# Remainders (Part I)

In this chapter, we are going to learn how to find remainders when a number or any expression is divided by a particular divisor. First of all, we should understand one thing that in all the problems related to finding out the remainder, we are just interested in finding out the remainder and not the quotient.

Grouping funda: Whenever we are dividing by a number, we make groups of that number and whatever cannot be clubbed into the groups, that remains as the remainder or left-over. In mathematical terms, the groups have been named as “Quotient” and left-overs as “Remainders”.  For example, when we divide 49 by 11, we get 4 groups (as 4 groups of 11 can be made) and left over of 5. We are getting remainder of 5, because we cannot make a group of 11 with 5. In mathematical terms, the groups that we got were actually the quotient and left over was remainder.

We can write 49 = 11 × 4 + 5.

We should understand this relation as it will be used extensively throughout this chapter.

Dividend = Divisor × Quotient + Remainder.

This can be used in the language like when a number say ‘m’divided by p leaves a remainder of r. We should be able to immediately frame equation like m = p × a + r, where a is the quotient which can be any whole number.

How to find the remainder:

Let us assume there are two numbers P and Q, which when divided by 7 leave remainders of 5 and 3 respectively. Find the remainder when (P + Q) is divided by 7.

We should be immediately able to frame the equation as P = 7 × a + 5 and Q = 7 × b + 3, where a and b can be any whole numbers.

When we add P + Q, we get P + Q = 7(a + b) + 8. Now, we have to find the remainder when this result is divided by 7. 7(a + b) is a multiple of 7 and any multiple of 7 when divided by 7 will always leave a remainder of zero and 8 divided by 7 will leave a remainder of 1. So, final answer is 1. If we talk in terms of group, we can easily refer to 7(a + b) as a number which is already divided into groups of 7 and any number which is divided into groups of 7 when divided by 7 will always leave a remainder of 0. So, our answer is just dependent on the summation of the remainders obtained from the two numbers. So, we could have just got the answer by adding up the remainder and not the entire number.

So, when numbers are being added and we have to find the remainder, we just need to add up the remainders obtained from the respective numbers.

What would be the remainder if (P – Q) is divided by 7?

Again, if we apply the standard method, we get P – Q = 7(ab) + 2. When this is divided by 7, we get a final remainder of 2, since 7(ab) is a multiple of 7, it will leave a remainder of 0, so our answer is dependent on direct subtraction of remainders.

So, when two numbers are being subtracted, the remainders from the respective numbers should be subtracted to give the final remainder.

What would have been the remainder if (P × Q) ÷ 7?

On multiplying P with Q, we get = 7a × 7b + 7a × 3 + 7b × 5 + 15, on observing we can say that the first three terms in the product are multiples of 7. So, when we divide them by 7, all three terms will leave us remainders of zero and our answer will be obtained by dividing 15 by 7. So, final answer is 1. Now, we should think what was 15? 15 was nothing but the product of the remainders obtained from the two numbers.

So, when the numbers are being multiplied, final remainder can be directly obtained by just directly multiplying the remainders from the respective numbers.

E.g. 1: What will be the remainder when 1235 × 1489 + 1378 – 1987 is divided by 6?

We will use the same logic what we just learnt. We will find out the individual remainders of each number by dividing them by 6 and then perform the same operation on the remainders which are being done on the numbers.

Remainders obtained from the four numbers are 5, 1, 4 and 1. Now we will perform the same operation on these remainders which was being done on their respective numbers.

Thus, 5 × 1+ 4 – 1 =9 – 1 = 8. But the remainder cannot be greater than the divisor, so we will have to further divide 8 by 6, (as one more group of 6 can be formed). So, the final answer will be 2.

E.g. 2: Y is a number which when divided by 7 leaves a remainder of 4. Then find the remainder when (Y2 – 5Y + 5) is divided by 7.

Now, Y2 = Y × Y; If Y is leaving a remainder of 4, then Y2 will leave a remainder of (4 × 4) as remainders are also multiplied if numbers are being multiplied. So, Y2 leaves a remainder of 16, but remainder obtained cannot be greater than the divisor which is 7 in this case. So, we will have to further divide 16 by 7 which will give us a remainder of 2.

And, 5Y will leave us a remainder of 5 × 4 = 20 as Y when divided by 7 leaves us a remainder of 4, but since remainder cannot be greater than the divisor, we will divide 20 by 7 which will lead us to remainder of 6.

Now, just plug-in the values of remainder obtained from each of the individual numbers to get the final answer. We get 2 – 6 + 5 = 1

E.g. 3: The number ‘a’ is exactly divisible by 5. The remainder after the division of the number ‘b’ by 5 is equal to 1 and the remainder after the division of the number ‘c’ by 5 is equal to 2. What will be the remainder if (3a +b – 3c) is divided by 5?

This problem is also based on the same logic. We can solve it either by assuming values or using the standard method. Let us see both the methods.

We can write a = 5x, b = 5y + 1 and c = 5z + 2 (where x, y and z can assume any whole number). Rather than proceeding ahead in the standard way, we can just assume any value of a, b and c which satisfies the condition.

