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The topic of remainders is an important one for the CAT exam. The questions from this topic often appear in CAT or other competitive exams. In this article we are going to discuss this topic in detail.

We know that Dividend = Divisor × quotient + remainder

The number which is to be divided is known as dividend and the number which divides the dividend is known as divisor. Let us take an example:

If we divide 23 by 5, then 23 is dividend and 5 is divisor. Now 23 = 5 × 4 + 3, so, 4 is the quotient and 3 is the remainder. Similarly, if 99 is divided by 7, then 99 = 7 × 14 + 1. So, 14 is the quotient and 1 is the remainder.

Negative Remainders: Generally, the remainder is always a non negative number but sometimes to solve the questions in an easy way, we can take the remainders as negative also. Negative remainders are simply the difference of the positive remainder and the divisor and the sign used is negative. Let us try to understand it with the help of an example. If we divide 23 by 5, the positive remainder is 3. Now the difference of 3 and 5 is 2, so the negative remainder when 23 is divided by 5 is – 2. Similarly, when 99 is divided by 7, the remainder is 1 or – 6.

Note: The negative remainder can be converted into the positive remainder by adding divisor to it.

Few Basic Rules to find the remainders when sum, product or the difference of the numbers is given:

I: When the sum of numbers is divided by same divisor D:

Let us consider that x/D gives remainder r1 and y/D gives remainder r2. Note that here the divisor is same in both the cases.

Now if we divide (x + y) by the same divisor D, the remainders will also get added. So, the final remainder will be r1 + r2. Three cases will arise here:

Case I: If r1 + r2 < D, then r1 + r2 will be the remainder

Case II: If r1 + r2 = D, then the remainder will be zero

Case III: If r1 + r2 > D, then divide r1 + r2 by D to get the final remainder.

II: When the product of numbers is divided by same divisor D:

Let us consider that x/D gives remainder r1 and y/D gives remainder r2. Note that here the divisor is same in both the cases.

Now if we divide (x × y) by the same divisor D, the remainders will also get multiplied. So, the final remainder will be r1 × r2. Three cases will arise here:

Case I: If r1 × r2 < D, then r1 × r2 will be the remainder

Case II: If r1 × r2 = D, then the remainder will be zero

Case III: If r1 × r2 > D, then divide r1 × r2 by D to get the final remainder.

Note: In the above discussion, we have taken two numbers x and y only, but this result is applicable for ‘n’ number of numbers. So, if we get ‘n’ different remainders by dividing ‘n’ different numbers with the same divisor D, then on dividing the sum/product of all those numbers with the same divisor D, the remainders will get added/multiplied and the above three cases will be applicable again.

III: When the difference of numbers is divided by same divisor D:

Let us consider that x/D gives remainder r1 and y/D gives remainder r2. Note that here the divisor is same in both the cases.

Now if we divide (x – y) by the same divisor D, the remainders will also get subtracted. So, the final remainder will be r1 – r2. Now in this case, if the difference is positive, then that will be our remainder and if the difference of r1 and r2 comes out to be negative then we convert that negative difference or negative remainder to the positive remainder by adding the divisor to it.

Let us discuss few examples:

1. Find the remainder of (13 × 12)/11.

Sol: One way of solving the above question is to find the product, which comes out to be 156 and when 156 is divided by 11, the remainder comes out to be 2. Other way of doing is to find the individual remainders. 13/11 gives the remainder 2 and 12/11 gives the remainder 1. Now, we can multiply the individual remainders as there is multiplication sign between the numbers. Hence, the final remainder is 2 × 1 = 2. As it is less than the divisor, so it is the final remainder.

2. Find the remainder of (18 × 19 × 20 × 21)/17.

Sol: When 18, 19, 20 and 21 are divided by 17, the individual remainders are 1, 2, 3 and 4 respectively. Now 1 × 2 × 3 × 4 = 24 which is greater than the divisor 17. Here the case III discussed above will be applicable i.e., we will divide 24 by 17 to get the final remainder. Therefore, 24/17 = 7 is the remainder for the given division.

3. Find the remainder of (82 × 83 × 84)/17.

Sol: When 82, 83 and 84 are divided by 17, the remainders are 14, 15 and 16. Now, as we can see that 14, 15 and 16 are 3, 2 and 1 less than the divisor, so we can take the negative remainders as – 3, – 2 and – 1. So, the final remainder will be (- 3) × (- 2) × (- 1) = – 6. To convert this negative remainder to positive, we will add the divisor to it. Therefore, the final remainder = – 6 + 17 = 11.

Let us now discuss some other type of questions on remainders.

I: When numerator and denominator are co-prime to each other.

4. Find the remainder when 2143 is divided by 7.

Sol: To solve such questions, we need to find out the power of 2 which when divided by 7 will give remainder as 1 or -1. We know that 23 = 8 which when divided by 7 gives the remainder 1. So, 2143/7 can be written as (23)47 × 22/ 7 = 147 × 4/7 = 4/7. So, the final remainder is 4.

