**Remainders: Part-4**

In this article, for calculating remainders we will use concept of cyclicity.

Letâ€™s take an example of questions based on concept of remainders.

__Example__

**Find the remainder when 47 ^{123} is divided by 7?**

**Solution:**

We will do this question step by step

Step 1: We divide 47 with 7 to get remainder as 5.

Step 2: Here in the question 47 is multiplied by itself 123 times. Now understand every time when we will divide 47 with 7 the remainder will be 5.

Step 3: Hence we can say finding remainder of 47

^{123}is equivalent to finding remainder of 5

^{123}when divided by 7.

So we start observing powers of 5.

5

^{2}gives remainder 4.

5

^{3}gives remainder 6.

5

^{4}gives remainder 2.

5

^{5}gives remainder 3.

5

^{6}gives remainder 1.

Hence after 6th power remainders will start repeating. So we have a cycle of 6 and thus we divide 123 by 6

i.e. 123= 6(20) + 3

Hence 5 ^{123}

-> 5 ^{6(20) +3 }

-> 5 ^{6(20) }.5 ^{3 }-> 1.6Â =6Â Â (As 5^{6} gives remainder 1 and 5^{3} gives remainder 6)

Hence the remainder of 47^{123} when divided by 7 is 6.

__Example__

**Find the remainder when 79 ^{644} is divided by 9 ?**

**Solution:**

We will do this question step by step

Step 1: We divide 79 with 9 to get remainder as 7.

Step 2: Here in the question 79 is multiplied by itself 644 times. Now understand every time when we will divide 79 with 9 the remainder will be 7.

Step 3: Hence we can say finding remainder of 79

^{644}is equivalent to finding remainder of 7

^{644}when divided by 9.

So we start observing powers of 7.

7

^{2}gives remainder 4.

7

^{3}gives remainder 1.

Hence after 3rd power remainders will start repeating. So we have a cycle of 3 and thus we divide 644 by 3 i.e. 644= 3(214) + 2

Hence 7

^{644Â }-> 7

^{3(214) + 2 }-> 7

^{3(214) }.7

^{2}-> 1.4Â = 4Â Â (As 7

^{3}gives remainder 1 and 7

^{2}gives remainder 4)

Hence the remainder of 79

^{644}when divided by 9 is 4.

*Â*

*Assignment:***QUESTIONS**

**1.** Find the remainder when 2^{57} is divided by 3?

a)0

b)1

c)2

d) None of these

**Solution:**

Answer â€“ c

So we start observing powers of 2.

2^{1} gives remainder 2.

2^{2} gives remainder 1.

2^{3} gives remainder 2.

2^{4} gives remainder 1.

Hence we can observe odd power of 2 gives remainder 2 and even power gives remainder 1.

Hence 2^{57} will give remainder 2 when divided by 3.

**2**. Find the remainder when 3^{91} is divided by 5 ?

a)1

b)2

c)3

d)4

**Solution:**

Answer-b

So we start observing powers of 3.

3^{1} gives remainder 3.

3^{2} gives remainder 4.

3^{3} gives remainder 2.

3^{4} gives remainder 1.

Hence after 4th power remainders will start repeating. So we have a cycle of 4 and thus we divide 91 by 4 and 3^{91} will give same remainder as 3^{3} i.e. 2.

Hence the remainder of 3^{91} when divided by 5 is 2.

**3.** What is the remainder when 15^{48} is divided by 4?

a)1

b)2

c)3

d)0

**Solution :**

Answer â€“ b

Finding remainder of 15^{48} is equivalent to finding remainder of 3^{48} when divided by 4.

So we start observing powers of 3.

3^{1} gives remainder 3.

3^{2} gives remainder 1.

3^{3} gives remainder 3.

3^{4} gives remainder 1.

Hence we can observe odd power of 3 gives remainder 3 and even power gives remainder 1.

Hence 3^{48} will give remainder 1 when divided by 4.

Hence the remainder of 3^{48} when divided by 4 is 1.

**4.** Find the remainder when 13^{48}is divided by 9?

a)1

b)2

c)3

d)4

**Solution :**

Answer- a

Finding remainder of 13^{48} is equivalent to finding remainder of 4^{48} when divided by 9.

So we start observing powers of 4.

4^{1} gives remainder 4.

4^{2} gives remainder 7.

4^{3} gives remainder 1.

Hence after 3rd power remainders will start repeating. So we have a cycle of 3 and we know 48 is divisible by 3 so answer will same as remainder of 4^{3}.

Hence the remainder of 13^{48} when divided by 9 is 1.

**5.** What is the remainder when 28^{71} is divided by 7?

a)1

b)2

c)3

d)0

**Solution :**

answer â€“ d

Since 7 is a factor of 28 hence remainder will be zero.