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

Explanation

We have, `ON+ON = GO` ....(i)
Clearly, `GO` is a two digit number whose maximum value can be `99`
the maximum value of `ON` can be `49`. So the maximum value of digit `O` can be `4`.
Since LHS of (i) is an even number so `GO` is also an even number.
Consequently, `O` can take even values only. Digit `O` can take values `2` or `4`.
Case I :when digit `O` lakes value 2. Substituting `O = 2` 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` N = 6` . when `N = 1` , equation (ii) gives
`2 xx 21=G2`
`=> G2 = 42`
`G= 4`
when` N = 6` equation (ii)gives
`2 xx 26 = G2`
`G= 5`
Thus we have,
`D = 2 ,G = 4` , and `N = 1`
or
`D=2, G = 5` and `N = 6`
Case II: when digit O Takes value `4` Substituting `O= 4` in equation(i) we get, `2xx4N = G4`
` 2 xx N` is either equal to `4` or, `2 xx N = 14` when 2N=4
`N = 2`
When `2N = 14`
`N = 7`
when `N = 2` , equation (iii) gives,
`2 xx 42 = 64` or, `84= G4 `
`G = 8`
when `N =7`,equation (iii) gives,
`2 xx 47 = 64` or
`G=9`
Thus we have
`O=4, 6=8 `and `N = 2`
`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`