We can take a = 5, b = 6 and c = 7.

Now, the question is asking to find out the remainder when (3a +b – 3c) is divided by 5. Now, just substitute values of a, b and c in the expression.

We get 15 + 6 – 21 = 0; When 0 is divided by 5, remainder will be 0 only. We will get the same answer if we follow the standard method.

We had a = 5x, so 3a will be 15x.

We also had b = 5y + 1

and c = 5z + 2. Then 3c becomes 15z + 6.

Again just substitute these values in the expression (3a +b – 3c), we get (5x + 5y + 1 – 15z – 6). On simplifying it we can write it as 5(x + y – 3z – 1). The expression which we got is a multiple of 5 and when any multiple of 5 is divided by 5, it leaves a remainder of 0.

Concept of Negative Remainders: We just saw whenever we are finding remainder; we are making groups of the divisor. For e.g., if 18 is divided by 7, remainder will be 4 as 2 groups of 7 can be formed and 4 will be left behind. Now, think in the reverse way. 4 is left over as we cannot make a group of 7 with 4. How many more are required to make a group of 7?

Answer should be that we need 3 more. So, we are lacking in 3 to make a group of 7, that’s why a negative remainder of –3 is also the same thing as positive remainder of +4, when a number is divided by 7.

Similarly, if 34 is divided by 8, remainder will be +2 as 4 groups of 8 can be formed. How many are we lacking to make one more group of 8? Answer is we need 6 more. So, answer is –6 or +2.

If the remainder obtained on dividing a expression by 15 is –7, we should be able to convert it into positive remainder. Think in terms of group formation that when we are dividing a number by 15 means that we are making group of 15. Our remainder obtained is –7, it means that we are actually lacking in 7, then how many do we actually have? Answer should be we have 8 (–7 + 15 = 8). So, we should be able to convert positive remainder into negative remainder and vice-versa.

If the remainder obtained on dividing a number by 23 is –18, how much is the positive remainder? Answer should be +5 as a remainder of –18 when being divided by 23 means that we are lacking in 18, and we have in actual leftover of 5, (–18 + 23 = 5.)

So, if the remainder comes in negative, to convert it into positive remainder we just need to add that particular divisor to the negative remainder, whatever resultant comes is the positive remainder.

Range of Remainders: When a number is divided by a particular divisor, the remainder obtained cannot be greater than or equal to the divisor. For e.g. if any number is divided by 9, remainder obtained can be in the range of 0 to 8. Remainder cannot be 9 or more. If remainder is greater than the divisor, then we will have to further divide it so that it becomes smaller than divisor. Grouping logic is also applied here. If the remainder is greater than the divisor, then some more groups can be formed. For e.g., if remainder obtained on dividing a number by 13 is 20. But, the remainder cannot be greater than 20, so we will have to further divide 20 by 13. One more group can be formed and then the final remainder will be 7.

Cyclicity of Remainders:

We have already learnt how the cyclicity of unit’s place in the previous chapter. Same logic applies for the cyclicity of remainder. We will revisit this logic with the help of examples.

Find the remainder when 7n is divided by 4.( n can take values 1, 2, 3, 4, ……..).

If n = 1, 7 ÷ 4 gives a remainder of 3.

If n = 2, 72 ÷ 4 gives a remainder of 1, (rather than dividing 72 = 49 by 4, we know 7 divided by 4 left a remainder of 3, so 72 will leave a remainder of 32. And 9 divided by 4 will leave a remainder of 1.This is a much better and easy way).

If n = 3, 73 ÷ 4 gives a remainder of 3, (Again rather than dividing 73 = 343 by 4, we know 72 divided by 4 left a  remainder of 1, then 73 = 72 × 71 = 1 × 7 = 7, and 7 divided by 4 will leave a remainder of 3). Remember we are using the funda of remainders being multiplied if the numbers are in multiplication.

If n = 4, 74 ÷ 4 gives a remainder of 1, (74 = 73 × 7 = 3 × 7 = 21 = 1, as 73 left a remainder of 3).

We see that when n assumes values 1, 2, 3, 4,……, the remainders are 3, 1, 3, 1. So, the remainders are following a cycle of 2. So, now we should be able to identify the logic that was used in Cyclicity chapter.

If n was 95, then the remainder would  be 3 as when n is odd the remainder is 3. If n was 96, then the remainder would be 1 as when n is even, remainder is 1. Using this approach, we can solve almost all the problems of Remainder.

E.g. 4: What is the remainder when 2843 is divided by 15?

This problem seems very difficult as 2843 is a very large number, but it can be solved very easily with the help of logic of cyclicity of remainders.

2843 = 28 × 28 × 28 × 28 × ……… × 2843 times.  We are dividing by 15 means we are making groups of 15. Every 28 divided by 15 will leave us a remainder of 13 and we will get 13 multiplied 43 times. Now, the question reduces to finding the remainder of 1343 divided by 15.

131 ÷ 15 leaves a remainder of 13.

