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If a 3-digit number 42a is divisible by 9, then the value of digit 'a' is:
9
1
6
3
3
For a number to be divisible by 9, the sum of its digits must be a multiple of 9. Simply add 4 and 2 to get 6, and find the smallest digit a such that 6+a equals 9.
A 3-digit number 42a is divisible by 9.
4+2+a=9k
For a number to be divisible by 9, the sum of its digits must be a multiple of 9. Simply add 4 and 2 to get 6, and find the smallest digit a such that 6+a equals 9.
Students often confuse the divisibility rule of 9 with the rule of 3, or fail to remember that a must be a single digit (0-9).
State the divisibility rule
A number is divisible by 9 if the sum of its digits is divisible by 9.
4+2+a=9n,┬аwhere┬аnтИИZ
Calculate the sum
Sum the known digits and express the condition for divisibility by 9.
6+a=9k
Solve for 'a'
For the smallest non-negative integer a such that 6+a is a multiple of 9, set k=1.
6+a=9тЯ╣a=3
D is correct because the sum of the digits 4+2+3=9, which is divisible by 9.
The divisibility rule for 9 is identical to the rule for 3 (sum of digits must be a multiple of 3). Recognizing this pattern helps in solving problems involving modular arithmetic quickly.
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