Composite Index:
E.g. 1: Find the remainder of (1822)10 divided by 7.
If we had to find the remainder of this question, it could have been solved with the logic learnt in the previous chapter, because we can rewrite this question as 18220 ÷ 7. We know to proceed ahead.
Now, this question can be solved based on the logic of the cycles learnt in the remainder chapter. We just have to assume the index to a variable as the index 2210 is a very large number whose value calculation is not feasible. And one more difference in this question with the questions done in the remainder chapter is that we will have to keep dividing the index by the divisor until it becomes smaller than the divisor.
Let n = 2210, so our question boils down to 18n ÷ 7.
Now, again we can reduce our question to 4n ÷ 7.
Now, we will have to find such cycle of 4n which when divided by 7 leaves us a remainder of either (+1) or (–1).
41 ÷ 7 = 4 ÷ 7 leaves a remainder of 4.
42 ÷ 7 = 16 ÷ 7 leaves a remainder of 2.
43 ÷ 7 = 8 ÷ 7 leaves a remainder of 1. (as 43 = 42 × 41 = 2 × 4 = 8 )
So, the remainders are following a cycle of 3.
Now, if the question was asking to find the remainder of 4100 ÷ 7, we were required to find the remainder of the 100th term. So, we would have divided 100 by 3 (as cycle is of 3). But in this question we are required to find the remainder of 2210 th term.
So, we will divide n = 2210 by 3 or 2210 ÷ 3.
2210 ÷ 3 can be reduced to 110 ÷ 3 = 1 ÷ 3 = remainder of 1.
Remainder of 1 means the answer will be the 1st term of the cycle. (Try to relate the logic with finding the remainder of 4100 ÷ 7; the cycle was of 3, so we would have divided 100 by 3 which would have given us a remainder of 1, which means that final remainder will be the first term of the cycle. So, answer would have been 4 only).
So, answer is 4.
So, the logic still remains the same, only difference is that the index here is pretty large, so we will have to keep dividing it until it becomes lesser than divisor. Let us see some more examples.
E.g. 2: What will be the remainder when 19n ÷ 16, where n = 2334 . .
Before finding the remainder the question can be further reduced to 3n ÷ 16, where n = 2334. Now, we will make cycle which leaves a remainder of (+1) or (–1).
We will start our cycle from the third term which is 33 as 31 and 32 are smaller than 16 and dividing 31 and 32 by 16 will yield the same values.
So, 33 ÷ 16 leaves a remainder of 11.
And 34 ÷ 16 = 33 ÷ 16 leaves a remainder of 1 (as 34 = 33 × 31 = 11 × 31 = 33).
So, the remainders are following a cycle of 4, but we need to find n = 2334th term, which means we will have to divide n by 4.
Now, n ÷ 4 = 2334 ÷ 4 = 334 ÷ 4 = (–1)34 ÷ 4 = 1 ÷ 4 =1 (is the remainder).
1 is the remainder means the answer is the first term of the cycle which is 31.
So, finally answer is 3.
E.g. 3: Find the remainder of 32n ÷ 9 where n = 3732.
We will reduce the question which becomes 5n ÷ 9. Now, we will frame the cycle.
52 ÷ 9 = 25 ÷ 9 leaves remainder of 7
53 ÷ 9 = 35 ÷ 9 leaves remainder of (–1) as (53 = 52 × 5 = 7× 5 = 35) .
If 53 leaves a remainder of (–1), then 56 will leave remainder of (+1). That means that remainders are having a cycle of 6.
So, our next step should be to find nth term which is 3732th term and for that we will have to divide n by 6 as cycle is of 6.
So, n ÷ 6 or 3732 ÷ 6 = 132 ÷ 6 = (a remainder of 1).
A remainder of 1 means the answer is the first term of the cycle which is 51. So, finally answer is 5.
E.g. 4: Find the remainder of 14n ÷ 11 where n = 1722 .
