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If \(\bar{4a3b}\) is divisible by 11 find all possible value of a + b. |
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Answer» The number is divisible by 11 if and only if the difference between the sum of the digits in the odd places and the sum of the digits in the even places is divisible by 11. (a + b) – (4 + 3) = 0 (a + b) – 7 = 0 or (a + b) – 7 = 11 a + b = 7 or a + b = 11 + 7 = 18 (a + b) – 7 cannot be equal to 22 because it a + b – 7 = 22 then a + b = 29 since a and b are digits their sum cannot be 29. |
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