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# Tests of Divisibility - Divisibility by 9 and 3

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They use only the one’s digit of the given number; they do not bother about the ‘rest’ of the digits. Thus, divisibility is decided just by the one’s digit. 10, 5, 2 are divisors of 10, which is the key number in our place value.
But for checking divisibility by 9, this will not work. Let us take some number say 3573. Its expanded form is:
3 × 1000 + 5 × 100 + 7 × 10 + 3
This is equal to 3 × (999 + 1) + 5 × (99 + 1) + 7 × (9 + 1) + 3
= 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 3)            ... (1)

We see that the number 3573 will be divisible by 9 or 3 if (3 + 5 + 7 + 3) is divisible by 9 or 3.
We see that 3 + 5 + 7 + 3 = 18 is divisible by 9 and also by 3. Therefore, the number 3573 is divisible by both 9 and 3.
Now, let  us consider the number 3576. As above, we get
3576 = 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 6)              ... (2)

Since (3 + 5 + 7 + 6) i.e., 21 is not divisible by 9 but is divisible by 3,
therefore 3576 is not divisible by 9. However 3576 is divisible by 3. Hence,
(i) A number N is divisible by 9 if the sum of its digits is divisible by 9. Otherwise it is not divisible by 9.
(ii) A number N is divisible by 3 if the sum of its digits is divisible by 3. Otherwise it is not divisible by 3.
If the number is ‘cba’, then, 100c + 10b + a = 99c + 9b + (a + b + c)
=9 (11c + b) + (a + b + c)
Hence, divisibility by 9 (or 3) is possible if a + b + c is divisible by 9 (or 3).

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