Last 2 digits | Algebra - Number Theory

Answer: 89

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Explanation :

Last 2 digits of (56489)⁴¹ will be same as last 2 digits of (89)⁴¹.

For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.

After that, we can use the property, last two digits of (…a1)^(…b) will be [Last digit of a×b] 1

Also, last two digits of (100 ± x)² or (50 ± x)² is same as last 2 digits of x².

Last two digits of (89)⁴¹ = Last two digits of 89 × 89⁴⁰
= Last two digits of 89 × (89²)²⁰
= Last two digits of 89 × (..21)²⁰
= Last two digits of 89 × 01.

Here, last two digits of 89² = (100 - 11)² = 11² = 21

Hence, 89.

Answer: 61

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Explanation :

For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.

After that, we can use the property, last two digits of (…a1)^(…b) will be [Last digit of a×b] 1

Also, last two digits of (100 ± x)² or (50 ± x)² is same as last 2 digits of x².

Observe that the last two digits of 59² = 9² = 81.

Therefore, the last two digits of 59⁶⁴ = last two digits of (9²)³² = 81³² = 61.

Hence, 61.

Answer: 2

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Explanation :

For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.

After that, we can use the property, last two digits of (…a1)^(…b) will be [Last digit of a×b] 1

Also, last two digits of (100 ± x)² or (50 ± x)² is same as last 2 digits of x².

The last two digits of 47⁷³ = last two digits of 47 × (472)³⁶ 
= 47 × (09)³⁶ 
= 47 × (09²)¹⁸
= 47× (81)¹⁸ 
= 47 × 41 
= 27

Hence, 2.

Answer: 84

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Explanation :

We know,

Last two digits of 2^{10×odd number} = 24, and

Last two digits of 2^{10×even number} = 76.

2⁵⁴ = (2^{50 + 4}) = 2^10×5 × 2⁴

Now, last 2 digits of 2^10×5 × 2⁴ = last two digits of (2^10×5) × last two digits of (2⁴)

= 24 × 16 = 84

Hence, 84.

Answer: 36

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Explanation :

Last 2 digits of 1456⁷² = last two digits of 56⁷². 

Now, 56⁷² = 8⁷² × 7⁷² = (2³)⁷² × 7⁷² = 2²¹⁶ × 7⁷².

Last two digits of 2²¹⁶ = 2^10×21+6 = last two digits of (2^10×21) × last two digits of (2⁶)
= 24 × 64 = 36.

Last two digits of 7⁷² = Last two digits of (7²)³⁶
= Last two digits of (49)³⁶
= Last two digits of (49²)¹⁸
= Last two digits of (01)¹⁸
= 01

∴ Last two digits of 1456⁷² = last two digits of (2²¹⁶) × last two digits of (7⁷²) = 36 × 01 = 36.

Hence, 36.

Answer: 96

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Explanation :

Now, 62⁴⁸ = 2⁴⁸ × 31⁴⁸

Last two digits of 2⁴⁸ = 2^10×4+8 = last two digits of (2^10×4) × last two digits of (2⁸)
= 76 × 2⁸ = 56.

Note: last two digits of 76 × 2^a = last two digits of 2^a. (a > 1)

Last two digits of 31⁴⁸ = 41

∴ Last two digits of 62⁴⁸ = last two digits of (2⁴⁸) × last two digits of (31⁴⁸) = 56 × 41 = 96.

Hence, 96.

Answer: 76

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Explanation :

Now, 54³⁸⁰ = 2³⁸⁰ × 27³⁸⁰

Last two digits of 2³⁸⁰ = 2^10×38 = 76.

Last two digits of 27³⁸⁰ = Last two digits of (3³)³⁸⁰
= Last two digits of (3)¹¹⁴⁰
= Last two digits of (3⁴)²⁸⁵
= Last two digits of (81)²⁸⁵
= 01

∴ Last two digits of 54³⁸⁰ = last two digits of (2³⁸⁰) × last two digits of (27³⁸⁰) = 76 × 01 = 76.

Hence, 76.

Answer: 16

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Explanation :

Now, 56²⁸³ = (2³)²⁸³ × 7²⁸³

Last two digits of 2⁸⁴⁹ = 2^10×84+9 = last two digits of (2^10×84) × last two digits of (2⁹) = last two digits of (2⁹) = 12.

Last two digits of 7²⁸³ = Last two digits of (7⁴)⁷⁰ × last two digits of (7³)
= Last two digits of (01)⁷⁰ × last two digits of (43)
= 01 × 43
= 43

∴ Last two digits of 56²⁸³ = last two digits of ((2³)²⁸³) × last two digits of (7²⁸³) = 12 × 43 = 16.

Hence, 16.

Answer: 92

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Explanation :

Now, 78³⁷⁹ = 2³⁷⁹ × 39³⁷⁹

Last two digits of 2³⁷⁹ = 2^10×37+9 = last two digits of (2^10×37) × last two digits of (2⁹) 
= 24 × 12 = 88

Last two digits of 39³⁷⁹ = Last two digits of (39²)¹⁸⁹ × last two digits of (39¹)
= Last two digits of (21)¹⁸⁹ × last two digits of (39¹)
= 81 × 39
= 59

∴ Last two digits of 78³⁷⁹ = last two digits of (2³⁷⁹) × last two digits of (39³⁷⁹) = 88 × 59 = 92.

Hence, 92.

Answer: 75

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Explanation :

The last two digits of 25⁷⁷ are always 25.

Last two digits of 83⁴¹ = An even number   3 

i.e. ten's digit will be an even number and unit's digit will be 3.

Note: Tens digit of any power of a number, whose units digit is odd and tens digit is even, is always even.

Thus, the last two digits of 25⁷⁷ × 83⁴¹ will be given by 25 × a3 (where a is even).

Hence, 75.