Is it unique?
Above is a cryptogram. Where each letter represents a distinct non-negative integer.
Find the prime number .
The answer is 19.
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2(100N + 10E + W) = 1000Y + 100E + 10A + R
Y is definitely 1 and R is definitely even. Since all 4 of Y, E, A & R are distinct, then Y + E + A + R ≥ 1 + 0 + 2 + 3 = 6, and thus, must be an odd prime. Because of that, E + R must be an even sum, and thus the even-odd parity of E & R must be the same.
Because E as summands in tens place will effect the sum E in hundreds, then E must either be small evens which doubles are less than 10 & has no carryover or large odds which doubles are ten or more & must have carryover. With this,
E = { 2 , 4 , 5 , 7 , 9 }
Respectively, keeping in mind to have the same parity for A,
A = { 4 , 8 , 1 , 5 , 9 }
Now, we can reject the third pair of (E,A) = (5,1) for reusing 1 for A with established Y = 1 and also the fifth pair of (E,A) = (9,9) for reusing 9 for both E & A.
For these new set of E and A after the cut-downs, E = { 2 , 4 , 7 } & A = { 4 , 8 , 5 } , we get a new set for possibility of N which must all be larger than or equal to 5.
N = { 6 , 7 , 8 }
Moving on to R that we know must be even, refer to set A to see whether the ones place did send a carryover or not, and make sure that no repeating or overlapping digits occur between both R & W when compared against N, E, A or Y.
N = { 6 , 7 , 8 }
E = { 2 , 4 , 7 }
A = { 4 , 8 , 5 }
R = { X , 6 , 2 }
Thus, we get no solutions for first possibility, a sum of 1 + 4 + 8 + 6 = 19 (which is definitely prime) for the second possibility and a sum of 1 + 7 + 5 + 2 = 15 (which is not prime) for the third possibility.
Answer = Y + E + A + R = 1 + 4 + 8 + 6 = 19