Question

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Solve the Cryptarithm
ON + ON + ON = GO or, 3 × ON = GO​

Answer 1

Solution :

We have,

$\rm \: 3 \times ON = GO \: .....(i)$

Since GO is a two digit number. So, it can have maximum value equal to 99

$\therefore \rm \: 3 \times ON = GO$

$\hookrightarrow \sf \: 3 \times ON \: is \: at \: most \: equal \: to \: 99$

$\hookrightarrow \sf \: ON \: is \: at \: most \: equal \: to \: 33$

$\hookrightarrow \sf \: O \: is \: at \: most \: equal \: to \: 3$

$\hookrightarrow \sf \: O \: is \: at \: most \: equal \: to \: 1, \: 2 \: or \: 3$

Case I

When O = 1

Putting O = 1 in (i) we get :

$\rm \: 3 \times 1N = G1.....(ii)$

$\hookrightarrow \sf \: 3 \times N = 1 \: or \: 3 \times N = \\ \sf \: two \: digit \: no \:with \: 1 \: at \: unit \: place$

$\hookrightarrow \large \boxed{ \sf \: N = 7}$

Putting N = 7 in (ii) we get :

$\sf \: 3 \times 17 = G1 \implies \: G1 = 51 \implies \: G = 5$

$\therefore \sf \: O = \boxed{1} \: G = \boxed{5} \: and \: N = \boxed{ 7}$

Case II

When O = 2

Putting O = 2 in (ii),we get

$\rm \: 3 \times 2N = G2....(iii)$

$\sf \: 3 \times N = two \: digit \: no \: with \: 2 \: at \: units \: place$

$\hookrightarrow \large \boxed{ \sf \: N = 4}$

Putting N = 4 in (iii),We get :

$\sf \: 3 \times 24 = G2 \implies \: G2 = 72 \implies \: G = 7$

$\therefore \sf \: O = \boxed{2} \: G = \boxed{7} \: and \: N = \boxed{ 4}$

Case III

When O = 3

Putting O = 3 in (i),we get

$\sf \: 3 \times 3N = G3$

$\hookrightarrow \sf \: 3 \times N = 3$

$\hookrightarrow \large \boxed{ \sf \: N = 1}$

Putting N = 1 in (iv),we get

$\sf \: 3 \times 31 =G3 \implies \: G3 = 93 \implies \: G = 9$

$\therefore \sf \: O = \boxed{3} \: G = \boxed{9} \: and \: N = \boxed{ 1}$

Thus, We have Following solution:

$\begin{array}{|c|c|c|}\cline{1-3}O & N & G\\ \cline{1-3}1 & 2 & 3 \\ \cline{1-3}5 & 7 & 9 \\ \cline{1-3}7 & 4 & 1 \\ \cline{1-3}\end{array}$