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ElitmusZone » Cryptarithmetic Multiplication 31

# Cryptarithmetic Multiplication 31 1. Value of T ? (a) 3 (b) 4 (c) 5 (d) 6
 2. Value of P? (a) 5 (b) 1 (c) 3 (d) 4
 3. Value of K ? (a) 2 (b) 8 (c) 0 (d) 1

1. Abhishek Jain

how to prepare for elitmus within two weeks.

1. Shubham Kumar

why u have not consider case 5. taking 4 as t ,we get W=22

1. Kanchana

So when you start replacing with the T=4, you will find that Q gets the value 4, and thats not possible! Cause T is already 4.

2. Suchit Sharma

In case 5, why you havn’t take T =4 and W = 2 ?? There is no problem is we take it .

1. Srujan K'mat

Because there is already ‘2’ mentioned in the question. So W can’t be ‘2’.
Also in the solution, I guess ‘2’ shouldn’t be there in the possible values of ‘W’.

1. Kanchana

I dont think taking W = 2 is an issue because 2 is already given. i replaced the values in case 5 in the main question, and found that Q is getting the value 4, not possible. Therefore move to next case.

2. akash

cause there is already value 2 given in the main equation

3. vikas solanki

one more way to solve this question.
we have possible values of
p= {5,1,3,4,} and
k = {2,8,0,1}
and if you will look closely you will find value of p cant be 1
and value of k cant be 0 and 1 (according to basic crypt rule)
so only possible set of values for p and k are
p={5,3,4}
k={2,8,1}
possible set of p and k =
{ (5,2),(5,8),(3,2),(3,8),(4,2),(4,8)}
so start hit and trial with these 6 cases and you will get the answer with second element i.e.. p=5 and k=8.

4. ayush

how did u get E=0 in the beginning ?