132 ÷ 15 = 169 ÷ 15 = a remainder of 4.

133 ÷ 15 = 132 × 13 = 4 × 13 = 52 ÷ 15 = a remainder of 7.

134 ÷ 15 = 133 × 13 = 7 × 13 = 91 ÷ 15 = a remainder of 1.

135 ÷ 15 = 134 × 13 = 1 × 13 = 13 ÷ 15 = a remainder of 13.

136 ÷ 15 = 135 × 13 = 13 × 13 = 169 ÷ 15 = a remainder of 4.

137 ÷ 15 = 136 × 13 = 4 × 13 = 52 ÷ 15 = a remainder of 7.

138 ÷ 15 = 137 × 13 = 7 × 13 = 91 ÷ 15 = a remainder of 1.

The remainders are following a cycle of 13, 4, 7, 1, 13, 4, 7, 1, …………………… when the power of 13 takes values starting from n = 1, 2, 3, 4, 5, 6, 7, 8, ……… So, the cyclicity of the remainders is 4.

And the question is asking remainder of 1343, what should we do? We will divide the power by the cycle. 43 ÷ 4 gives us a remainder of 3.

It means that the remainder of 1343 will be same as that of 133 (third term of the cycle) which is 7.

You should have observed in both the examples we have used the remainder obtained in the previous stage to multiply with the new number coming in the next stage. That makes our calculation easier. And our objective in all such remainder problems should be to find such power of that particular number which leaves us with a remainder of 1, because after that we would not be going ahead with calculations as the cycle starts repeating itself.

E.g. 5: What is the remainder when 1947 is divided by 7?

1947 has to be divided by 5, we can reduce this problem to 547 ÷ 7. Now follow the approach learnt.

5 ÷ 7 leaves a remainder of 5.

52 ÷ 7 leaves a remainder of 4.

53 ÷ 7 leaves a remainder of 6, (53 = 52 × 5 = 4 × 5 = 20 ÷ 7 = 6)

54 ÷ 7 leaves a remainder of 2, (54 = 53 × 5 = 6 × 5 = 30 ÷ 7 = 2)

55 ÷ 7 leaves a remainder of 3, (55 = 54 × 5 = 2 × 5 = 10 ÷ 7 = 3)

56 ÷ 7 leaves a remainder of 1, (56 = 55 × 5 = 3 × 5 = 15 ÷ 7 = 1)

We don’t need to go any further as we got a remainder of 1, and cycle will start repeating itself. The remainders have a cycle of 6, so we will divide 47 by 6 which gives us a remainder of 5. That means that remainder of 547 when divided by 7 will be same as that of the fifth term of the cycle (i.e. 55) which happens to be 3.

Significance of negative remainder:

We could have solved the previous example faster with the help of negative remainder of (–1). If we can just find out which cycle gives us remainder of (–1), then by doubling the cycle we can get the cycle which will give us the positive remainder of (+1). This approach will help solve the problem in half the iteration. Earlier, our objective was to find such cycle which will give us a remainder of (+1), but now we will redefine our objective. We should find such cycles which give us a remainder of (–1) and after that we can solve these problems comfortably.

Let us see how to use it. In the previous example, 53 when divided by 7 gave a remainder of 6. We should have stopped there only and thought that if a number when divided by 7 leaves a remainder of 6, it means that it will also leave a remainder of (–1). And if 53 leaves us a remainder of (–1), then if we take one more group of 53, it will also leave a remainder of (–1).

Thus, 53 × 53 = (–1) × (–1) = 1; So, 56 will leave us a remainder of 1.We could have solved from here on with the cyclicity of remainder logic. Remainders are following a cycle of 6, so divide 47 by 6, we get 5 leftover which means 55 (fifth term of the cycle) will be left behind. Now, you will think that we have found our remainders up till 53 only, so to find remainder of 55 we will have to do it again. No, we can split 55 into 53 × 52 as we know the remainders for 53 and 52.

55 ÷ 7 = (53 × 52) ÷ 7 = [(–1) × 4] ÷ 7 = [–4] ÷ 7 = a remainder of 3. (We have used the remainder obtained from 53 and 52 to find the final remainder of 55).

E.g. 6: Find the remainder of 1499 divided by 15.

If we are not looking for a remainder of (–1), this problem will involve a lot of calculations. So, we should be able to identify and apply wherever (–1) remainder is found because it saves times and makes our calculations efficient.

1499 means 14 × 14 × 14 × 14 × ……………………………..×14 99 times. Every 14 divided by 15 either leaves us a remainder of 14 or (–1). And 14 is written 99 times, so we will get (–1) 99 times. So, now our question reduces to finding the remainder of (–1)99 ÷ 15, which can be done orally.

(–1)odd power always results in (–1). And (–1) ÷ 15 will leave us a remainder of 14. (Remember the negative remainder funda).

E.g. 7: Find the remainder of 498 divided by 25.

We will start the cycle so to find the remainder of either (–1) or (+1).

42 ÷ 25 leaves us a remainder of 16.

43 ÷ 25 leaves us a remainder of 14.

