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The remainder when $'m+n'$ is divided by $12$ is $8$, and the remainder when $'m-n'$ is divided by $12$ is $6$. If $m>n$, then what is the remainder when $'mn'$ is divided by $6$?

REMAINDER  : 1
m + n = 12*x + 8

m -  n = 12*x + 6

This implies n=1 and m = 12*x +7

m*n = 12*x + 7

So, m*n/6 leaves remainder 1.

@ank73811

m + n = 12*x + 8
m -  n = 12*x + 6

why do you think 12 will divide both x times only?

didn't notice that thank you

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

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

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 $\textbf{1}$

but we can solve this with m+n=8 and with m-n=6 with least value of m and n?
edited
Yes. But I derived a general case proof, to actually answer the question. There are many other ways to save time by taking cases or substituting options.

m+n = 8

m-n = 6

So, by evaluating , we will get 2m=14

so, m=7, n=1

mn/6 = 7*1/6 = 1 is remainder

But sir then why getting values by this😅

@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.

edited

@afroze ok will keep in mind

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