REMAINDERS

 Question 1:

What is the remainder when the product 1998 × 1999 × 2000 is divided by 7?

[1] 0

[2] 1

[3] 2

[4] 4

Answer & Solution

Option # 4

The remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively. Hence the final remainder is the remainder when the product 3 × 4 × 5 = 60 is divided by 7. Therefore, remainder = 4

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Question 2:
What is the remainder when 22004 is divided by 7?

[1] 4

[2] 1

[3] 2

[4] 6

Answer & Solution

Option # 2

22004 is again a product (2 × 2 × 2... (2004 times)). Since 2 is a number less than 7 we try to convert the product into product of numbers higher than 7. Notice that 8 = 2 × 2 × 2. Therefore we convert the product in the following manner- 22004 = 8668 = 8 × 8 × 8... (668 times).

The remainder when 8 is divided by 7 is 1. Hence the remainder when 8668 is divided by 7 is the remainder obtained when the product 1 × 1 × 1... is divided by 7. Therefore, remainder = 1

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Question 3:
What is the remainder when 22006 is divided by 7?

Answer & Solution

Correct Answer: 4

This problem is like the previous one, except that 2006 is not an exact multiple of 3 so we cannot convert it completely into the form 8x. We will write it in following manner- 22006 = 8668 × 4.

Now, 8668 gives the remainder 1 when divided by 7 as we have seen in the previous problem. And 4 gives a remainder of 4 only when divided by 7. Hence the remainder when 22006 is divided by 7 is the remainder when the product 1 × 4 is divided by 7. Therefore, remainder = 4

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Question 4:
What is the remainder when 2525 is divided by 9?

Answer & Solution

Correct Answer: 7

Again 2525 = (18 + 7)25 = (18 + 7)(18 + 7)...25 times = 18K + 725

Hence remainder when 2525 is divided by 9 is the remainder when 725 is divided by 9.

Now 725 = 73 × 73 × 73.. (8 times) × 7 = 343 × 343 × 343... (8 times) × 7.

The remainder when 343 is divided by 9 is 1 and the remainder when 7 is divided by 9 is 7.

Hence the remainder when 725 is divided by 9 is the remainder we obtain when the product 1 × 1 × 1... (8 times) × 7 is divided by 9. The remainder is 7 in this case. Hence the remainder when 2525 is divided by 9 is 7.

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Question 5:
What the remainder when 296 is divided by 96?

Answer & Solution

Correct Answer: 64

The common factor between 296 and 96 is 32 = 25.

Removing 32 from the dividend and the divisor we get the numbers 291 and 3 respectively.

The remainder when 291 is divided by 3 is 2.

Hence the real remainder will be 2 multiplied by common factor 32.

Remainder = 64

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Question 6:
Find the remainder when 752 is divided by 2402.

Answer & Solution

Correct Answer: 2401

752 = (74)13 = (2401)13 = (2402 – 1)13 = 2402K + (−1)13 = 2402K −1.

Hence, the remainder when 752 is divided by 2402 is equal to −1 or 2402 – 1 = 2401.

Remainder = 2401.

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Question 7:
What is the remainder when 3444 + 4333 is divided by 5?

Answer & Solution

Correct Answer: 0

The dividend is in the form ax + by. We need to change it into the form an + bn.

3444 + 4333 = (34)111 + (43)111. Now (34)111 + (43)111 will be divisible by 34 + 43 = 81 + 64 = 145. Since the number is divisible by 145 it will certainly be divisible by 5. Hence, the remainder is 0.

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Question 8:
What is the remainder when (5555)2222 + (2222)5555 is divided by 7?

Answer & Solution

Correct Answer: 0

The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively. Hence, the problem reduces to finding the remainder when (4)2222 + (3)5555 is divided by 7.

Now (4)2222 + (3)5555 = (42)1111 + (35)1111 = (16)1111 + (243)1111. Now (16)1111 + (243)1111 is divisible by 16 + 243 or it is divisible by 259, which is a multiple of 7. Hence the remainder when (5555)2222 + (2222)5555 is divided by 7 is zero.

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Question 9:
202004 + 162004 – 32004 − 1 is divisible by:

[1] 317

[2] 323

[3] 253

[4] 91

Answer & Solution

Option # 2

We can arrange the term as:

${20}^{2004}+{16}^{2004}-3^{2004}-1=\left({20}^{2004}-3^{2004}\right)+\left({16}^{2004}-1^{2004}\right).$

We know that $a^n-b^n$ is divisible by (a+b) and (a-b) when n is odd.

