Fill in the blank with smallest digit to make $. . . . . . . 6724$ divisible by $3 .$

Open in App

Solution

Since as we know that sum of all digits of a number is divisible by $3$ then the number is divisible by $3 .$
The sum of all digits of a number $x 6724$ is $x + 6 + 7 + 2 + 4 = 19 + x$ where $x$ is the smallest digit.
When we $x = 2$ then $19 + 2 = 21$ which is divisible by $3 .$
Hence, the smallest digit to be placed in blank space is $2$ then digit $2$ to make $. . . . . . . 6724$ divisible by $3 .$

Suggest Corrections

Similar questions

Fill in the blanks with the smallest digit to make the number divisible by $9$ :

$923___93$

Fill in the blank with smallest digit to make $1462$ ………. divisible by $3 .$

Fill in the blank with smallest digit to make $189 . . . . 44$ divisible by $3 .$

Fill in the blank with smallest digit to make $701 . . . . . 9308$ divisible by $11 .$

Fill in the blank with smallest digit to make $6000 . . . . . 3$ divisible by $3 .$

Join BYJU'S Learning Program