------------
                1 1 9 P
             1 2 P 2
          1 6 H A
        ----------------
          1 9 E P 1 P
Always try to check whether the value you have just put is correct or not. Let's check it. The value we just put is 9 for N. The last multiplication:
                  W H 2
                      * 9
               ---------
                1 6 H A
Don't you think W should be 1 for this case to be true. But we already have O = 1. So we must reject N = 9. We got N = 9 because we put Y = 2.
Let's put another value for Y and proceed further. Let's put Y = 4 and proceed:
                  W H 4
                 *N 6 T
         ---------------
                1 1 N P
             1 4 P 4
          1 6 H A
        ----------------
          1 N E P 1 P
Now look at the following multiplication:
                  W H 4
                      * 6
               ---------
                1 4 H 4
What do u think the value of W will be ?? Will not it be 2 ? 
Remember we had found four even: Y(4), U(6), P and A.We found the last one also W(2)
                  2 H 4
                 *N 6 T
         ---------------
                1 1 N P
             1 4 P 4
          1 6 H A
        ----------------
          1 N E P 1 P
Now apply some common sense and guess what will be the value of T ?
                  2 H 4
                     * T
               ---------
                1 1 N P
If I am not wrong you just thought it would be 5.
                  2 H 4
                 *N 6 5
         ---------------
                1 1 N P
             1 4 P 4
          1 6 H A
        ----------------
          1 N E P 1 P
So we are here:
                  2 H 4
                 *N 6 5
         ---------------
                1 1 N 0
             1 4 0 4
          1 6 H A
        ----------------
          1 N E 0 1 0
We had found all five even digits: Y(4), U(6), P(0), A and W(2) 
So A must be 8.
                  2 H 4
                 *N 6 5
         ---------------
                1 1 N 0
             1 4 0 4
          1 6 H 8
        ----------------
          1 N E 0 1 0
From the multiplication:
                   2 H 4
                      * 6
               ---------
                1 4 0 4
We can easily say, H will be 3
                  2 3 4
                 *N 6 5
         ---------------
                1 1 N 0
             1 4 0 4
          1 6 3 8
        ----------------
          1 N E 0 1 0
Now you can easily solve it to get the following:
                  2 3 4
                *7 6 5
         ---------------
                1 1 7 0
             1 4 0 4
          1 6 3 8
        ----------------
          1 7 9 0 1 0
We had two option for O(1, 3). We selected O = 1 and proceeded. Further we had three option for Y(2, 4, 8). First we tried with Y = 2 but we failed, then we put Y = 4 and we got up to the final result. What if Y = 4 was not correct ? We had to put Y = 8 and proceed further. If we had no solution for even Y = 8 then we would have backtrack again and started off with O = 3 and all the possible values of Y(2, 4 and 8) and would have tried again. Sometimes the problem may go lengthy but with lots of practice you can achieve a very good speed and can solve it within 10-15 minutes, which I think is worth.





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