Question

Find the last two digits of (right most) of 3^999.

Answer 1

@cinar @noided

Answer 2

67

Answer 3

3^999=440690273160268878963485086584048121988474010917382722554973456075609532448901633180259437950202687321303259232290860785316984860700206303955114241752651224675873408399440267959338258076321613758130133372529539347042982605207698146020522057684695558163502059375160114801849018132346298605821789418305378740276756187926194096742805466102629298972852134694966312536457747390615453312898505588339646862703020142029890479621367604783461882915721944003538122044057700922967618406667

Answer 4

Thank you Sir

Answer 5

I'm going to use modulo notation.

We need to find

3^999 (mod 100)

By Euler's Theorem, since gcd(3, 100) = 1 we see that 3^(phi(100)) = 1 (mod 100)

phi(100) = phi(2^2 * 5^2) = phi(2^2) * phi(5^2) = 2^1 (2 - 1) * 5^1 (5 - 1)
phi(100) = 2 * 20 = 40

Thus, 3^40 = 1 (mod 100).

Let 3^999 = x (mod 100). Multiply both sides by 3. Then

3^1000 = 3x (mod 100)
(3^40)^25 = 3x (mod 100)
1^25 = 3x (mod 100)
1 = 3x (mod 100)

Note that if x = -33, then 1 = -99 (mod 100) which is true since 100 | (1 - -99).

Note that -33 = 67 (mod 100) since 100 | (-33 - 67).

Answer: 67

Answer 6

Thank you Sirr