Unit's Digit | Algebra - Number Theory

Answer: Option B

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Explanation :

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

2¹ = 2, 
2² = 4, 
2³ = 8, 
2⁴ = 16, 
2⁵ = 32, 
2⁶ = 64 
…

Thus, it can be seen that the unit’s digit of 2ⁿ 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 2⁵⁵ ≡ 2^4×13+3 ≡ 2³ = 8.

Hence, option (b).

Answer: Option D

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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).

Answer: Option B

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Explanation :

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

8¹ = 8,
8² = 4,
8³ = 2,
8⁴ = 6, 
8⁵ = 8, 
8⁶ = 4,
8⁷ = 2,
8⁸ = 6,
… 

Thus, it can be seen that the unit’s digit of 8ⁿ 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 8²⁵⁰⁰ is same as the last digit of 8⁴ = 6.

Hence, option (b).

Answer: Option B

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Explanation :

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

3¹ = 3,
3² = 9,
3³ = 7,
3⁴ = 1, 
3⁵ = 3, 
3⁶ = 9,
3⁷ = 7,
3⁸ = 1,
… 

Thus, it can be seen that the unit’s digit of 3ⁿ 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 3⁹⁹, which be same as unit’s digit of 3^4×24+3 = 3³ = 7.

Hence, option (b).

Answer: 1

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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.

Answer: 8

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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.

Answer: Option C

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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 2⁹ ≡ last digit of 2¹ = 2.

Last digit of 3⁷³ ≡ last digit of 3¹ = 3.

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

Hence, option (c).

Answer: Option A

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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 39⁹⁸ = last digit of 9⁹⁸ = last digit of 9² = 1.

Now, last digit of 72¹²¹ = last digit of 2¹²¹ = last digit of 2¹ = 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).

Answer: Option A

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Explanation :

Here, Unit’s digit of 6824¹⁷⁹³ will be an even number.

Also, Unit’s digit of 3125³¹⁷ 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).

Answer: 3

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Explanation :

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

7¹ = 7,
7² = 9,
7³ = 3,
7⁴ = 1,
7⁵ = 7,
7⁶ = 9,
7⁷ = 3,
7⁸ = 1,
…

Dividing 385 by 4 and the remainder is 1.

Thus, the last digit of 7³⁸⁵ is equal to the last digit of 7¹ i.e. 7.

Similarly, unit’s digit of 3²⁷⁸ = 3² = 9.

Therefore, unit’s digit of (7³⁸⁵ × 3²⁷⁸) is unit’s digit of product of digit at unit’s place of 7³⁸⁵ and 3²⁷⁸ = 7 × 9 = 3. 

Hence, 3.