44 ÷ 25 leaves us a remainder of 6 (44 = 43 × 4 = 14 × 4 =56 ÷ 25 = remainder of 6)

45 ÷ 25 leaves us a remainder of 24 or (–1).

Since 45 divided by 25 leaves a remainder of (–1), then 410 when divided by 25 will leave a remainder of +1. So, the remainders are following a cycle of 10.

So, to find the remainder of 498, we need to divide 98 by 10 (since the cycle is of 10). The required term will be the 8th term (98  ÷  10,  remainder of  8  ) of the cycle. But we do not know the 8th term of the cycle, we just know the first 5 terms of the cycle.

So, to find the remainder of 48, we can break 48 into two parts as 45 × 43.

45 gave us a remainder of (–1) and 43 gave us a remainder of 14. So, 48 = 45 × 43 = (–1) × 14 = (–14). So, the remainder when 498 is divided by 25 comes out to be (–14), but we want a positive answer, final answer is (–14) + 25 = 11. (Use of negative remainder funda).

E.g. 8: A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

Let us assume that number is P and the divisor be X.

So, P = Xa + 24, (Where a is the quotient which can be any whole number). –(i)

Now, 2P = Xb + 11, (where b is the quotient which can be any whole number). –(ii)

If we multiply first equation by 2, we will get 2P = 2Xa + 48. –(iii)

Now the second equation and the third equation can be equated as both of them are equal to 2P.

On equating, we get Xb + 11 = 2Xa + 48. On rearranging,

X(b – 2a) = 37. Now, we have to find the value of the divisor which is X.

We came across such type of question in “number of factor” chapter. Whenever two natural numbers or integers are in product, it becomes an application of number of factors.

In this case also X and (b – 2a) are in product which is equal to 37. Now we should see that in how many ways 37 can be written as the product of two natural numbers.

Answer should be just 1 way as 37 is a prime number. So, only way is 1 × 37. But still one problem persists that which value X will take?

X cannot take 1 as remainder obtained when the number is divided by the divisor (X) is 24, so the divisor has to be greater than the remainder. We have already done this part in reverse logic that remainders have to be less than the divisors.

So, X has to be greater than 24, that solves our problem. So, the divisor‘s value is 37.

E.g. 9: Find the remainder when (1523 + 2323) is divided by 19.

There are two ways of solving it.

First Approach (Using negative remainder):

1523 ÷ 19 = 15 × 15 × 15 × 15 × 15 × …………….. × 1523 times. Every 15 divided by 19 leaves a remainder of 15 or (–4). So, our question reduces to finding the remainder of (–4)23 ÷ 19.

And 2323 ÷ 19 = 23 × 23 × 23 × 23 × 23 × …………….. × 2323 times. Every 23 divided by 19 leaves a remainder of 4. So, the question reduces to finding the remainder of 423 ÷ 19.

Thus, (1523 + 2323) can be written as [(–4)23 + 423] ÷ 19, the expression in the bracket gets cancelled out, so the remainder is 0.

Second Approach:

We would have come across this identity but would not have paid attention to the important deduction of it.

an + bn = always divisible by (a + b) if n is odd.

For e.g., a3 + b3 = (a + b) (a2ab + b2).

Similarly, a5 + b5 = (a + b) (a4 –…………..+ b4).

And also, a7 + b7 = (a + b) (a6 –…………+ b7). And so on.

All these terms like (a3 + b3), (a5 + b5), (a7 + b7) and so on are multiples of (a + b). That is why they are completely divisible by (a + b).

We can apply the same logic here as the power given in the question is same for both the numbers and is odd.

So, we can solve it orally like, (1523 + 2323) = (15 + 23) (1522 –……………..+2322).

(1523 + 2323) = 38 × (1522 –……………..+2322).

So, (1523 + 2323) is a multiple of 38, so it will be completely divisible by 38 or 19 or 2 or 1 (i.e. all the factors of 38), and the answer is 0.

## Exercise:

Find the remainder in each of the following:

1. 3250 ÷ 9.

2. 2498 ÷ 25.

3. 24199 ÷ 13.

4. 14907 ÷ 11.

5. 6994 ÷ 13.

6. 39150 ÷ 17.

7. 3193 ÷ 244.

8. 799 ÷ 31.

9. 5147 ÷ 7.

10. 25135 ÷ 17.

11. (35555 + 42222) ÷ 7.

12. (91 + 92 + 93 + …. + 914) is divided by 6.

13. (47101 + 23101) is divided by 70.

14. 2536 ÷ 601.

15. (1421 × 1423 × 1425 × 1427) ÷ 12.

16. A number when divided by 437 leaves a remainder of 200. So, find the remainder when the same number is divided by 23.

17. A number ‘Y’ when divided by 5 leaves remainder of 4. Find the remainder when (Y4 + Y3 + 2Y2 – 3Y + 2) is divided by 5.

18. 162000 ÷ 11.

19. 2494 ÷ 19.

20. 2699 ÷ 23.

