Test Index

TGTET Paper 1 Exam 23 Jul 2017 Paper

© examsnet.com
Question : 91 of 150
Marks: +1, -0
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 = 3+2+1+7+2+7=223 + 2 + 1 + 7 + 2 + 7 = 22.
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 = 24−22=224 - 22 = 2.
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:
2
The correct option is C.
© examsnet.com
Go to Question: