**Remainders: Part-3**

**Problem Type 1: What will be the remainder when p + q +r +… is divided by d**

__Example__

**What will be the remainder when 63 + 67 +81 is divided by 11 ?**

**Solution**: Instead we add up all numbers, lets do it separately

63 when divided by 11 gives 5 as quotient and 8 as remainder.

67 when divided by 11 gives 6 as quotient and 1 as remainder.

81 when divided by 11 gives 7 as quotient and 4 as remainder.

Now the remainders are 8, 1 and 4 respectively.

By adding all the remainders 8+1+4=13, the sum of remainders is greater than the divisor

Therefore again the sum of remainders is divided by the divisor i.e. by 11

On dividing 13 by 11,quotient will be 1 and 2 will be the remainder

Hence the remainder on dividing 63 +67 +81 by 11 will be 2.

**Tooltip :**

Now the question arises that by dividing 53 sticks in groups of 5 we are left with 3 extra sticks. Then how many more sticks should be added to this bundle so that the left out sticks are again 3?

Answer

If we want left out sticks to be 3 again then we should add minimum 5 more sticks or

10,15,20, ….., 5n.

But remember the additional sticks should be multiple of 5.

53 when divided by 5 gives 10 as quotient and 3 as remainder and adding multiple of 5 is equivalent to adding remainder zero every time.

Let’stake an example to understand this

**Problem Type 2:When the remainder is same on dividing a number ‘a’ by different divisors.**

__Example__

What is the minimum two-digit number which when divided by 3, 4 and 5 leaves 2 as the remainder each time?

**Solution:**

The question would have been easy if we were asked about minimum number, which is completely divisible by three numbers 3,4 and 5 then we can say it should be least common multiple of 3,4 and 5 i.e. 60. Again 60 is the number which is completely divisible by 3,4,5 but if we want 2 remainder in each case then add 2 in 60 i.e. 62 will be the minimum two digit number which when divided by 3, 4 and 5 leaves 2 as the remainder each time.