Q2.    (i) Write a 3-digit number $abc$ as $100a + 10b + c$                                                             $\\= 99a + 11b + (a - b + c)\\ = 11(9a + b) + (a - b + c)$ If the number $abc$ is divisible by 11, then what can you say about $(a - b + c)$Is it necessary that $(a - b + c)$  should be divisible by 11?

let the number abc be 132

Here a = 1, b = 3 and c = 2

132= 100*1 + 10*3 + 2 = 99 + 11*3 + (1 - 3 + 2)

= 11(9*1+3) + (1 - 3 + 2 )

if number is divisible by 11 then (a - b + c )  must be divisible by 11.

as in above case of number 132 the a - b + c = 1 -3 + 2 = 0 whichis divisible by 11.

Hence we conclude ( a - b + c ) should be divisible by 11 if abc is divisible by 11.

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