1. 7                         2. 19                       3. 2                         4. 9                         5. 9                         6. 2

7. 27                       8. 8                         9. 4                         10. 15                    11. 0                       12. 0

13. 0                       14. 1                       15. 9                       16. 16                    17. 2                       18. 1

19. 17                    20. 1

## Explanations:

Question 1.

3250 ÷ 9 can be reduced to 550 ÷ 9. Now, we need to frame cycles for 3250 which divided by 9 leave a remainder of (–1) or (+1).

53 ÷ 9 leave remainder of (–1), then 56 will leave a remainder of (+1). So, remainders are following a cycle of 6. Now, we need to find 550 will be which term of the cycle.

50 divided by 6 leaves remainder of 2, that means 550 will be the second term of the cycle i.e. 52. So, 52 divided by 9 leave remainder of 7.

Question 2.

We have already seen cycle of 4n divided by 25.

45 divided by 25 leaves remainder of (–1), so 410 will finally leave remainder of (+1). If 410 leaves remainder of (+1) when divided by 25, then 220 will also remainder of (+1) when divided by 25. So, the cycle is of 20.

We need to find 2498 will be which term of the cycle. For that we need to divide 498 by 20 which leaves remainder of 18. That signifies that the remainder of 2498 will be same as that of 218. So, we need to find the remainder of 218 ÷ 25, since the dividend cannot be greater than divisor.

Now, 218 ÷ 25 can be further broken down into (210 × 28) ÷ 25.

210 leaves remainder of (–1) and 28 = 256 which divided by 25 leaves remainder of +6. Just substitute these values.

So, (210 × 28) ÷ 25 = [(–1) × 6] ÷ 25 = (–6) ÷ 25 = 19.

Question 3.

24199 ÷ 13 can be reduced to 11199 ÷ 13. Now, we make cycles.

111 ÷ 13 gives a remainder of 11.

112 ÷ 13 gives a remainder of 4.

113 ÷ 13 gives a remainder of 5 as (113 = 112 × 11 = 4 × 11 = 44 ÷ 13 gives remainder of 5).

114 ÷ 13 gives remainder of 3 as (114 = 113 × 11 = 5 × 11 = 55 ÷ 13 gives remainder of 3).

115 ÷ 13 gives remainder of 7 as (115 = 114 × 11 = 3 × 11 = 33 ÷ 13 gives remainder of 7).

116 ÷ 13 gives a remainder of 12 or –1 (116 = 115 × 11 = 7 × 11 = 77 ÷ 13 gives remainder of 12).

If 116 gives remainder of (–1), then 1112 will leave remainder of (+1). That means remainder are following a cycle of 12. Now, we need to find 11199 will be which term of the cycle. For that we will have to divide 199 by 12, which will leave remainder of 7. The remainder of 7 means that the 11199 is the seventh term of the cycle i.e. 117.

So, 117 ÷ 13 can be split further as [116 × 11] ÷ 13 = [(–1) × 11] ÷ 13 = [(–11)] ÷ 13 = 2.

So, final remainder is 2 when 11199 ÷ 13.

Question 4.

14907 ÷ 11 can be reduced to 3907 ÷ 11.

3n when divided by 11 leaves remainder of 3, 9, 5, 4, 1, 3, 9, 5, ….. as n takes values of 1, 2, 3, 4, 5, 6, 7, 8, ……

The cycle is of 5 and we need to find 3907 is which term of the cycle. 3907 is the 2nd term of the cycle as [907 ÷ 5 leaves remainder of 2].

So, remainder of 3907 ÷ 11 is same as that of 32 i.e. 9.

Question 5.

6994 ÷ 13 can be written in reduced form as 494 ÷ 13. Now, we can make cycles of 4 quickly as we know the powers of 4 orally.

43 ÷ 13 = 64 ÷ 13 leaves remainder of 12 or (–1). So, 46 will leave remainder of (+1). That means the remainder have a cycle of 6. So, 494 will be the 4th term of the cycle as (94 ÷ 6 = remainder of 4).

Therefore remainder of 494 ÷ 13 = 44 ÷ 13 = [43 × 41 ] ÷13 = [(–1) × 4] ÷ 13 = (–4) ÷ 13 =  remainder of 9.

Question 6.

39150 ÷ 17 can be reduced to 5150 ÷ 17. Now, we will frame cycles.

52 ÷ 17 = 25 ÷ 17 leave remainder of 8.

53 ÷ 17 = 40 ÷ 17 leave remainder of 6 as (53 = 52 × 5 = 8 × 5 = 40).

54 ÷ 17 = 30 ÷ 17 leave remainder of 13 as (54 = 53 × 5 = 6 × 5 = 30).

55 ÷ 17 = 65 ÷ 17 leave remainder of 14 as (55 = 54 × 5 = 13 × 5 = 65).

56 ÷ 17 = 70 ÷ 17 leave remainder of 2 as (56 = 55 × 5 = 14 × 5 = 70).

57 ÷ 17 = 10 ÷ 17 leave remainder of 10 as (57 = 56 × 5 = 2 × 5 = 10).