Again we will reduce the sum to 3n ÷ 11 where n = 1722. Now, we construct the cycle which satisfies our requirement. We should start our cycle with 33 as 31 and 32 divided by 11 will result in the same value.
33 ÷ 11 = 27 ÷ 11 leaves a remainder of 5.
34 ÷ 11 = 15 ÷ 11 leaves a remainder of 4 as (34 = 33 × 31 = 5 × 3 = 15).
35 ÷ 11 = 12 ÷ 11 leaves a remainder of 1 as (35 = 34 × 31 = 4 × 3 = 12).
So, remainders are following a cycle of 5.
So, we need to find 1722 will be which term of the cycle? For that we need to divide 1722 by 5.
1722 ÷ 5 = 222 ÷ 5. In all the three problems which we did, we got the answer in this step. But in this question, we will have to do one more step as the remainder 222 is greater than 5. In the previous questions, we were getting either 1 some power or (–1)some power. But, here we will find the remainder of 222 divided by 5.
For finding the remainder of 222 ÷ 5, we will have to again frame the cycle again.
We know powers of 2 orally, so we can frame the cycles quickly.
24 ÷ 5 gives a remainder of 1. So remainder are having a cycle of 4. So, we need to divide 22 by 4 which gives us that remainder of 222 will be the second term of the cycle which is 22.
Now, we got the remainder of 1722 as 4 which means that final remainder of 3n will be the fourth term of the cycle.
Fourth term of the cycle was 34 which divided by 11 gave us remainder of 4.
So, in this question we need to frame cycles two times since the remainder obtained in the first stage is greater than the divisor which is not possible.
E.g. 5: Find the remainder of 38n ÷ 11, where n = 3339 .
We reduce the question to 5n ÷ 11, where n = 3339.
Now, 52 ÷ 11 leaves a remainder of 3.
53 ÷ 11 leaves a remainder of 4 as (53 = 52 × 5 = 3 × 5 = 15 ÷ 11 = 4)
54 ÷ 11 leaves a remainder of 20 ÷ 11 which is equal to 9.
55 ÷ 11 leaves a remainder of 45 ÷ 11 which is equal to 1.
The remainders have a cycle of 5, so now we need to find 3339 is which term of the cycle?
So, now we have to find the remainder of 3339 ÷ 5, which can be reduced as 339 ÷ 5.
In this problem also, we will again have to frame the cycle again as the remainder obtained 339 is greater than the divisor i.e. 5.
Now cycle of 339 which on divided by 5 leaves a remainder of (–1) or (+1).
32 ÷ 5 leaves a remainder of 4 or (–1). If 32 leaves a remainder of (–1), then 34 will leave a remainder of (+1). So, the remainder are following a cycle of 4, so the remainder of 339 will be [39 divided by 4; a remainder of 3] the third term of the cycle. So, the remainder of 339 is 33 which is 27. 27 can be further divided by 5, so finally remainder of 2.
So, remainder of 339 is 2 and it will be same for 3339.
That means the remainder of 5n will be the 2nd term of 5n, which happens to be 3.
So, final answer is 3.
E.g. 6: Find the remainder of 31n ÷ 9, where n = 3131 .
31n divided by 9 leaves remainder of 4n.
4n when divided by 9, when n takes values starting from n = 1, 2, 3,4, …….. will leave remainders 4, 7, 1, 4, ……… So, remainders have a cycle of 3.
Now, we need to find 3131th term of the cycle, which will be the term equal to the remainder when 3131 is divided by 3.
Now, 3131 ÷ 3 will leave a remainder of 131 ÷ 3 , which means a final remainder of 1. That means 3131th term will be the 1st term of the series which is 4.
This problem is different from the previous two as we got a remainder of 1 some power which made our work easy. If we would have got 2 some power or 3 some power or 4 some power except 1 some power, we would have to find the remainder once again as (2, 3, 4) some power would had been greater than the divisor. And we have learnt that the remainder is always smaller than the divisor.
