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A number is divisible by 9 if

  • A.the sum of its digits is divisible by 9
  • B.its last digit is 9
  • C.it is divisible by 3 and by 6
  • D.its last two digits are divisible by 9

Answer: D

Correct answer: the sum of its digits is divisible by 9

A number is divisible by 9 exactly when the sum of its digits is divisible by 9 - and the test may be applied repeatedly until a single digit remains. For 4,617 the digits sum to 18, which is divisible by 9, so 4,617 is too. The divisibility rules worth having ready form a set. By 2: the last digit is even. By 3: the digit sum is divisible by 3. By 4: the last two digits form a multiple of 4. By 5: the last digit is 0 or 5. By 6: divisible by both 2 and 3. By 8: the last three digits form a multiple of 8. By 10: the last digit is 0. By 11: the alternating sum of the digits is 0 or a multiple of 11. The third option fails because divisibility by 3 and 6 together only guarantees divisibility by 6 - the two share a factor, so their product is not implied.

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