58 ÷ 17 = 50 ÷ 17 leave remainder of 16 or (–1) as (58 = 57 × 5 = 10 × 5 = 50).

If 58 divided by 17 leaves remainder of (–1), then 516 divided by 17 will leave remainder of (+1). Now, 5150 will be 6th term of the cycle as (150 ÷ 16 leaves remainder of 6).

So, the remainder of 5150 is the 6th term of the cycle which is 2.

Question 7.

The remainder of 3193 ÷ 244 can be calculated orally as we know the power of 3 orally.

35 = 243 and 243 divided by 244 will leave remainder of (–1), so 310 will leave remainder of +1. So, 3193 will be the 3rd term of the cycle which is 33.

So, remainder of 3193 ÷ 244 is same as that of 33 ÷ 244 which is 27.

Question 8.

To find the remainder of 799 ÷ 31, we need to frame cycles of 7n.

72 ÷ 31 leave remainder of 18.

73 ÷ 31 = 126 ÷ 31 leave remainder of 2 as (73 = 72 × 7 = 18 × 7 = 126).

74 ÷ 31 = 14 ÷ 31 leave remainder of 14 as (74 = 73 × 7 = 2 × 7 = 14).

75 ÷ 31 = 98 ÷ 31 leave remainder of 5 as (75 = 74 × 7 = 14 × 7 = 98).

76 ÷ 31 = 35 ÷ 31 leave remainder of 4 as (76 = 75 × 7 = 5 × 7= 35).

77 ÷ 31 = 28 ÷ 31 leave remainder of 28 as (77 = 76 × 7 = 4 × 7 = 28).

78 ÷ 31 = 196 ÷ 31 leave remainder of 10 as (78 = 77 × 7 = 28 × 7 = 196).

79 ÷ 31 = 70 ÷ 31 leave remainder of 8 as (79 = 78 × 7 = 10 × 7 = 70).

710 ÷ 31 = 56 ÷ 31 leave remainder of 25 as (710 = 79 × 7 = 8 × 7 = 56).

711 ÷ 31 = 175 ÷ 31 leave remainder of 20 as  (711 = 710 × 7 = 25 × 7 = 175).

712 ÷ 31 = 140 ÷ 31 leave remainder of 16 as (712 = 711 × 7 = 20 × 7 = 140).

713 ÷ 31 = 112 ÷ 31 leave remainder of 19 as (713 = 712 × 7 = 16 × 7 = 112).

714 ÷ 31 = 133 ÷ 31 leave remainder of 9 as (714 = 713 × 7 = 19 × 7 = 133).

715 ÷ 31 = 63 ÷ 31 leave remainder of 1 as (715 = 714 × 7 = 9 × 7 = 63).

So, remainders are following a cycle of 15 and 799 will be the 9th term of the cycle.

Therefore the remainder of 799 will be same as that of 79 ÷ 31 which we have already calculated as 8.

Question 9.

5147 ÷ 7 can be reduced as 247 ÷ 7. We can frame the cycles of 2 orally.

23 = 8 ÷ 7 will leave remainder of 1. So, 247 will be the 2nd term of cycle as remainders are having cycle of 3. (47 ÷ 3 = remainder of 2).

So, remainder of 247 ÷ 7 = 22 ÷ 7 = 4.

Question 10.

25135 ÷ 17 can be reduced to 8135 ÷ 17. Now, cycle of 8n can be formed as n takes values of 2, 3, 4, 5 and so on.

82 ÷ 17 = 64 ÷ 17 will leave remainder of 13.

83 ÷ 17 = 104 ÷ 17 will leave remainder of 2 as (83 = 82 × 8 = 13 × 8 = 104).

84 ÷ 17 = 16 ÷ 17 will leave remainder of 16 or (–1) as (84 = 83 × 8 = 2 × 8 = 16).

If 84 divided by 17 leaves remainder of (–1), then 88 divided by 17 will leave remainder of (+1). So, remainders have cycle of 8. 8135 will be 7th term of the cycle as 135 ÷ 8 = remainder of 7.

So, 87 ÷ 17 can be split into (84 × 83) ÷ 17 as we know the remainders of (84) and (83); [(–1) × 2] ÷ 17 = (–2) ÷ 17 =  final remainder of 15.

Question 11.

Remainder of (35555 + 42222) ÷ 7 can be found individually. We can rewrite this question as [(35555 ÷ 7) + (42222 ÷ 7)].

We will use the logic of remainders being added when the numbers are added.

So, first of all we will find the remainder of [(35555 ÷ 7)] and then add with the remainder of [(42222 ÷ 7)].

Remainder of 35555 ÷ 7 can be found orally as we know the powers of 3.

33 ÷ 7 will leave remainder of (6) or (–1), then 36 will give us remainder of (+1). That means remainders have a cycle of 6. Now, we need to find 35555 will be which term of the cycle and for that we need to divide 5555 by 6. 5555 divided by 6 gives us remainder of 5, that means 35555 will be the 5th term of the cycle.

So, final remainder is 35 ÷ 7 which can be split into [33 × 32] ÷ 7. 33 when divided by 7 gave remainder of (–1) and 32 when divided by 7 will give remainder of 2.