Finding the Remainder when Dividend and Divisor have some common factor:
What is the remainder when we divide 30 by 8?
The correct answer is 6, but many of us will say that (30 ÷ 8 ) can be reduced to (15 ÷ 4) which will give us remainder of 3. This method also seems correct, but the answer is not matching.
What is the flaw?
We can relate it to the funda of grouping. The original question asked us to divide the number 30 by 8, and whenever we are dividing by 8, we are making groups of 8. But in the alternative method, we cancelled 2 from dividend and divisor and in the process we are dividing by 4. When we are dividing by 4, we are making groups of 4 and not of 8. That’s why the answer is not matching.
So, if we want to solve a remainder problem by cancelling out the common terms, after finding out the remainder we need to multiply the remainder obtained with the term which we cancelled earlier. Then only we will get the correct answer.
In this case also, to get the right answer, we need to multiply the remainder obtained from (15 ÷ 4) with 2 to get the correct answer which is 6.
E.g. 7: Find the remainder of 17193 divided by 51?
As soon as we see this question, we should observe that divisor and dividend has some common factor. We can solve orally if we cancel that common term.
We can write [17 × 17192] ÷ [17 × 3].
Common term is 17, so the question boils down to finding the remainder of 17192 ÷ 3.
Every 17 divided by 3 gives a remainder of (+2) or (–1).
Again the question reduces to (–1)192 ÷ 3, which gives a remainder of 1 as (–1)even number = 1.
So, the remainder obtained is 1, but do not forget to multiply 17 to the remainder found.
So, final answer is 17.
E.g. 8: Find the remainder when 4100 is divided by 100?
The dividend and the divisor have 4 in common, so we can cancel them, the question reduces to finding out the remainder of 499 by 25.
45 = 1024 (we should be well-conversed with the powers of 2 up till 210 as they can make calculations ridiculously easy) which when divided by 25 leaves a remainder of 24 or (–1). If 45 leaves a remainder of (–1), then 410 will leave a remainder of (+1). That means the remainders have a cycle of 10.
We need to find 499 will be which term of the cycle? For that we need to divide 99 by 10. We get remainder as 9. That means 499 will be the 9th term of the cycle.
But, we know remainder up till 45 only, and we have to find the value of 49 when divided by 25. We can split 49 as 45 × 44. 45 leaves a remainder of (–1) when divided by 25 and 44 divided by 25 leaves remainder of 6.
So, remainder of 49 = 45 × 44 = (–1) × 6 = (–6).
Remainder of ( –6) when divided by 25 means a positive remainder of 19.
Again do not forget to multiply 4 to the remainder obtained from 499.
So, final answer is not 4 but 76.
E.g. 9: Find the remainder of 65n ÷ 39 where n = 3980 .
This problem is of composite index, so we need to assume the index 3980 as n. So, our question reduces to 65n ÷ 39. Now, we can observe that the dividend and the divisor have 13 as the common factor. We need to cancel them.
We can write the question as [65 × 65n –1] ÷ (13 × 3). If we cancel out 13, we get [5 × 65n –1] ÷ 3.
5 divided by 3 leaves a remainder of 2 and [65n –1] divided by 3 leaves a remainder of (–1) n –1. Whenever the remainder comes as (–1), our objective is to find whether the power is odd/even.
So, we need to determine whether n – 1 is odd/even.
So, n – 1 = 3980 – 1 =odd – odd = even, (as 3980 is an odd number). So, the power is even.
Now, the final remainder of [5 × 65n –1] ÷ 3 is [2 × (–1)even power] = 2 × 1 = 2. Again, do not forget to multiply the common term (i.e. 13) to the remainder obtained from [5 × 65n –1] ÷ 3.
Thus, the final answer is not 2 but 26.
E.g. 10: Find the remainder of 85n ÷ 51, where n = 1953 .
