Question

If 4^101+6^101 is divided by 25,the remainder is

Answer 1

Answer:10

Step-by-step explanation:

(4^101 + 6^101)/25

(4^101 + 6^101)/25

=(5–1)^101 + (5+1)^101

=(5^101–101*5^100+…..-(101C2).25+101.5–1)+(5^101+101*5^100+…..+(101C2).25+101.5+1)

= 25K + 101*10

(4^101 + 6^101)/25

= K + 101*10/25

= K + 40 + 10/25

So the balance is 10

Answer 2

Answer:

If 4¹⁰¹+6¹⁰¹ is divided by 25, the remainder is 10.

Step-by-step explanation:

We have 4¹⁰¹ and 6¹⁰¹.

4¹⁰¹ can be written as (5 - 1)¹⁰¹ and 6¹⁰¹ can be written as (5 + 1)¹⁰¹.

Using binomial expansion, we get

4¹⁰¹ = (5 - 1)¹⁰¹

4¹⁰¹ = ¹⁰¹C₀5¹⁰¹ - ¹⁰¹C₁5¹⁰⁰ + ¹⁰¹C₂5⁹⁹ - ... + ¹⁰¹C₉₈5³ - ¹⁰¹C₉₉5² + ¹⁰¹C₁₀₀5¹ ...(i)

6¹⁰¹ = (5 + 1)¹⁰¹

6¹⁰¹ = ¹⁰¹C₀5¹⁰¹ + ¹⁰¹C₁5¹⁰⁰ + ¹⁰¹C₂5⁹⁹ + ... + ¹⁰¹C₉₈5³ + ¹⁰¹C₉₉5² + ¹⁰¹C₁₀₀5¹ ...(ii)

Now, if we add equations (i) and (ii), all the even power terms of 5 will get canceled. Then,

4¹⁰¹ + 6¹⁰¹ = 2 × [ ¹⁰¹C₀5¹⁰¹ + ¹⁰¹C₂5⁹⁹ + ... + ¹⁰¹C₉₈5³ + ¹⁰¹C₁₀₀5¹]

All the terms except the last term have 5² as a factor. Therefore, all the terms are divisible by 25.

So, the remainder will be given as (2×¹⁰¹C₁₀₀5¹) mod 25.

Remainder = (2×¹⁰¹C₁₀₀5¹) mod 25

Remainder = (2 × 101 × 5) mod 25

Remainder = 1010 mod 25

Remainder = 10

Therefore, the remainder for the given operation is 10.

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