Solve the cryptarithm : ` O N +\ O N + O N + O N = G O `

Explanation

We have,
`ON + ON = GO or 2 xx ON = GO ------------ (i)`
Clearly, `GO` is a two-digit number whose maximum value can be `99`. Therefore, the maximum value of ON can be `49`. So, the maximum value of digit `O` can be `4`.
Since the LHS of (i) is an even number. So, `GO` is also an even number.
Consequently, `O` can take even values only.
Therefore, Digit `O` can take values `2` or `4`.
Case `1`: When digit `O` takes value `2`.
Substituting `2` in place of digit `O` in equation (i), we get
`2 xx 2N = G2 `----------------- (ii)
`=>` Multiplication of `2` and `N` must be either `2` or a two digit number between `10` and `19` having `2` at ones place.
`=> N = 1` or `6`.
When `N = 1`, equation (ii) gives
`2 xx 21 = G2 => 42 = G2 => G = 4`.
When `N = 6`, equation (ii) gives
`2 xx 26 = G2 => 52 = G2 => G = 5`.
Thus, we have
`O = 2, G = 4` and `N = 1` or, `O = 2,G = 5, N = 6`.
Case `2`: When digit `O` takes value `4`.
Putting `O = 4` in (i), we get
`2 xx 4N = G4 `------------- (iii)
`=> 2×N` is either equal to `4` or `2 xx N = 14`
`=> N =2` or `7`
When `N = 2`, equation (iii) gives
`2 xx 42 = G4 => 84 = G4 => G = 8.`
When `N = 7`, equation (iii) gives
`2 xx 47 = G4 => 94 = G4 => G = 9`.
Thus, we have
`O = 4, G = 8` and `N = 2` or `O = 4, G = 9` and `N = 7`.
Hence, the solutions of the given cryptarithms are :
i) `O = 2, G = 4, N = 1`
ii) `O = 2, G = 5, N = 6`
iii) `O = 4, G = 8, N = 2`
iv) `O = 4, G = 9, N = 7`