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Since, 33 = 11 x 3
So, 5731x7 is divisible by both 11 and '3'.
Since, the given number divisible by 3, sum of all the digits in the given number should be divisible by '3'.
So, (23 + x) should be a multiple of 3
So, 'x' can be 1, 4 or 7
Since, the given number divisible by 11, the difference of the sum of the digits at even place and sum of the digits at odd place should be zero or a multiple of 11
So, (5 + 3 + x) - (7 + 1 + 7) = 0 or multiple of 11
Or, (8 + x - 15) should be 0 or multiple of 11
So, 'x' will be 7
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