Question

what are the last 2 digits of 7^2018?? pls answer asap i will mark brainliest

Answer 1

Answer:

Last two digits of 2²⁰¹⁸ = 49

Step-by-step explanation:

Hii mate ^_^

For doing these types of questions,

You have to remember last two digits formula for 7.

Last two digits of

7⁴ⁿ⁺¹=07

7⁴ⁿ⁺²=49

7⁴ⁿ⁺³=43

7⁴ⁿ⁺⁴=7⁴ᵃ=01

where a and n are integers.

Here, given,

7²⁰¹⁸

=7²⁰¹⁶⁺²

=7⁴ⁿ⁺²     where n=504

So, from last 2 digits formula, we get that

Last 2 digits of 7²⁰¹⁸ = 49

Hence,

Last two digits of 2²⁰¹⁸ = 49

HOPE IT HELPS,

PLEASE THANK , FOLLOW AND MARK AS BRAINLIEST.

Answer 2

Answer:

49

Step-by-step explanation: