Revision Notes
1. Divisibility by 2n i.e. by 2, 4, 8, 16, ….
When a given number is divided by 2n, the remainder will be same as the remainder when just the last n digits is divided by 2n.
If the last n digits, when divided by 2n leaves a remainder of 0 i.e. is completely divisible, then the number will also be completely divisible by 2n.
2. Divisibility by 5n i.e. by 5, 25, 125, ….
Same as the rule of 2n.
When a given number is divided by 5n, the remainder will be same as the remainder when just the last n digits is divided by 5n.
If the last n digits, when divided by 5n leaves a remainder of 0 i.e. is completely divisible, then the number will also be completely divisible by 5n.
3. Divisibility by 3 and 9
When a given number is divided by 3 (or 9), the remainder will be same as the remainder when just the sum of digits of the number is divided by 3 (or 9).
If the remainder when the sum of digits is divided by 3 (or 9) is 0 i.e. is completely divisible, then the number will also be completely divisible by 3 (or 9).
This rule cannot be extended to higher powers of 3 i.e. it wont work for 27.
4. Divisibility by 11
For any given number,
starting with the unit’s digit, sum all the alternate digits moving leftwards, say the sum = U
Starting with the ten’s digit, sum all the alternate digits moving leftwards, say the sum = T
When the number is divided by 11, the remainder will be same as (U – T). Now, the remainder when the divisor is 11 has to be a value from 0 to 10. If (U – T) is more than 10, then just subtract (or keep subtracting) 11 till we get a value 10 or less, this will be the remainder; and (U – T) could also be negative to start with … do not change the order of subtraction, it has to be (U – T) and not (T – U) …. if (U – T) is negative, then add 11 (or keep adding 11) till we get a value from 0 to 10, this will be the remainder.
It directly follows that if (U – T) is any value out of {…, −33, −22, −11, 0, 11, 22, 33, …..} i.e. 0 or any multiple of 11, even negatives are fine, then the remainder will be 0 i.e. the number is completely divisible by 11.
5. Divisibility checks for composite numbers.
Say we need to check if any number is divisible by a divisor which is a composite number. Write the divisor, the composite number, as a product of co-prime numbers …
e.g. 24 has to be written as 3 × 8 and not as 2 × 12 or 2 × 3 × 4;
60 will be written as 3 × 4 × 5
… say, the composite divisor = a × b × c × ….
A number will be divisible by the composite divisor if and only if the number is divisible by EACH of a, b, c, …
To use these rules, we should know the divisibility rules of a, b, c … So, this can be used only in cases where a, b, c, … are 2, 3, 4, 5, 8, 9 or 11. But then only these get asked i.e. check divisibility by 6, 12, 15, 18, 24, 36, 72, 88, 99.
We cannot use the above logic to find the remainder when divided by the composite divisor, we can only check if the number is divisible or not divisible.
6. When you solve more difficult questions on remainders, you may come across finding remainder when divided by 77 or 91 or 143 or 1001. All these are linked with 1001.
1001 = 7 × 11 × 13
And 77, 91 and 143 are 7×11, 7×13, 11×13 respectively.
Next, any digit, say a, written 6 times i.e. the number aaaaaa can be worked upon as …
aaaaaa
= aaa000 + aaa
= aaa × 1000 + aaa
= aaa × (1000 + 1)
= aaa × 1001.
Thus, for now, it is enough to realise that any number of the form aaaaaa is divisible 1001 and hence by any factor of 1001 including 77, 91 and 143 or even 7, 11 and 13.
7. Must Know Questions from the Exercise
Q. 4, 5, 6 each having numbers with two unknown digits and divisibility by 88, 24 and 99.
This doesn’t mean you dont solve the other questions, every question in the exercise is tricky, even if they may not be very exam-like, they help you build problem solving and need to be solved. If you learn something of value from them, mark them as well, to revisit them.
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