If the number 583_437 is completely divisible by 9, then the smallest whole number in the place of the blank digit will be:
- A4
- B5
- C3
- D6
Solution & Step-by-step Explanation
Let the missing blank digit be denoted by x. The number is 583x437.
According to the divisibility rule of 9, a number is completely divisible by 9 if the sum of its digits is a multiple of 9.
Sum of the digits:
S=5+8+3+x+4+3+7
S=30+x
For the number to be divisible by 9, (30+x) must be a multiple of 9.
The smallest multiple of 9 greater than or equal to 30 is 36.
Therefore:
30+x=36
x=36−30=6
Thus, the smallest whole number in place of the blank is 6.
According to the divisibility rule of 9, a number is completely divisible by 9 if the sum of its digits is a multiple of 9.
Sum of the digits:
S=5+8+3+x+4+3+7
S=30+x
For the number to be divisible by 9, (30+x) must be a multiple of 9.
The smallest multiple of 9 greater than or equal to 30 is 36.
Therefore:
30+x=36
x=36−30=6
Thus, the smallest whole number in place of the blank is 6.