# PE 3 - Unit's Digit | Numbers

**PE 3 - Unit's Digit | Numbers**

What is the digit in the unit’s place of 2^{55}?

- A.
2

- B.
8

- C.
1

- D.
4

Answer: Option B

**Explanation** :

Let’s look at the last digit of powers of 2.

2^{1} = 2,

2^{2} = 4,

2^{3} = 8,

2^{4} = 16,

2^{5} = 32,

2^{6} = 64

…

Thus, it can be seen that the unit’s digit of 2^{n} is repeating after every 4 terms in the pattern of 2, 4, 8 and 6.

In general, last digit of 2^{4n+a} will be same as the last digit of 2^{a}.

Hence, the last digit of 2^{55} ≡ 2^{4×13+3} ≡ 2^{3} = 8.

Hence, option (b).

Workspace:

**PE 3 - Unit's Digit | Numbers**

Find the unit’s digit of 61 × 62 × 63 × ... × 69.

- A.
4

- B.
3

- C.
2

- D.
0

Answer: Option D

**Explanation** :

This is a product of all the integers from 61 to 69.

In this product, there is one 5 (at units place of 65) and at least one even number (at units place of 62, 64, 66 or 68).

When 5 is multiplied with any even number, it always ends with 0. So, the unit’s digit is 0.

Also, when 0 is multiplied with any number, it always ends with 0. So, the unit’s digit is 0.

Hence, option (d).

Workspace:

**PE 3 - Unit's Digit | Numbers**

Find the last digit of (1008)^{2500}.

- A.
8

- B.
6

- C.
4

- D.
2

Answer: Option B

**Explanation** :

Let’s look at the last digit of powers of 8.

8^{1} = 8,

8^{2} = 4,

8^{3} = 2,

8^{4} = 6,

8^{5} = 8,

8^{6} = 4,

8^{7} = 2,

8^{8} = 6,

…

Thus, it can be seen that the unit’s digit of 8^{n} is repeating after every 4 terms in the pattern of 8, 4, 2 and 6.

Hence, last digit of 8^{4n+a} will be same as the last digit of 8^{a}.

i.e., to find the last digit, divide the required power by 4 and find the remainder.

So, when 2500 is divided by 4, it gives 0 as the remainder. Therefore, the last digit of 8^{2500} is same as the last digit of 8^{4} = 6.

Hence, option (b).

Workspace:

**PE 3 - Unit's Digit | Numbers**

Find the last digit of (173)^{99}.

- A.
9

- B.
7

- C.
1

- D.
3

- E.
4

Answer: Option B

**Explanation** :

Let’s look at the last digit of powers of 3.

3^{1} = 3,

3^{2} = 9,

3^{3} = 7,

3^{4} = 1,

3^{5} = 3,

3^{6} = 9,

3^{7} = 7,

3^{8} = 1,

…

Thus, it can be seen that the unit’s digit of 3^{n} is repeating after every 4 terms in the pattern of 3, 9, 7 and 1.

Hence, last digit of 3^{4n+a} will be same as the last digit of 3^{a}.

i.e., to find the last digit, divide the required power by 4 and find the remainder.

So, unit’s digit of (173)^{99} will be units digit of 3^{99}, which be same as unit’s digit of 3^{4×24+3} = 3^{3} = 7.

Hence, option (b).

Workspace:

**PE 3 - Unit's Digit | Numbers**

What is the right most non-zero digit of 13890000^{13890000}?

Answer: 1

**Explanation** :

The required answer is nothing but the last digit of 9^{13890000}.

We know that the cyclicity of 9 is of 2 terms i.e.,

Last digit of 9^{odd number} = 9, and

Last digit of 9^{even number} = 1.

Here 13890000 is an even number, hence, last digit of 9^{13890000} will be 1.

Hence, 1.

Workspace:

**PE 3 - Unit's Digit | Numbers**

The last digit of the expression:

541 × 847 × 373 × 929 × 261 × 783 × 104 × 656 is

Answer: 8

**Explanation** :

Unit’s digit of multiplication of more than one number depends on the multiplication of unit’s digits of individual numbers.