Again, this problem is of composite index, we will assume the index1953 as n. Now, we have to find the remainder of 85n ÷ 51. The divisor and the dividend have 17 as the common factor, so we need to cancel them. After cancelling the question further reduces to
[5 × (85n – 1)] ÷ 3.
5 divided by 3 leaves a remainder of 2 and (85n – 1) divided by 3 gives a remainder of 1. In this question we are not bothered about whether the index (n – 1) is even or odd as (+1)even/odd is always 1.
So, the remainder from [5 × (85n – 1)] ÷ 3 is 2 (1 × 2).
But, the final answer will be 17 multiplied with 2 as 17 was the common term which we cancelled.
So, answer is 34.
To find the unit’s place, last two digits, last three digits of a number:
We have already learnt about finding the unit’s place of a digit in the Cyclicity chapter. There is an alternate method for finding the unit’s place of a number.
To find the unit’s place, divide the number by 10 and the remainder obtained is the digit at the unit’s place.
To find the last two digits of a number, divide the number by 100 and the remainder obtained will be the last two digits of that number. This method is particularly handy if the number has a common factor of 2, 4, 6, 8 or 5. Because, if the number has common factor of 2, 4, 6, 8 or 5, then when dividing by 100, the common term will get cancelled out and the remainder can be found very easily. Apart from this method, we will also learn “Binomial Theorem” to find the last two digits.
Similarly, we can extrapolate the logic for finding the last three digits of a number, we just need to divide the number by 1000 and the remainder obtained will be the answer.
Why does this logic work?
The explanation is pretty simple. Let’s say we want to find the unit’s digit of 78.
You will say the 8 is the unit’s digit of 78. Yes, but what is the underlying logic behind it?
78 = 7 × 10 + 8 × 1. When we divide this number by 10, the first part which is 7 × 10 gives remainder of 0, but the second part which is 8 will leave remainder of 8. So, whenever we divide a number by 10, the ten’s place, hundred’s place, thousand’s place and so on will leave a remainder of 0 since they have place values of 10, 100, 1000 and so on. So, the remainder will be the digit at the unit’s place since it has a place value of 1.
Same logic also follows for finding the last two digits and last three digits.
E.g. 11: Find the unit’s place of 235.
First Method (Cyclicity):
We will divide the power 35 by 4 as the digit 2 has a cyclicity of 4. We get a remainder of 3, it means that unit’s place of 235 will be same as third term of the cycle i.e. the unit’s place of 23 which is 8.
Second Method:
We will find the remainder of 235 divided by 10 that will give the unit’s place of 235.
235 ÷ 10 has a common term of 2 in both the dividend and the divisor. Let us cancel it first. So, now the question reduces to finding the unit’s place of 234 divided by 5. Now, we will frame the cycle.
The remainder obtained when 2n is divided by 5 when n takes the values 1, 2, 3, 4 is respectively 2, 4, 3 and 1. That means the remainders have a cycle of 4.
So, the unit’s place of 234 will be same as that of 2nd term of the cycle i.e. 22 which is 4. (We divided 34 by 4 as cycle is of 4). But, the question is asking to find the unit’s place of 235 and not of 234. So, we will have to multiply the common term which we cancelled in the divided and the divisor.
So, the remainder is 4 × 2 = 8, and 8 is the digit at the unit’s place of 235.
E.g. 12: Find the digit at the ten’s place of 1580.
If we have to find the digit at the ten’s place of 1580, we need to find the last two digits of 1580, that will solve our purpose.
To find the last two digits of 1580, we need to divide 1580 by 100 and find its remainder.
The dividend and the divisor have 25 as the common factor which we will have to cancel first. So, 1580 ÷ 100 can be rewritten as (152 × 1578) ÷ (52 × 4).
On cancelling, we get [9 × 1578] ÷ 4.
9 divided by 4 leaves a remainder of 1.
1578 divided by 4 leaves a remainder of (–1)78 which is equal to 1.
So, the overall remainder of [9 × 1578] ÷ 4 is 1.
