Problem Type 1: What will be the remainder when p + q +r +… is divided by d
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.
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?
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.
What is the minimum two-digit number which when divided by 3, 4 and 5 leaves 2 as the remainder each time?
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.