Maths and Logic Puzzles by Gordon Burgin

u-can-do

Solving a Logic Problem

by G R Burgin

The Puzzle

Solve the following equation if N, O & G are single digit numbers
and NO & GN are double digit numbers. Find the values for N, O & G.

NO + NO + NO + NO = GN

The Solution

N can only be equal to either 1 or 2; otherwise G is a two digit number. (i.e., N + N + N + N = G or N x 4 = G)
N has to be an even number (i.e., O + O + O + O = N or O x 4 = N)!
Therefore, N = 2, since 4 times any number will always result in an even number.
Since O is still unknown, G (G = N x 4) can only be equal to either 8 or 9;

8 if there is no carry over from the Equation O x 4 = N (where N x 4 = G or = 8)

9 if there is a carry over from the Equation O x 4 = N (where (N x 4) +1 = G or = 9).
Solving for O without a carry over:

((Nx10) + O) x 4 = 82
(20 + O) x 4 = 82
80 + (4 x O) = 82
4 x O = 2
O = 0.5 (not valid, must be a whole number)
Solving for O with a carry over:

[((Nx10) + O) x 4] = 92
(20 + O) x 4 = 92
80 + (4 x O) = 92
4 x O = 12
O = 3 (valid answer)
Now check the solution:

NO + NO + NO + NO = GN

23 + 23 + 23 + 23 = 92

Contact: mathspuzzles@gordonburgin.com