So, the remainder of 1580 ÷ 100 is not 1 but 25 as we cancelled out 25 from the dividend and the divisor.
Thus the last two digits of 1580 are 25 and the digit at the ten’s place is 2.
Exercise:
Find the remainder in each of the cases:
1. 39n ÷ 8 where n = 9980.
2. 98n ÷ 11 where n = 9898 .
3. 18n ÷ 13 where n = 3332.
4. 37n ÷ 13 where n = 2310 .
5. 45n ÷ 17 where n = 4779.
6. 37n ÷ 15 where n = 4880.
7. 45n ÷ 60 where n = 2360.
8. 72n ÷ 54 where n = 3060.
9. 76n ÷ 95 where n = 1520.
10. 25n ÷ 9 where n = 1520.
11. 18n ÷ 11 where n = 9798.
12. 25n ÷ 15 where n = 9798.
13. Find the last two digits of 4900.
14. Find the digit at the ten’s place of 16100.
15. Find the digit at the ten’s place of 35200.
16. Find the last two digits of 28300.
Answer key
1. 7 2. 1 3. 5 4. 11 5. 14 6. 1 7. 45 8. 0 9. 76 10. 1 11. 8 12. 10 13. 76 14. 7 15. 2 16. 68
Explanations:
Question 1:
We have to find the remainder of 39n ÷ 8 where n = 9980.
The question can be reduced as 7n ÷ 8 or (–1)n ÷ 8. When we get the remainder as (–1), our concern should be to find out whether the power is odd/even. The power given is 9980 which is an odd number. So, n is odd.
So, our question becomes (–1)odd power ÷ 8 = (–1) ÷ 8 = Remainder of 7.
Question 2:
98n ÷ 11 where n = 9898, 98 divided by 11 leaves remainder of (–1) and again our concern is whether the power is odd or even. Power is n = 9898, which is an even number.
So, the question becomes (–1)even power ÷ 11 = Remainder of 1.
Question 3:
18n ÷ 13 where n = 3332. The question can be reduced as 5n ÷ 13. Now, we need to frame cycles for 5n as n takes values 1, 2, 3, ……
When n takes values 1, 2, 3, 4………, then remainders take the values 5, (–1), 8, 1, and so on. Remainders are having a cycle of 4, so now we need to find 5n is which term of the cycle. For that we will need to divide the power 3332 by 4.
3332 ÷ 4 leaves remainder of 132 i.e. 1. So, remainder of 5n is the first term of the cycle.
Final remainder is 5.
Question 4:
37n ÷ 13 where n = 2310 can be reduced to 11n ÷ 13. Now, we will have to frame cycles.
112 ÷ 13 = 121 ÷ 13 leaves remainder of 4.
113 ÷ 13 = 44 ÷ 13 leaves remainder of 5 as (113 = 112 × 11 = 4 × 11 = 44).
114 ÷ 13 = 55 ÷ 13 leaves remainder of 3 (114 = 113 × 11 = 5 × 11 = 55).
115 ÷ 13 = 33 ÷ 13 leaves remainder of 7 (115 = 114 × 11 = 3 × 11 = 33).
116 ÷ 13 = 77 ÷ 13 leaves remainder of 12 or (–1) as (116 = 115 × 11 = 7 × 11 = 77).
If 116 divided by 13 leaves remainder of (–1), then 1112 divided by 13 will definitely leave remainder of (+1). That means remainder have cycle of 12.
Now, we need to find 11n is which term of the cycle, for that we will have to divide n by 12.
So, n ÷ 12 can be written as 2310 ÷ 12 = (–1)10 ÷ 12 = Remainder of 1. Remainder of 1 means 11n is the first term of the cycle.
So, 111 = 11 is the final remainder.
Question 5:
45n ÷ 17 where n = 4779 can be reduced to 11n ÷ 17 for which we will have to frame cycles.
112 ÷ 17 = 121 ÷ 17 leaves remainder of 2.
