# Remainder Theorem

If a number, when divided by 3 leaves remainder 1, the number can be written as 3x + 1.

This can also be written as N = 1(mod 3).

This means that the number N when divided by 3 leaves remainder 1.

If we have to find the remainder when a number of high powers is used, this concept is helpful.

**Illustration 15:**

What is the remainder when 725 is divided by 6?

Notice that 7 can be written as (6 + 1).

So the sum becomes (6 + 1)25/6.

If we open the brackets by using binomial expansion, all the terms except the last will have 6 in them.

We can write this as 6x + 1, or 1(mod 6).

Hence the remainder obtained on dividing the number by 6 is 1.

In general, using remainder theorem, we try to reduce the number to the form: (ax + 1)n/a.

In all such numbers, the remainder will be 1.

# Negative remainde

Remember that two remainders can be obtained when dividing.

For instance, when 21 is divided by 5, we get remainder 1 and –4.

This is useful when finding the nearest number that is divisible by another number.

**Illustration 16**

What is the least three-digit number that is divisible by 7?

We know that the least three-digit number is 100.

When 100 is divided by 7, we get 7 ́14 = 98 and remainder 2.However, this will not give us the required answer.

In this case, we have to use the negative remainder, which is –5.

So the least three digit number that is divisible by 7 will be 100+5 = 105.

# Time Savers - Some useful points to remember

1. The tenth’s digit of 3n is always even for any natural n.

2. The sum of five successive whole numbers is always divisible by 5.

3. The square of an odd number when divided by 8 leaves a remainder of 1.

4. The product of 3 consecutive natural numbers is divisible by 6.

5. The product of 3 consecutive natural numbers, the first of which is an even number is divisible by 24.

6. If a number is not a multiple of 3, then its square divided by 3 leaves a remainder of 1.

7. The last digit of the product of any nine consecutive numbers is always 0.

8. The sum of a two-digit number and a number formed by reversing its digits is divisible by 11. E.g. 28 + 82 = 110, which is divisible by 11.

9. A three-digit number where all the digits are same is divisible by 37.

10. 10n –7 is divisible by 3 for any natural n.

11. Σn = n(n+1)/2. Where Σn is the sum of first n natural numbers.

12. Σn2 = [n(n+1)(2n+1)]/6. Where Σn2 is the sum of first n perfect squares

Sum of first n nos = *n(n+1)/2.
*Sum of first n squares *[n(n+1)(2n+1)]/6. *Sum of first n cubes *n2(n+1)2/4*

13. Σn3 = n2(n+1)2/4 = (Σn)2 Where Σn3 is the sum of first n perfect cubes.