# Q2.    (ii) Write a 4-digit number $abcd$ as $1000a + 100b + 10c + d$                                                     $\\= (1001a + 99b + 11c) - (a - b + c - d)\\ = 11(91a + 9b + c) + [(b + d) - (a + c)]$ If the number $abcd$ is divisible by 11, then what can you say about $[(b + d) - (a + c)]$ ?

If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11.

let the number be 1089

here a = 1, b = 0, c = 8 and d = 9

1089 = 1000*1 + 100*0 + 10*8 + 9

= (1001*1 + 99*0 + 11*8) + [(0 + 9) - (1 + 8)]

= 11(91*1 + 9*0 + 8) + [ 9 - 9 ]

here  [ (b + d) - (a + c) ] = [9 - 9 ] = 0 which is divisible by 11.

hence If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11.

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