113 ÷ 17 = 22 ÷ 17 leaves remainder of 5 as (113 = 112 × 11 = 2 × 11 = 22).
114 ÷ 17 = 55 ÷ 17 leaves remainder of 4 as (114 = 113 × 11 = 5 × 11 = 55).
115 ÷ 17 = 44 ÷ 17 leaves remainder of 10 as (115 = 114 × 11 = 4 × 11 = 44).
116 ÷ 17 = 110 ÷ 17 leaves remainder of 8 as (116 = 115 × 11 = 10 × 11 = 110).
117 ÷ 17 = 88 ÷ 17 leaves remainder of 3 as (117 = 116 × 11 = 8 × 11 = 88). .
118 ÷ 17 = 33 ÷ 17 leaves remainder of 16 or (–1) as (118 = 117 × 11 = 3 × 11 = 33).
If 118 divided by 17 leaves remainder of (–1), then 1116 will leave remainder of +1. So, remainders are having a cycle of 16. Now, we need to find 11n will be which term of the cycle and so n will have to be divided by 16.
So, n ÷ 16 = 4779 ÷ 16 = (–1)79 ÷ 16 = (–1) ÷ 16 = remainder of 15.
Remainder of 15 means 11n is the 15th term of the cycle. So, final remainder is 1115 ÷ 17.
1115 ÷ 17 can be further split into [118 × 117] ÷ 17 = [(–1) × 3] ÷ 17 = (–3) ÷ 17 = remainder of 14.
So, answer is 14.
Question 6:
37n ÷ 15 where n = 4880 can be reduced to 7n ÷ 15.
72 ÷ 15 = 49 ÷ 15 leaves remainder of 4.
73 ÷ 15 = 28 ÷ 15 leaves remainder of 13.
74 ÷ 15 = 91 ÷ 15 leaves remainder of 1.
So, remainders are following cycle of 4 and we need to know 7n is which term of the cycle. For that we will have to divide n by 4.
So, n ÷ 4 = 4880 ÷ 4 = Remainder of 0.
That means 7n will be the last term of the cycle which is the 4th term. So, final remainder is 74 ÷ 15 which is 1.
Question 7:
45n ÷ 60 where n = 2360.
On observing the question, we can say that the divisor and the dividend have something in common. So, that common factor will have to be cancelled first, after that we can proceed in our standard way.
We can write the question as [45 × 45n – 1] ÷ 60; 15 is the common factor and after cancelling 15 we get [3 × 45n – 1] ÷ 4.
45 divided by 4 leaves remainder of 1 and 1 raised to any power is always 1.
So, the remainder obtained is [3 × 1] = 3. But the final remainder will be 45 as we cancelled out 15, so we will have to multiply it back to the remainder obtained i.e. {15 × 3 = 45}.
Question 8:
72n ÷ 54 where n = 3060.
Again some factors are common and on cancelling them we get, [72 × 72n – 1] ÷ 54.
[4 × 72n – 1] ÷ 3.
Now, 72 any power divided by 3 will leave remainder of 0 as 72 is a multiple of 3. So, final remainder is 0 as 0 multiplied with any number is 0.
Question 9:
76n ÷ 95 where n = 1520.
We can re-write the question as [76 × 76n – 1] ÷ 95. 19 is the common term.
On cancelling, we get [4 × 76n – 1] ÷ 5.
Now, 76any power divided by 5 will leave remainder of 1. So, overall remainder is {4 × 1} = 4.
But, the final remainder will be 4 × 19 = 76 as we cancelled 19 as the common term.
Question 10:
25n ÷ 9 where n = 1520. We can reduce this problem to 7n ÷ 9 and make cycles of it.
72 ÷ 9 = 49 ÷ 9 leaves remainder of 4.
73 ÷ 9 = 28 ÷ 9 leaves remainder of 1 as (73 = 72 × 7 = 4 × 7 = 28).
So, remainders have cycle of 3, now we need to find 7n is which term of the cycle for which n needs to be divided by 3.
