Question

2. State the divisibility tests for 3,5 and 9 with an example for each.

Ask by Greene Reed. in Australia
Mar 18,2025

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Answer

A number is divisible by 3 if the sum of its digits is divisible by 3. For example, 123 is divisible by 3 because \(1 + 2 + 3 = 6\), and 6 is divisible by 3. A number is divisible by 5 if it ends with 0 or 5. For example, 45 is divisible by 5 because it ends with 5. A number is divisible by 9 if the sum of its digits is divisible by 9. For example, 729 is divisible by 9 because \(7 + 2 + 9 = 18\), and 18 is divisible by 9.

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Beyond the Answer

To test if a number is divisible by 3, simply add up all the digits of the number; if the sum is divisible by 3, so is the number. For example, with the number 123: 1 + 2 + 3 = 6, which is divisible by 3, so 123 is divisible by 3. For divisibility by 5, a number is divisible if it ends in 0 or 5. Take, for example, the number 45, as it ends in 5, this means it is divisible by 5. To check for divisibility by 9, add all the digits together just like with 3. If the sum is divisible by 9, then the original number is as well. For instance, with 729: 7 + 2 + 9 = 18, and 18 is divisible by 9, hence 729 is divisible by 9.

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