Cryptarithmetic Puzzles

Cryptarithmetic problems are mathematical puzzles in which the given alphabet are replaced by a single digit(unique) number.

Here we need to assign each alphabet a unique single digit number.So as we see the second example. The word SEND and MORE are of 4 alphabet.So the maximum value both of them can be assigned is 9999.So in this case if we assume the maximum value of SEND and MORE then we will get the sum 19998.So we observe that the fifth digit must be 1 because all the digit present here are of single digit and single digit will always give a carry 1.And once the value of alphabet is assigned it can’t be assigned to other alphabets since the numeric value of each alphabet is unique.

Theorem 1:- Each alphabet must have unique single digit numeric value

In Column 5
So,here M is the carry obtained by the addition of S and M. And from theorem 1 they both must a single unique digit number.So the carry can only be 1.

Therefore, M=1

In Column 4
S + 1 =10 + O | Here 10 is added in O because O lies between 10–19 => S=9+O — — — — (i)
1+S+1 =10 + O | If there is carry from column 3 => S=8+O — — — — (ii)
From equation (i) O = 0 & (ii) O = 0,1
According to theorem 1,each alphabet must have unique digit.So O = 0(Zero).

In Column 3
E + 0 = N
From theorem 1
E != N
So there must be a carry in column 3.
1 + E = N –(iii)
Assume E lies between 0>=E>=9. [0,9]
So it sum will give a result 10 if E is 9. But N cant be 0 as O is already 0.So E = [0,9).
As column 3 will never give a carry.
So eqn(i) is true.
Therefore , S = 9.

In Column 2
N+R=10+E — — — — (iv)
1+N+R=10+E — — — — (v)
Substitution of value of N from eqn(iii) in eqn(iv) gives
R = 9 (Not Possible. As S is already assigned 9)
Substitution of value of N from eqn(iii) in eqn(v) gives
R = 8
Therefore , R = 8.

In Column 1
D + E = 10 + Y
Since in Column-2 equation having carry is correct.
So ,
D + E = 10,11,12,13,14,15,16,17,18,19
Number already assigned 0,1,8,9
So number or sum of two digit resulting D+E ,have last digit 0,1,8,9 are discarded.
So out of them only 12,13 are the only choice.
12 = 7 + 5
13 = 7 + 6
So among them 7 is common. Either D or E must be 7.
If E = 7 then N = 8 (Not Possible)
Therefore, D=7 . Then E is either 5 or 6.
If E is 5 then N = 6. (It even satisfies all the equation)
And add the value of D + E , the last digit of their sum gives the value of Y.

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