Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3. Find _6724.
Hint: Find the sum of digits of the given number. Apply natural numbers from 0-9 in the blank space. A number is divisible by 3, if the sum of the digits are divisible by 3. Substitute different values in the sum and check divisibility.
Complete step-by-step answer: We know that a number is divisible by 3, if the sum of the digits are divisible by 3. We have to find the missing number in _6724. The sum of the digits _6724 = _+6+7+2+4 = _+19. Now we have to check different values for the blank digit and check if the number is divisible by 3. Sum of digit = _+19 Let us put 0 to 9 digits in the blank space. Let’s put variable ‘x’ in the blank space for easy calculation. $\therefore $Sum of digit= x+19 Let’s put x=0, sum = 19+0=19 x=1, sum=19+1=20 x=2, sum=19+1=21 x=3, sum=19+1=22 x=4, sum=19+1=23 x=5, sum=19+1=24 x=6, sum=19+1=25 x=7, sum=19+1=26 x=8, sum=19+1=27 x=9, sum=19+1=28 From the above possibilities, x=2, sum=21, is divisible by 3. x=5, sum=24, is divisible by 3. x=8, sum=27, is divisible by 3. $\therefore $The smallest digit in blank space = 2. The greatest digit in blank space =8. Therefore the no. formed are 26724 and 86724
Note: Here we used to find a single digit to fit the blank space, that’s why the natural nos from 0-9 were considered.