Things to write in a solution
Above shows a cryptarithm such that each letter represent a distinct single non-negative integers with and are non-zero. Find the value of the 6-digit number: .
The answer is 908721.
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1 solution
Since in the 2nd and the 3th row of the addition there is that M+I->O and A+M->0 therefore for the same quantity M corresponding by addition with I and A which are distinct the same quantity it means that at least in one of the rows there is a carry and since in the first row of the addition A+M gives a different result than A+M in the third it means that there is necessarily a carry in the third row such that the result in the third is modified. Also , observe that since A+M -> F and in the third A+M -> O , O = F + 1 (O must be A+M which is F plus a carry which in addition can't be greater than 1) in the second row where there is that M+I -> O there is no carry since that would imply also that M+I+1 = O therefore making M+I = M+A which is not possible. By knowing this we can find out the values for O and F such that they are consistent with the conditions found here. Since M+I -> O and F (that is A+M) + 1 -> O , F and O must be at a distance of 1 therefore there must be a sum of M+A which is smaller than 10 and a sum of M+I which is bigger than 10 such that the final resulting numbers (O and F) in their corresponding rows are consecutive which is possible just when O = 0 and F=9 since for setting any other value for one of them (say O) the corresponding value for the other wouldn't be at a distance of 1 as a result of the fact that when M+A < 10 and M+I > 10 for any M the last number of the sum M+A will be above M and in the second sum , M+I , it will be smaller their being the closest for the values of the minimum value in the first of M+1 and and the maximum in the second of M+9 , that is of M-1 therefore being at a minimum difference of 2. Therefore O = 0 and F=9. From here we can check cases such for the first rows there is one number (M) which added with two other gives 10 and 9 (A and I) and for the next 2 rows there is also another number (L) which added with two other different numbers (E and L) gives two other different numbers (P and R) though such an approach would be pretty annoying and I'm pretty convinced that a nicer way is.