Any number that is divisible by 3 is also divisible by 9.

Find a counterexample to show that the conjecture is false.
18
27
45
48

$$48\ is\ divisible\ by\ 3\ but\ isn’t\ divisible\ by\ 9:\\\\48:3=16\\\\48:9=5\frac{3}{9}=5\frac{1}{3}\\\\Other\ examples:\\\\51:3=17;\ 51:9=5\frac{2}{3}\\\\6:3=2;\ 6:9=\frac{2}{3}\\\vdots$$

For a number to be divisible by 9 the sum of the digits is also divisible by 9.
For the examples given 48 (4+8 = 12) (1 + 2 = 3) therefore 48 is not divisible by 9 as the sum of the digits is 3.

For a number to e divisible by 3 the sum of the digits is also divisible by 3. (eg 48 (4+8 = 12), (1 + 2 = 3) therefore 48 is divisible by 3

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