Therefore, ${20}^{2004}-3^{2004}$ is divisible by 17 and ${16}^{2004}-1^{2004}$ is divisible by 17

Hence the complete expression is divisible by 17 .

Similarly, ${20}^{2004}+{16}^{2004}-3^{2004}-1=\left({20}^{2004}-1^{2004}\right)+\left({16}^{2004}-3^{2004}\right).$

Now ${20}^{2004}-1^{2004}$ is divisible by 19 and ${16}^{2004}-3^{2004}$ is divisible by 19 .

 Hence the complete expression is also divisible by 19 .

Hence the complete expression is divisible by $17\times 19=323$.

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Question 10:
Find the remainder when 537 is divided by 63.

Answer & Solution

Correct Answer: 5

5 and 63 are coprime to each other, therefore we can apply Euler’s theorem here.

63 = 32 × 7 $\Rightarrow$ $\varphi \left(63\right)=63(1-\frac{1}{3})(1-\frac{1}{7})=36$

Therefore, Remainder $\left[\frac{5^{37}}{63}\right]$

=Remainder $\left[\frac{5^{36}\times 5}{63}\right]=5$

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Question 11:
What is the remainder when n7 – n is divided by 42?

Answer & Solution

Correct Answer: 0

Since 7 is prime, n7 – n is divisible by 7. n7 – n = n(n6 – 1) = n (n + 1)(n – 1)(n4 + n2 + 1). Now (n – 1)(n)(n + 1) is divisible by 3! = 6. Hence n7 – n is divisible by 6 x 7 = 42. Hence the remainder is 0.

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Question 12:
Find the remainder when 5260 is divided by 31.

Answer & Solution

Correct Answer: 1

31 is a prime number therefore f(N) = 30. 52 and 31 are prime to each other. Therefore, by Fermat’s theorem:

Remainder $\left[\frac{{52}^{30}}{31}\right]=1$

$\Rightarrow$ Remainder $\left[\frac{{52}^{60}}{31}\right]=1$

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Question 13:
Find the remainder when 40! is divided by 41.

Answer & Solution

Correct Answer: 40

By Wilson’s theorem, we can see that 40! + 1 is divisible by 41

$\Rightarrow$ Remainder$\left[\frac{40!}{41}\right]=41-1=40$

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Question 14:
Find the remainder when 39! is divided by 41.

Answer & Solution

Correct Answer: 1

In the above problem, we saw that the remainder when 40! is divided by 41 is 40.

$\Rightarrow$ 40! = 41k + 40 $\Rightarrow$ 40 × 39! = 41k + 40. The R.H.S. gives remainder 40 with 41 therefore L.H.S. should also give remainder 40 with 41. L.H.S. = 40 × 39! where 40 gives remainder 40 with 41. Therefore, 39! should give remainder 1 with 41.

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Question 15:
What will be the remainder when N = 1010 + 10100 + 101000 +........... + 1010000000000 is divided by 7?

Answer & Solution

Correct Answer: 5

By Fermat’s Little Theorem 106 will give remainder as 1 with 7.

Remainder $\left[\frac{{10}^{10}}{7}\right]=$ Remainder $\left[\frac{{10}^6\times {10}^4}{7}\right]$

=Remainder $\left[\frac{{10}^4}{7}\right]=$ Remainder $\left[\frac{3^4}{7}\right]=4$

Similarly, all the other terms give remainder of 4 with 7. Therefore, total remainder = 4 + 4 + 4… (10 times) = 40.

Remainder of 40 with 7 = 5

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Question 16:
Find the remainder when 8643 is divided by 132.

Answer & Solution

Correct Answer: 116

Note that here 8 and 132 are not co-prime as HCF (8, 132) = 4 and not 1. Therefore, we cannot apply Euler’s theorem directly.

Remainder $\frac{8^{643}}{132}]=$ Remainder $\left[\frac{2^{1929}}{132}\right]=4\times$ Remainder $\left[\frac{2^{1927}}{33}\right]$

Now we can apply Euler’s theorem.

$\varphi \left(33\right)=33(1-\frac{1}{3})(1-\frac{1}{11})=20$

$\Rightarrow$ Remainder $\left[\frac{2^{20}}{33}\right]=1$

= Remainder $\left[\frac{2^{1927}}{33}\right]=$ Remainder $\left[\frac{2^7}{33}\right]=29$

$\Rightarrow$ Real remainder $=4\times 29=116$