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$m + n = 12k_1 + 8 ------------------(1)$

$m - n = 12k_2 + 6 --------------- (2)$

Observe here $k_1$ will be greater than or equal to $k_2$

Adding $(1)\ and\ (2)$

$2m = 12(k_1+k_2+1) + 2$

let $k_1+k_2+1 = k_3$

$m = 6(k_3) + 1$

Similarly by subtracting $(2)$ from $(1),$

$n = 6(k_4) + 1$

Now $mn = 36(k_3)(k_4) + 6(k_3) + 6(k_4) + 1$

$= 6(k_5) + 1$

Clearly now $mn$ when divided by $6$, will leave remainder $\textbf{1}$

$m - n = 12k_2 + 6 --------------- (2)$

Observe here $k_1$ will be greater than or equal to $k_2$

Adding $(1)\ and\ (2)$

$2m = 12(k_1+k_2+1) + 2$

let $k_1+k_2+1 = k_3$

$m = 6(k_3) + 1$

Similarly by subtracting $(2)$ from $(1),$

$n = 6(k_4) + 1$

Now $mn = 36(k_3)(k_4) + 6(k_3) + 6(k_4) + 1$

$= 6(k_5) + 1$

Clearly now $mn$ when divided by $6$, will leave remainder $\textbf{1}$

0 votes

@gatecse no sir in this qsn we can't say it before solving

wl have to solve like

m+n=12x+8

m-n=12x+6

I was just telling Mansi that her approach is wrong to solve this typ of qsn.

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