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

Explanation

We have,
`3xxON=`Since `GO` is a two-digit number so it can have a maximum value equal to `99`.
`3xxON = GO`
`3xxON` is at most equal to `99`
`ON` is almost equal to `33`
`O` is at most equal to `3`
`O` is equal to `1`or `2` or `8` [`O` can not be zero as `ON` is a two digit number]
Case I: `O = 1`
Putting `O=1` In equation (i)
`3xxN=1` or, `3xxN` = a two-digit number having `O` at the units place.
Putting `N= 7` in `10`, use get
`3xx17 = G1`
`O= 1 , G=5 `and `N = 7`
Case II When `D = 2`
Putting `O = 2` in (ii) we get,
`3xx2N = G2`- (iii)
`3N` = a two digit number having `2` at units place.
`N = 4`
Putting `N=4` in (iii), weget
`3xx24=G2`
or `G2 = 72`
`O = 7`
`D = 2 =7` and `N = 4`
Case III:
When `O = 3`
Putting `b = 3` in (i) we get,
` N =1`
Thus, we have following solutions of equation
(1) O= 1 ,G = 5 , N = 7