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Clearly ‘abc’ will be divisible by ‘5’ if and only if c = 5 or c = 0.

Similarly, for 4-digit number abcd in general form.
abcd = 1000a + 100b + 10c + d = 5(200a + 20b + 2c) + d = 5x + d, where x = 200a + 20b + 2c

So, abcd will be divisible by 5 if its units digit d = 0 or d = 5
Think and Answer
Thus, we can conclude that if the units digit of a number
is 5 or 0, then the number is divisible by 5. A number when divided by 5, leaves a
remainder 1, what are its units digit.
Divisibility Test for 3 and 9

We know that a number is divisible by 3, only if the sum of its digits is divisible by 3. Also, a number
is divisible by 9, only if the sum of its digits is divisible by 9.
For example, the number 3447 is divisible by 3 as the sum of A number divisible by 9 is also
its digits = 3 + 4 + 4 + 7 = 18 is divisible by 3 and the number Note: divisible by 3 but a number
5121 is divisible by 9 as 5 + 1 + 2 + 1 = 9 is divisible by 9. divisible by 3 may or may not
be divisible by 9.
The number 5121 is also divisible by 3.

We can write the general form of a 3-digit number abc as:
abc = 100a + 10b + c = (99 + 1) a + (9 + 1) b + c

= 99a + a + 9b + b + c
= (99a + 9b) + (a + b + c) = 9(11a + b) + (a + b + c) = 9x + (a + b + c), where x = 11a + b

Hence, abc will be divisible by 9 if (a + b + c) is divisible by 9.

Further, we can write (99a + 9b) + (a + b + c) = 3(33a + 3b) + (a + b + c) = 3x + (a + b + c),
where x = 33a + 3b, and if (a + b + c) is divisible by 3, then the number abc must be divisible by 3.

Similarly, we can write the general form of a 4-digit number abcd as
abcd = 1000a + 100b + 10c + d = (999 + 1) a + (99 + 1) b + (9 + 1) c + d

= 999a + a + 99b + b + 9c + c + d
= (999a + 99b + 9c) + (a + b + c + d) = 9(111a + 11b + c) + (a + b + c + d)

= 9x + (a + b + c + d), where x = 111a + 11b + c

Hence, abcd will be divisible by 9 if (a + b + c + d) is divisible by 9.
Further, we can write (999a + 99b + 9c) + (a + b + c + d) = 3(333a + 33b + 3c) + (a + b + c + d)
= 3x + (a + b + c + d),

where x = 333a + 33b + 3c, and if (a + b + c + d) is divisible by 3, then the number abcd must be
divisible by 3.

Thus, we can conclude that if the sum of digits of a given number is divisible by 9, then the
number is divisible by 9. And if the number being divisible by 9, then it must be divisible by 3.

Teacher’s Any number is divisible by 4, if its last two digits is divisible by 4 and it is divisible by 8, if its last three digits is
Tip divisible by 8. If any number is divisible by both 2 and 3, it is divisible by 6 also. Take some examples and verify it.

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