Three Posts you have to check before continuing reading this:
Elitmus pH Test Pattern & Syllabus
Tips For Elitmus Test
Tips To Solve Cryptarithmetic Puzzles
The puzzle that we will be solving today looks something like this,
W H Y * N U T
--------------------
O O N P
O Y P Y +
O U H A + +
--------------------
O N E P O P
Follow the steps to solve the above puzzle:
Step 1: The term (W H Y)* U = O Y P Y, we can guess that U=6 and Y=2,4,8. Whenever there are more than one occurence of the same character in these kind of problems, Y is usually 4(No all the times though). So let us take y=4 as proceed to see if the assumption is correct.
W H 4 * N 6 T
--------------------
O O N P
O 4 P 4 +
O 6 H A + +
--------------------
O N E P O P
Step 2: Looking at the sum term, O+6=N, and also since there is no carry from this sum to the next term, we conclude that N<=9, and as a result O could be 1/2/3. Considering O=1 as proceeding we get,
W H 4 * N 6 T
--------------------
1 1 N P
1 4 P 4 +
1 6 H A + +
--------------------
1 N E P 1 P
Step 3: Now looking at the term N+4=1, we can easily guess that N=7, since there is no carry being carried to this term from P(which is the previous term). Rewriting with N=7 would yield,
W H 4 * 7 6 T
--------------------
1 1 7 P
1 4 P 4 +
1 6 H A + +
--------------------
1 7 E P 1 P
Step 4: Looking at the product term (W H 4)*7, we can guess that, A=8. And since the last term in all the sum term of individual products (i.e. "1 1 7 P", "1 4 P 4", "1 6 H A") is always 1, the minimum value W can take is 2. So considering W=2, we get
2 H 4 * 7 6 T
--------------------
1 1 7 P
1 4 P 4 +
1 6 H 8 + +
--------------------
1 7 E P 1 P
Step 5: Looking at the product term, (2 H 4)* T = 1 1 7 P, we can guess that T=5, since (2 H 4)*T = 1 1 7 P. And, further it is safe to assume P=0, since (2 H 4)*T(=5). Rewriting we get,
2 H 4 * 7 6 5
--------------------
1 1 7 0
1 4 0 4 +
1 6 H 8 + +
--------------------
1 7 E 0 1 0
Step 6: Now, (2 H 4)*5=1 1 7 0, we find H to be 3, and hence E turns out to be 9 (1+4+3+carry_of_1=E). This gives us the solution of,
2 3 4 * 7 6 5
--------------------
1 1 7 0
1 4 0 4 +
1 6 3 8 + +
--------------------
1 7 9 0 1 0
which is the required result.
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