Question

What smallest digit be written in the blanks space of the numbers --------- 6724 so that the numbers formed is divisible by 3 ? (a) 3 (b) 4 (c) 2 (d) 1

Answer 1

Answer:

2

Step-by-step explanation: As the divisibility test of 3 is ' The sum of the digits in the number given have to be divisible by 3....if it is,the whole no is divisible by 3'...Therefore,when we add 1 to the sum of all digits of 6724,that is 19,we get the no as 20...and 20 is not divisible by 2...so move on to the next smallest no which is 2...when 2 is added to 19, we get the ans as 21,which is divisible by 3...

Therefore,the ans is option C, ans is 2