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TGTET Paper 1 Exam 23 Jul 2017 Paper
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Question : 91 of 150
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The smallest number that must be added to 321727, so that the resultant exactly divisible by 3 is ______
Solution:
Concept:A number is divisible by 3 if the sum of its digits is divisible by 3.Explanation:First, let's find the sum of the digits of the given number 321727.Sum of digits = .Now, we need to find the smallest number to add to 321727 so that the new sum of digits is divisible by 3. We look for the next multiple of 3 greater than 22.The multiples of 3 are ..., 18, 21, 24, 27, ...The next multiple of 3 after 22 is 24.To get a sum of 24, we need to add a number to 22.Number to add = .So, if we add 2 to the original number, the sum of its digits will be 24, which is divisible by 3. Therefore, the resultant number will be exactly divisible by 3.Answer:2The correct option is C.
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