So, multiply the unit’s digits of the expression given, i.e., 1 × 7 × 3 × 9 × 1 × 3 × 4 × 6 = 13608.

Therefore, 8 will be at the unit’s place.

Note: We don’t have to multiply all last digits and calculate the answer as 13608. Successively multiply the unit’s digits of previous answer with next number.

For e.g., last digit of 7 × 3 × 9 = last digit of (7 × 3) × 9 = 1 × 9 = 9.

Hence, 8.

Workspace:

**PE 3 - Unit's Digit | Numbers**

What is the last digit of the number obtained by dividing 6^{73} by 4^{32}?

- A.
2

- B.
4

- C.
6

- D.
9

- E.
8

Answer: Option C

**Explanation** :

$\frac{{6}^{73}}{{4}^{32}}=\frac{{2}^{73}\times {3}^{73}}{{\left({2}^{2}\right)}^{32}}=\frac{{2}^{73}\times {3}^{73}}{{2}^{64}}=\frac{{2}^{9}\times {3}^{73}}{1}$

Last digit of 2^{9} ≡ last digit of 2^{1} = 2.

Last digit of 3^{73} ≡ last digit of 3^{1} = 3.

So, the required answer is the last digit of 2 × 3 = 6.

Hence, option (c).

Workspace:

**PE 3 - Unit's Digit | Numbers**

Find the last digit of the following expression.

(75)^{424} + (39)^{98} + (72)^{121}

- A.
8

- B.
6

- C.
5

- D.
3

Answer: Option A

**Explanation** :

Unit digit of (75)^{242} = 5.

Power of any number with 5 at the unit’s place will always have a 5 at its units place, irrespective of the positive natural number power.

Also, last digit of 39^{98} = last digit of 9^{98} = last digit of 9^{2} = 1.

Now, last digit of 72^{121} = last digit of 2^{121} = last digit of 2^{1} = 2.

So, basically after every 4^{th} power, the units digit repeats itself.

Thus, the digit at the unit’s place is 5 + 1 + 2 = 8.

Hence, option (a).

Workspace:

**PE 3 - Unit's Digit | Numbers**

What is the unit digit in (6824)^{1793} × (3125)^{317} × (6681)^{491}?

- A.
0

- B.
2

- C.
3

- D.
5

Answer: Option A

**Explanation** :

Here, Unit’s digit of 6824^{1793} will be an even number.

Also, Unit’s digit of 3125^{317} will always by 5.

Hence, Unit’s digit of 5 × even number will always be 0.

⇒ Unit’s digit of (6824)^{1793} × (3125)^{317 }× (6681)^{491 }= 0.

**Alternately,**

Unit’s digit of powers of 4 have a cyclicity of 2 terms i.e., unit’s digit follows the pattern of 4 and 6.

⇒ Unit’s digit of 4^{odd number} = 4, and

⇒ Unit’s digit of 4^{even number} = 6.

Unit digit in (6824)^{1793 }= Unit digit in (4)^{1793 }= 4

Also,

Unit digit in (3125)^{317} = Unit digit in (5)^{317 }= 5

Unit digit in (6681)^{491} = Unit digit in (1)^{491 }= 1

Required digit = Unit digit in (4 × 5 × 1) = 0.

Hence, option (a).

Workspace:

**PE 3 - Unit's Digit | Numbers**

The digit in the unit’s place of the number 7^{385} × 3^{278} is

Answer: 9

**Explanation** :

The cyclicity of unit’s digit of powers of 7 is as follows:

7^{1} = 7,

7^{2} = 9,

7^{3} = 3,

7^{4} = 1,

7^{5} = 7,

7^{6} = 9,

7^{7} = 3,

7^{8} = 1,

…

Dividing 385 by 4 and the remainder is 1.

Thus, the last digit of 7^{385} is equal to the last digit of 7^{1} i.e. 7.

Similarly, unit’s digit of 3^{278} = 3^{2} = 9.

Therefore, unit’s digit of (7^{385} × 3^{278}) is unit’s digit of product of digit at unit’s place of 7^{385} and 3^{278} = 1 × 9 = 9.

Hence, 9.

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