So, [33 × 32] ÷ 7 = [(–1) × 2] ÷ 7 = [(–2)] ÷ 7 = remainder of 5.

Similarly remainder of 42222 ÷ 7 can be found out orally as we know the powers of 4. The remainder will be 2 as remainders are following cycle of 3.

Finally we can add up the remainders which we obtained individually from [(35555 ÷ 7)] and [(42222 ÷ 7)]. On adding up the remainders we get 7 which can be further divided by 7 as remainder has to be less than the divisor.

Thus, the final remainder is 0.

Question 12.

Again we can solve (91 + 92 + 93 + …. + 914) is divided by 6 as the previous question.

9 divided by 6 gives a remainder of 3, 92 divided by 6 gives remainder of 3 and remaining terms divided by 6 will also leave remainders of 3 only.

So, we can rewrite this question as (3 + 3 + 3 + 3 + …………….+ 3 written 14 times) ÷ 6 = 42 ÷ 6 = remainder of 0.

Question 13.

We have solved this type of problem in solved examples.

(47101 + 23101) is divided by 70. On seeing this question, we observe that the powers are odd and are same for both the numbers.

So, we can write (47101 + 23101) as  = (47 + 23) (47100 –…………….+23100) = 70 × (47100 –…………….+23100).

This means that (47101 + 23101) is a multiple of 70 and any multiple of 70 divided by 70 will always leave remainder of 0.

We could have solved this problem by separating them.

(47101) ÷ 70 will leave remainder of [(–23)101] ÷ 70, this can further rewritten as [–23101] ÷ 70 (as a negative number raised to an odd power always results in negative number).

Now, the original question (47101 + 23101) ÷ 70 becomes [(–23101 + 23101)] ÷ 70. Both the terms in the bracket add up to zero, so the final remainder will be 0.

Question 14.

2536 ÷ 601 can be solved by framing cycles for 2536.

252 ÷ 601 = 625 ÷ 601 leaves remainder of 24.

253 ÷ 601 = 600 ÷ 601 leaves remainder of 600 or (–1) as [253 = 252 × 25 = 24 × 25 = 600].

So, if 253 divided by 601 leaves remainder of (–1), then 256 will give us remainder of +1. So, remainders are following cycle of 6.

Now, we need to find 2536 is which term of the cycle. 36 divided by 6 gives remainder of 0, that means 2536 will be the last term of the cycle. Remainder for the last term of the cycle is 1 as [256 ÷ 601 leaves remainder of 1].

Question 15.

(1421 × 1423 × 1425 × 1427) ÷ 12 can be solved by finding out remainders individually and then multiplying the remainders.

1421 divided by 12 leaves remainder of 5, similarly 1423, 1425 and 1427 divided by 12 leave remainder of 7, 9 and 11 respectively. Since the numbers are being multiplied, respective remainders will also be multiplied.

So, we can write (5 × 7 × 9 × 11) ÷ 12 = [ (35) × (99)] ÷ 12 = [(–1) × 3] ÷ 12 = (–3) ÷ 12 will give remainder of 9. {35 divided by 12 leaves remainder of (–1) and 99 divided by 12 leaves remainder of 3}.

Question 16.

Number can be framed as 437a + 200, where a can be any whole number.

Now, this number has to be divided by 23, but 437 is a multiple of 23. So, it will leave remainder of 0. Our answer is dependent on 200 ÷ 23.

So, final remainder is 16 (200 divided by 23 will give us remainder of 16).

Question 17.

Y divided by 5 leaves remainder of 4. This question can be solved either by assuming values or by the standard method. We will see both the methods.

Y = 5a + 4, where a is any whole number. So, we can assume value for Y as 4 when a = 0.

Now, the question asks for the remainder of (Y4 + Y3 + 2Y2 – 3Y + 2) is divided by 5.

Just substitute the value of Y = 4 in the equation and find the remainder.

(44 + 43 + 2 × 16 – 3 × 4 + 2) ÷ 5

Now, divide each of the terms by 5 and find their individual remainders and perform the same operation on remainders whichever is being done on the numbers.

So, remainders obtained are (1 + 4 + 2 – 2 + 2) ÷ 5 will give us remainder of 2.

Standard Method: Y = 5a + 4, where a is any whole number. Now plug the value of Y in the expression given in the question.

We get [(5a + 4)4 + (5a + 4)3 + 2(5a + 4)2 – 3(5a + 4) + 2] ÷ 5.

We will find out the remainders individually and add or subtract them depending on whichever operation is being performed.

(5a + 4) is a multiple of 5, so it will leave remainder of 4 when divided by 5. So, if (5a + 4) ÷ 5 gives us remainder of 4 then (5a + 4)4 ÷ 5 will give us remainder of 44. We can plug this value in the expression. Similarly we can find values of remainders for other terms also. We will get the same values which we got in the first method.

Question 18.

162000 ÷ 11 can be reduced to 52000 ÷ 11. Now we can frame cycles of remainder.