So, n ÷ 3 = 1520 ÷ 3 = Remainder of 0 means 7n will be the last term of the cycle as all the cycles are complete.
So, final remainder is 73 ÷ 9 = 1.
Question 11:
18n ÷ 11 where n = 9798.
Again this question can be reduced to 7n ÷ 11 and for which we will have to make cycles.
72 ÷ 11 = 49 ÷ 11 leaves remainder of 5.
73 ÷ 11 = 35 ÷ 11 leaves remainder of 2 as (73 = 72 × 7 = 5 × 7 = 35).
74 ÷ 11 = 14 ÷ 11 leaves remainder of 3 as (74 = 73 × 7 = 2 × 7 = 14).
75 ÷ 11 = 21 ÷ 11 leaves remainder of 10 or (–1) as (75 = 74 × 7 = 3 × 7 = 21).
If 75 divided by 11 leaves remainder of (–1), then 710 divided by 11 will leave remainder of (+1). That means the cycle is of 10 and to find out 7n is which term of the cycle, we will have to divide n by 10.
So, n ÷ 10 = 9798 ÷ 10 = 798 ÷ 10. Since 798 is greater than the divisor 10, so we will have to frame cycles again for 798.
72 ÷ 10 = 49 ÷ 10 leaves remainder of (–1), that means 74 ÷ 10 will leave remainder of 1. Now, we will have to divide the power 98 by 4 to find out which term is 798 of the cycle.
So, 98 divided by 4 leaves remainder of 2, that means 798 is the 2nd term of the cycle.
So, remainder of 798 is 72 ÷ 10 = 9.
So, remainder of 9 means that 7n was the 9th term of the original cycle.
So, to find the original remainder we will have to divide 79 by 11.
Remainder of 79 ÷ 11 = [75 × 74] ÷ 11 = [(–1) × 3] ÷ 11 = (–3) ÷ 11 leaves remainder of 8.
So, answer is 8.
Question 12:
25n ÷ 15 where n = 9798.
Numerator and denominator have 5 as the common factor which we will have to cancel first.
[25 × 25n – 1] ÷ 15, after reducing we get [5 × 25n – 1] ÷ 3.
25 any power divided by 3 will always leave remainder of 1 and 5 divided by 3 will leave remainder of 2. So, overall remainder of [5 × 25n – 1] ÷ 3 = [2 × 1] = 2.
But, the final remainder will be 10 as we have to multiply 5 with the remainder obtained i.e. 2.
Question 13:
We have learnt to find the last two digits, we need to divide 4900 by 100.
So, 4900 ÷ 100 can be reduced to 4899 ÷ 25 as 4 is the common factor in dividend and the divisor.
So, now to find the remainder of 4899 ÷ 25, we will have to frame the cycles.
We know 45 ÷ 25 leaves remainder of (–1), then 410 divided by 25 will leave remainder of +1. So, remainders have a cycle of 10. We need to find out 4899 will be which term of the cycle and for that we will have to divide 899 by 10.
So, 899 divided by 10 leaves remainder of 9 which means 4899 is the 9th term of the cycle.
So, remainder of 4899 ÷ 25 = 49 ÷ 25 which can be further split into [45 × 44] ÷ 25. 45 divided by 25 leaves remainder of (–1) and 44 divided by 25 leaves remainder of 6.
So, overall remainder of [45 × 44] ÷ 25 = [(–1) × 6] ÷ 25 = (–6) ÷ 25 = remainder of 19.
But, the final remainder is 19 × 4 = 76 (as we cancelled 4).
So, the last two digits are 76.
Question 14:
We will divide 16100 by 100 to find the last two digits.
16100 ÷ 100 can be reduced to [16 × 1699] ÷ 100 = [4 × 1699] ÷ 25. We need to frame the cycles for 1699 divided by 25.
162 ÷ 25 = 256 ÷ 25 leaves remainder of 6.
