Question

Write the smallest and the greatest digits in the blank space of each of the following numbers so that the number formed is divisible by 3.
a) \[\_6724\]
b) \[4765\_2\_\].

Answer 1

Hint: To solve this problem we should know the rule of divisibility by 3.
A number is divisible by 3, if the sum of digits is divisible by 3.
Ex. 45 is divisible by 3 because \[4+5=9\] is divisible by 3.
Complete step-by-step answer:
We know
A number of is divisible by 3, if the sum of all the digits is divisible by 3
a) \[\_6724\]
Now, the number would be divisible by 3 if the sum of all the digits is divisible by 3.
Sum of digits =\[\_+6+7+2+4\]
=\[\_+19\]
By hit and trial method -
If we add 2 then the digit divisible by 3\[(19+2=21)\]
2 is the smallest digit
\[26724=2+6+7+2+4=21\]
Thus 2 is the smallest digit.
If we add 8 then the digit is divisible by 3 \[(19+8)=27\]
\[86724=8+6+7+2+4=36\]
Thus, 8 is the largest number
b) \[4765\_2\]
Now, the number would be divisible by 3 if the sum of all the digits is divisible by 3
Sum of digits = \[4+7+6+5+\_+2\]
= \[24+\_\]
By hit and Trial method,
As the sum of the digits is already divisible by 3
Thus 0 is the smallest number.
If we add 0 then the digit is divisible by 3\[(24+0)=0\]
\[476502=4+7+5+0+2=24\]
If we add 9 then the digit is divisible by 3\[(24+9=33)\]
\[476592=4+7+6+5+9+2=33\]
Thus, 9 is the largest number
Note: In the (b) part the sum of the digits is divisible by adding 0 as the smallest number. Some children get confuse that 0 will be considered or not
Thus, 0 will be considered.