Solve the cryptarithm on +on+on+on = go or 4 × on = go
Answers
Step-by-step explanation:
We have,
ON + ON = GO or 2 × ON = GO ------------ (i)
Clearly, GO is a two digit number whose maximum value can be 99. Therefore, 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.
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× 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
2×21 = G2 => 42 = G2 => G = 4.
When N = 6, equation (ii) gives
2×26 = 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 × 4N = G4 ------------- (iii)
≈> 2×N is either equal to 4 or 2×N = 14
≈> N = 2 or 7
When N = 2, equation (iii) gives
2×42 = G4 => 84 = G4 => G = 8.
When N = 7, equation (iii) gives
2×47 = G4 => 94 = G4 => G = 9.
Thus, we have
O = 4, G = 8 and N = 2 or O = 4, G = 9 and N = 7.