CAT4MBA

(121212......300 Times)/99 -  What is the remainder?

Solution:

If I want to find the remainder when a number is divided by 99, then I start from the unit's digit of the number, take two digits at a time and then move towards the leftmost digit finding the sum of all those two digit numbers obtained and then find the remainder when that sum is divided by 99.

Example:

What is the remainder when the number 1234567 is divided by 99?

The we first start from the right hand side of the number taking two digits at a time and find their sum.

So, we get: 67 + 45 + 23 + 01 = 136.

Now, when 136 is divided by 99, the remainder is 37.

So, the remainder when 1234567 is divided by 99, the remainder is 37.

We use the similar method for finding the remainder when the number 121212......300 times is divided by 99.

Note that if we take two digits at a time, starting from the right hand side, then everytime we will get the number 12 and since the given number is a 300 digit number, 12 will be added 150 times and so the sum obtained will be (12) * (150) = 1800 and this number, when divided by 99, the remainder will be 18, which is the required remainder.

Hence, when the number 1212121212 ............. 121212 (300 times) is divided by 99, the remainder is 18.

Properties of H.C.F. and L.C.M.

 Property 1:

The greatest number that divides each of a, b and c leaving the same remainder in each case is H.C.F. of |a – b|, |b – c|, |c – a|.

Example 1: The greatest number that will divide 132, 100 and 36 leaving the same remainder in each case is:
a) 8 b) 24 c) 32 d) 64

Solution:

In the given example, a = 132, b = 100 and c = 36

|a – b| = |132 – 100| = 32
|b – c| = |100 – 36| = 64
|c – a| = |36 – 132| = 96

Hence the required number = H.C.F. of 32, 64 and 96 which is equal to 32.

Hence 32 is the greatest number that divides each of the numbers 132, 100 and 36 to
leave the same remainder in each case and it can be verified that the remainder is 4 in
each case.

Example 2: Find the largest number that will divide 171, 333 and 576 to leave exactly
the same remainder in each case.
a) 9 b) 162 c) 27 d) 81

Solution:

In the given example, a = 171, b = 333 and c = 576

|a – b| = |171 – 333| = 162
|b – c| = |333 – 576| = 243
|c – a| = |576 – 171| = 405

Hence the required number = H.C.F. of 162, 243 and 405 which is equal to 81.

Hence 81 is the greatest number that divides each of the numbers 171, 333 and 576 to
leave the same remainder in each case and it can be verified that the remainder is 9 in
each case.

Property 2:

The greatest number that divides each of a, b and c to leave remainders p, q and r respectively is H.C.F. of (a – p), (b – q), (c – r).

Example 1: Find the greatest number that will divide 887 and 1061 leaving remainders 7 and 5 respectively.
a) 32 b) 16 c) 176 d) 18

Solution:

In the given example, a = 887, b = 1061, p = 7 and q = 5

a – p = 887 – 7 = 880
b – q = 1061 – 5 = 1056

Hence the required number = H.C.F. of 880 and 1056 which is equal to 176.

Hence 176 is the greatest number that divides each of the numbers 887 and 1061 leaving remainders 7 and 5 respectively.

Example 2: Find the greatest number that will divide 114, 148 and 182 leaving remainders 2, 4 and 6 respectively.
a) 32 b) 16 c) 8 d) 28

Solution:

In the given example, a = 887, b = 1061, p = 7 and q = 5

a – p = 114 – 2 = 112
b – q = 148 – 4 = 144
c – r = 182 – 6 = 176

Hence the required number = H.C.F. of 112, 144 and 176 which is equal to 16.

Hence 16 is the greatest number that divides each of the numbers 114, 148 and 182 leaving remainders 2, 4 and 6 respectively.

Property 3:

The least number which when divided by a, b, c leaves the same remainder ‘r’ in each case is {L.C.M. of a, b and c} + r.

Example 1: Find the least number which when divided by 16, 18 and 20 leaves a remainder 1 in each case.
a) 359 b) 361 c) 719 d) 721

Solution:

In the given example, a = 16, b = 18, c = 20 and r = 1

L.C.M. of 16, 18 and 20 = 720

Hence the required number is 720 + 1 = 721.

Example 2: Find the least number which when divided by 2, 3, 4, 5, 6, 7, 8, 9 and 10 leaves a remainder 1 in each case.
a) 2159 b) 2561 c) 2519 d) 2521

Solution:

The given property holds for any set of values of a, b and c.

Hence in this example, the required number is
{L.C.M. of 2, 3, 4, 5, 6, 7, 8, 9 and 10} + 1 = 2520 + 1 = 2521

Property 4:

The least number which when divided by each of a, b and c leaves remainders p, q and r respectively, such that a – p = b – q = c – r = k (say), is {L.C.M. of a, b and c} – k.

(Note that in these type of problems, the remainders are such that the difference between the divisors and the remainders is the same in each case. This will become easier to understand once you go through the following two examples)

Example 1: Find the least number which when divided by 7, 8, 9 and 10 leaves 5, 6, 7, and 8 as remainders respectively.
a) 2518 b) 2522 c) 2528 d) None of these

Solution:

In the given example, a = 7, b = 8, c = 9, d = 10, p = 5, q = 6, r = 7, s = 8 and k = 2

Note that the difference between the divisors a, b, c, d and their respective remainders p, q, r and s is the same in each case and this difference = 2.

Hence the required number is {L.C.M. of 7, 8, 9 and 10} – 2 = 2520 – 2 = 2518

Example 2: 20, 25, 30, 36 and 48 divides a number. The respective remainders obtained are found to be 15, 20, 25, 31 and 43. Find the least such number
a) 3600 b) 3605 c) 3595 d) None of these

Solution:

Similar to the above question, note that the difference between the divisors and their respective remainders is the same in each case and this difference = 5.

Hence the required number is {L.C.M. of 20, 25, 30, 36 and 48} – 5 = 3600 – 5 = 3595