52 ÷ 11 = 25 ÷ 11 leave remainder of 3.

53 ÷ 11 = 15 ÷ 11 leave remainder of 4 as (53 = 52 × 5 = 3 × 5 = 15).

54 ÷ 11 = 20 ÷ 11 leave remainder of 9 as (54 = 53 × 5 = 4 × 5 = 20).

55 ÷ 11 = 45 ÷ 11 leave remainder of 1 as (55 = 54 × 5 = 9 × 5 = 45).

So, the remainders are following cycle of 5 and 52000 will be the 5th term of the cycle as (2000 divided by 5 leaves remainder of 0).

So, the remainder of 52000 ÷ 11 is the fifth term which is 1.

Question 19.

2494 ÷ 19 can be reduced as 594 ÷ 19. Now, we can frame cycles of 594.

52 ÷ 19 = 25 ÷ 19 will leave remainder of 6.

53 ÷ 19 = 30 ÷ 19 will leave remainder of 11 as (53 = 52 × 5 = 6 × 5 = 30).

54 ÷ 19 = 55 ÷ 19 will leave remainder of 17 as (54 = 53 × 5 = 11 × 5 = 55).

55 ÷ 19 = 85 ÷ 19 will leave remainder of 9 as (55 = 54 × 5 = 17 × 5 = 85).

56 ÷ 19 = 45 ÷ 19 will leave remainder of 7 as (56 = 55 × 5 = 9 × 5 = 45).

57 ÷ 19 = 35 ÷ 19 will leave remainder of 16 as (57 = 56 × 5 = 7 × 5 = 35).

58 ÷ 19 = 80 ÷ 19 will leave remainder of 4 as (58 = 57 × 5 = 16 × 5 = 80).

59 ÷ 19 = 20 ÷ 19 will leave remainder of 1 as (59 = 58 × 5 = 4 × 5 = 20).

So, the remainders are following a cycle of 9 and 594 will be the 4th term of the cycle which is 17.

Question 20.

2699 ÷ 23 can be further reduced to 399 ÷ 23. We will frame cycles for 399.

33 ÷ 23 = 27 ÷ 23 will leave remainder of 4.

34 ÷ 23 = 12 ÷ 23 will leave remainder of 12 as (34 = 33 × 3 = 4 × 3 = 12).

35 ÷ 23 = 36 ÷ 23 will leave remainder of 13 as (35 = 34 × 3 = 12 × 3 = 36).

36 ÷ 23 = 39 ÷ 23 will leave remainder of 16 as (36 = 35 × 3 = 13 × 3 = 39).

37 ÷ 23 = 48 ÷ 23 will leave remainder of 2 as (37 = 36 × 3 = 16 × 3 = 48).

38 ÷ 23 = 6 ÷ 23 will leave remainder of 6 as (38 = 37 × 3 = 2 × 3 = 6).

39 ÷ 23 = 18 ÷ 23 will leave remainder of 18 as (39 = 38 × 3 = 6 × 3 = 18).

310 ÷ 23 = 54 ÷ 23 will leave remainder of 8 as (310 = 39 × 3 = 18 × 3 = 54).

311 ÷ 23 = 24 ÷ 23 will leave remainder of 1 as (311 = 310 × 3 = 8 × 3 = 24).

So, remainders are following cycle of 11 and 399 will be the 11th term of the cycle or the last term of the cycle and the remainder will be 1 as (99 divided by 11 leave remainder of 0 which means the last term of the cycle).

Hence, the remainder is 1.

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### 66 Responses to Remainders (Part I)

1. Siddhi Satpute says:

1.7
2.19
3.2
4.9
5.9
6.2
7.27
8.8
9.4
10.15
11.0
12.0
13.0
14.1
15.9
16.16
17.2
18.1
19.17
20.1

2. vinal_gandhi says:

1.7
2.19
3.2
4.9
5.9
6.2
7.27
8.8
9.4
10.15
11.0
12.0
13.0
14.1
15.9
16.16
17.2
18.1
19.17
20.1

3. Pals says:

1. 7
2. 19
3. 2
4. 9
5. 9
6. 2
7. 27
8. 8
9. 4
10. 15
11. 0
12. 0
13. 0
14. 1
15. 9
16. 16
17. 2
18. 1
19. 17
20. 1

4. Pranati says:

1.7
2.19
3.2
4.9
5.9
6.2
7.27
8.8
9.4
10.15
11.0
12.0
13.0
14.1
15.9
16.16
17.2
18.1
19.17
20.1

5. Shifa says:

done..

6. shreyd says:

1.7
2.19
3.2
4.9
5.9
6.2
7.27
8.8
9.4
10.15
11.0
12.0
13.0
14.1
15.9
16.16
17.2
18.1
19.17
20.1

7. tanu.agrawal says:

done

8. Pranay says:

done.

9. ganesh says:

Completed..

10. Vinayak says:

done

done

done….completed

13. Aishwarya says:

done

14. saket93 says:

done..!!!

15. pulkitmittal says:

done

16. Amol Pawar says:

done.