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Find the sum of all the possible values of (a + b), so that the number 4a067b is divisible by 11.
21
11
16
5
Correct Option is 1- 21
The divisibility by 11 rule states that if the difference between the sum of the digits at the odd and even places equals 0 or divisible by 11, then the number is divisible by 11. The Sum of any two digits is not more than 18. Calculation: Sum of the digits at the odd places = (4 + 0 + 7) = 11 Sum of the digits at the even places = (a + 6 + b) = (a + b + 6) Let the difference be 11k where K is an integer. So, 11 - (a + b + 6) = 11k Now, taking k = 0 we get, 11 - (a + b + 6) = 0 = (a + b) = 5 Now, we're taking k = -1 since (a + b) can't be negative and more than 18, 11 - (a + b + 6) = -11 = (a + b) = 16 Now, the sum of all the possible values of (a + b) = 5 + 16 = 21 The sum of all the possible values of (a + b), so that the number 4a067b is divisible by 11 is 21.
By: santosh ProfileResourcesReport error
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