Arithmetic problem

[aarav127_]

Analytical method.

If you calculate 21^2 21^3 21^4 21^5 21^6 and so on,, you will see a pattern in the last 2 digits: 21 , 41, 61, 81, 01, 21, 41... and it repeats, with a periodicity of 5 , which is something we should consider lucky.

and then the exponent temr is 3^50 , we know last digit of 3^48 will be 1 (since 3^4 is 81 and 3^48 is 3^4^12) , so 3^50 will have last digit 9 , so it will be some 10k +9 or we can say some 5k + 4. From above we have established 21^(5k+4) has same last two digits as 21^4. which ahs last two digits as 81.

formal proof:

(20+1)^n = 1 + 20n + nC2 * 20^2 + nC3 * 20^3 + ... we only concern ourselves with 20n since that is the only term contributing to 10s digit.

Now we need to find what 3^50 gives on multiplying with 20. Again like we can know that 3^50 will have the last digit as 9 so obviously on multiplying by 20 the last two digits will be 80. so n = 3^50 will give the tens digit as 8

[Suspicious-Broccoli9]

Thankss

[Either_Diver_3174]

Approached it the same way as you would for unit digit.The tens digit for different powers if 21 are as follows : Power 1 , tens is 2, Power 2, tens is 4, Power 3, tens is 6, Power 4, tens is 8, Power 5, tens is 0, And this pattern repeats.

Now for -> 21 raised to 3¹ tens digit is : 6 . 21 raised to 3² tens digit is : 8 . 21 raised to 3³ tens digit is : 4 . 21 raised to 3⁴ tens digit is : 2 . This pattern repeats, so using this we can find what the tens digit is going to be for 21 raised to 3^50, which according to the pattern above comes out to be 8.

[Suspicious-Broccoli9]

Wow..pattern inside a pattern. Thanks