5. Find the remainder when 380 is divided by 82.

Sol: We know that 34 = 81 and 81 divided by 82 will leave remainder – 1.

Therefore, 380/82 = (34)20/82 = (- 1)20/82 = 1/82 = 1.

So, the required remainder is 1.

6. Find the remainder when 4124 is divided by 13.

Sol: We have 43 = 64 and 64 divided by 13 will leave remainder 12 or – 1.

So, we have 4124/13 = (43)41 × 4/13 = (- 1)41 × 4/13 = – 4/13 = – 4. Now, we can convert the negative remainder – 4 into positive by adding the divisor 13 to it. Therefore, the required remainder is 13 – 4 = 9.

II: When numerator and denominator are not co-prime to each other.

7. Find the remainder when 5124 is divided by 875.

Sol: First step here is to make the numerator and denominator co-prime.

Here 875 = 53 × 7. So, let’s cancel out 53 from both numerator and denominator. We will be left with 5121/7. Now, this question can be solved by the method discussed in the previous part.

We have 53 = 125 which when divided by 7 gives remainder – 1.

5121/7 = (53)40 × 5/7 = (- 1)40 × 5/7 = 5.

Now, 5 is the remainder for 5121/7, but out question is 5124/875. So, here we need to multiply the remainder 5 by 53 (which we cancelled earlier).

Therefore, the final remainder will be 5 × 53 = 625.

8. Find the remainder when 287 is divided by 68.

Sol: Let’s make numerator and denominator co-prime.

68 = 22 × 17. So, let’s cancel out 22 and we are left with 285/17.

We know that 24 = 16, which when divided by 17 will give remainder – 1.

Now, 285/17 = (24)21 × 2/17 = (- 1)21 × 2/17 = – 2/17 = – 2

Hence, – 2 + 17 = 15 is the remainder for 285/17.

Now, we need to multiply 15 by 22 to get the final remainder.

Therefore, the final remainder is 15 × 22 ­= 60.

Euler Number: Euler number of a number N is the number of natural numbers less than N and co-prime to N. E.g., Euler number of 15 will be 8, because there are 8 natural numbers (1, 2, 4, 7, 8, 11, 13, 14) which are less than 15 and are co-prime to 15. Using the same logic, Euler number of any prime number will always be one less than that prime number.

Therefore, if ‘p’ is a prime number the E(p) = p – 1.

Euler number of 7 is 6 and Euler number of 19 is 18.

Euler number of N is represented by E(N) and it can be calculated by formula,

E(N) = N × (1 – 1/a) × (1 – 1/b) × (1 – 1/c)…, where a, b, c,….. are different prime factors of N.

Let us calculate the E(15).

We have 15 = 3 × 5

So, E(15) = 15 × (1 – 1/3) × (1 – 1/5) = 15 × 2/3 × 4/5 = 8.

Let us find E(80). We have, 80 = 24 × 5

Now, E(80) = 80 × (1 – 1/2) × (1 – 1/5) = = 80 × 1/2 × 4/5 = 32.

Let us now discuss the Euler’s theorem to find the remainders.

Euler’s Theorem: If M and N are co-prime numbers then ME(N)/N always leave remainder 1.

Let us apply the Euler theorem to find the remainders.

9. Find the remainder when 5123 is divided by 7.

Sol: Here 5 and 7 are co-prime, so we can apply the Euler’s theorem.

Now E(7) = 6, therefore, 56/7 will leave remainder 1.

Now 5123/7 = (56)23 × 53/7 = 1 × 125/7.

The remainder when 125 is divided by 7 is 6, therefore the answer to the original problem is 6.

10. Find the remainder when 1398 is divided by 120.

Sol: Here, 13 and 120 are co-prime to each other.

120 = 23 × 3 × 5 i.e., prime factors of 120 are 2, 3 and5.

E(120) = 120 × (1 – 1/2) × (1 – 1/3) × (1 – 1/5) = 120 × 1/2 × 2/3 × 4/5 = 32.

Therefore, by Euler’s theorem, 1332/120 will leave remainder 1.

Now 1398/120 = (1332)3 × 132/120 = 1 × 169/120 = 49

Therefore, the remainder is 49.

11. Find the remainder when 11483 is divided by 420.

Sol: We have 420 = 22 × 3 × 5 × 7

Therefore, E(420) = 420 × (1 – 1/2) × (1 – 1/3) (1 – 1/5) × (1 – 1/7) = 420 × 1/2 × 2/3 × 4/5 × 6/7 = 96

So, 1196/420 will leave remainder 1.

Now, 11483/420 = (1196)5 × 113/420 = 1 × 1331/420 = 71

Therefore, 71 is the required answer.

This completes our discussion on remainders. It is an important concept if you are preparing for the CAT. In the next article we will learn some other new concept.

Till then Happy learning!!!