49 = 7²

Euler's Totient function for 49 :

Φ(49) = 49 (1 - 1/7) = 49 (6/7) = 42

From Euler's Totient Theorem , if a is a positive integer and m is a positive integer relatively prime to a, then a^Φ(m) ≡ 1 (mod m)


So, since 6 and 8 are both relatively prime to 49, then 6⁴² ≡ 1 (mod 49) and 8⁴² ≡ 1 (mod 49)


For a₈₃ = 6⁸³ + 8⁸³,

Since 6⁴² ≡ 1 (mod 49) and 8⁴² ≡ 1 (mod 49),

then (6⁴²)² ≡ 1² (mod 49) and (8⁴²)² ≡ 1² (mod 49)
(Multiplication/power property)

So 6⁸⁴ ≡ 1 (mod 49) and 8⁸⁴ ≡ 1 (mod 49)


Next,

6⁸³ = 6-¹ (6⁸⁴) and 8⁸³ = 8-¹ (8⁸⁴)

Then 6⁸³ ≡ 6-¹ (mod 49) and 8⁸³ ≡ 8-¹ (mod 49)
(Multiplication property)


Then,

Let 6-¹ ≡ k (mod 49) and 8-¹ ≡ k (mod 49) such that 1 ≡ 6k (mod 49) and 1 ≡ 8j (mod 49)

We realise that 49 = 48 + 1 and 48 is a multiple of 6. So we need 5 × 49 + 1, such that it leaves a remainder of 1 (mod 49), and is a multiple of 6, since five '48's are a multiple of 6 and those 5 '1's can be combined with the remainder 1 to make a 6.

i.e 5 × 49 + 1
= 5 × (48 + 1) + 1
= 5 × 48 + 5 + 1
= 5 × 6 × 8 + 6
= 6 × (5 × 8 + 1)
= 6 × 41

So k = 41 .
Likewise, we can find 49 × 7 + 1 = 344 = 43 × 8 such that j = 43


So 6⁸³ ≡ 41 (mod 49) and 8⁸³ ≡ 43 (mod 49)


Lastly , Addition :


(6⁸³ + 8⁸³) ≡ (41 + 43) (mod 49)
(6⁸³ + 8⁸³) ≡ 84 (mod 49)
But 84 ≡ 35 (mod 49) since 84 = 49 + 35

So (6⁸³ + 8⁸³) ≡ 35 (mod 49)


The remainder is 35.

2011²⁰¹¹

Since you want to find the hundreds digit, we should divide 2011^2011 by 1000 and then see what is the remainder (it will be 3 digit or less)

Then you can tell the hundreds digit from there.


Now 2011 is prime.
2011 and 1000 are coprime .

Compute the Euler's totient function for 1000 first.

1000 = 2³ × 5³ so there are only two types of prime factors


Φ(1000) = 1000 × (1 - 1/2) × (1 - 1/5)
= 1000 × ½ × 4/5
= 400

Then, we can state that :

2011⁴⁰⁰ ≡ 1 (mod 1000)

It then follows that 2011²⁰⁰⁰ ≡ 1 (mod 1000) as well since 2011²⁰⁰⁰ = (2011⁴⁰⁰)⁵
(Multiplication or exponential property)

So we only need to look at 2011¹¹ to identify the remainder.

We can simplify this :

Since 2011 ≡ 11 (mod 1000),

then 2011⁵ ≡ 11⁵ ≡ 161051 ≡ 51 (mod 1000)

(2011⁵)² = 2011¹⁰ ≡ 51² ≡ 2601 ≡ 601 (mod 1000)

Then 2011¹¹ ≡ (601 × 11) ≡ 6611 ≡ 611 (mod 1000)


Combine this with 2011²⁰⁰⁰ ≡ 1 (mod 1000),


2011²⁰¹¹ = 2011²⁰⁰⁰ × 2011¹¹ ≡ (1 × 611) ≡ 611 (mod 1000)

The remainder is 611, so the hundreds digit is 6