163 ÷ 25 = 96 ÷ 25 leaves remainder of 21 as (163 = 162 × 16 = 6 × 16 = 96).
164 ÷ 25 = 336 ÷ 25 leaves remainder of 11 as (164 = 163 × 16 = 21 × 16 = 336).
165 ÷ 25 = 176 ÷ 25 leaves remainder of 1 as (165 = 164 × 16 = 11 × 16 = 176).
So, remainders are following cycle of 5, we need to find 1699 is which term of the cycle. We divide 99 by 5 which gives remainder of 4. We can conclude that 1699 is the fourth term of the cycle.
So, remainder of 1699 ÷ 25 = 164 ÷ 25 = 11.
So, overall remainder of [4 × 1699] ÷ 25 = [ 4 × 11] ÷ 25 = 44 ÷ 25 = 19.
So, final remainder of 16100 ÷ 100 is [19 × 4] = 76 as we cancelled out 4.
So, last two digits of the number given is 76 and the digit at the ten’s place is 7.
Question 15:
To find the digit at the ten’s place of 35200, we will have to find out the last two digits first. And for finding last two digits, we will need to divide 35200 by 100.
35200 ÷ 100 can be reduced to [35 × 35 × 35198] ÷ 100 = [7 × 7 × 35198] ÷ 4 = [49 × 35198] ÷ 4 as 25 is common factor in the dividend and the divisor.
Now, 49 divided by 4 leaves remainder of 1 and 35198 divided by 4 leaves remainder of (–1)198 which is equivalent to 1.
So, overall remainder of [49 × 35198] ÷ 4 = [1 × 1] ÷ 25 = remainder of 1. But, the final remainder is 25 as 25 was cancelled as the common term.
So, last two digits are 25 and the digit at the ten’s place is 2.
Question 16:
Last two digits of 28300 can be found out by dividing the number by 100.
So, 28300 ÷ 100 can be reduced to [28 × 28299] ÷ 100 = [7 × 28299] ÷ 25 as {4 is common factor in dividend and the divisor}.
28299 divided by 25 can be further reduced to 3299 ÷ 25. Now, we will have to frame cycles for 3299.
33 ÷ 25 leaves remainder of 2.
34 ÷ 25 = 6 ÷ 25 leaves remainder of 6 as [34 = 33 × 3 = 2 × 3 = 6].
35 ÷ 25 = 18 ÷ 25 leaves remainder of 18 as [35 = 34 × 3 = 6 × 3 = 18].
36 ÷ 25 = 54 ÷ 25 leaves remainder of 4 as [36 = 35× 3 = 18 × 3 = 54].
37 ÷ 25 = 12 ÷ 25 leaves remainder of 12 as [37 = 36 × 3 = 4 × 3 = 12].
38 ÷ 25 = 36 ÷ 25 leaves remainder of 11 as [38 = 37 × 3 = 12 × 3 = 36].
39 ÷ 25 = 33 ÷ 25 leaves remainder of 8 as [39 = 38 × 3 = 11 × 3 = 33].
310 ÷ 25 = 24 ÷ 25 leaves remainder of 24 or (–1) as [310 = 39 × 3 = 8 × 3 = 24].
If 310 ÷ 25 leaves remainder of (–1), then 320 will leave remainder of +1. So, the remainders are following cycle of 20 and we need to find out 3299 is which term of the cycle. For that we will have to divide 299 by 20 which will give remainder of 19. So, we can conclude that 3299 is the 19th term of the cycle.
So, 319 ÷ 25 can be split into [310 × 39] ÷ 25.
310 divided by 25 leaves remainder of (–1) and 39 divided by 25 leaves remainder of 8.
So, overall remainder of 319 ÷ 25 = [(–1) × 8] ÷ 25 = (–8) ÷ 25 = remainder of 17.
But, the final remainder will be 17 × 4 = 68 as 4 was the common term cancelled.
So, the last two digits of 28300 is 68.