Question

What is the remainder when 7 raised to power 2018 and is divided by 25

Answer 1

Hi there!
Here's the answer:

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Remainder of [7^(2018)] ÷ 25 :

For every expression, there comes attached a specific cyclicity of remainders.

To find cyclicity, we keep finding remainders until some remainder repeats itself.

7^(2018) ÷ 25

No. ÷ 25 :
7____7²___7³___7^4___(7^5)

Remainder:(mod 25)
7___-1____-7____1_____(7)

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^^^ To calculate remainder when a No. is divided by 25, last 2 digits are enough & apply negative remainder concept for simplicity

REFER :

7^1 = 7
7² = 49 = -1 (mod 25)
7³ = -1×7 = -7 (mod 25)
7^4 = -7 × 7 = -49 = 1 (mod 25)

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7^5 gives same remainder as 7, when divided by 25 & this cycle continues.
•°• Cyclicity = 4.

So any power of 4 or a multiple of 4 will give a remainder 1.

Express 7^2018 separately as 7^(4x+y)
i.e., 7^(4x) × 7^(y)


•4 × 504 = 2016
=> 2018 = 2016 + 2

•°•
7^(2018) ÷ 7 = [7^(504) ÷ 25 ] × [7² ÷25 ]

• Rem of{ 7^(2018) ÷ 25}
= Rem of{ 7^(504) ÷ 25 } × Rem of{ 7² ÷ 7 }

¶¶¶ This is according to remainder theorem,

>>> The product of any 2 or more than 2 no.s has the same remainder when divided by any Natural No., as the product of their remainders.

=> Rem of{ 7^(2018) ÷ 25 }
= 1× (-1)
= -1
= 24 (mod 25)

•°• Final remainder = 24.

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Hope it helps