what are the last 2 digits of 7^2800 or what is $$7^{2800} \mod 100$$ ?
$$7^{2800} \mod 100 \quad | \quad 2800 = 2^4*175\\
= 7^{ 16 * 175 } \mod 100 \\
= ( 7^{16} )^{175} \mod 100 \quad | \quad 7^{16}= (7^4)^4\\
= ( (7^4)^4 )^{175} \mod 100 \quad | \quad 7^{4}= (7^2)^2\\
= ( ( (7^2)^2 )^4 )^{175} \mod 100 \quad | \quad 7^2 = 49\\
= ( ( (49)^2 )^4 )^{175} \mod 100 \quad | \quad 49^2 = 2401\\
= ( ( 2401 )^4 )^{175} \mod 100 \quad | \quad 2401 \mod 100 = 1 \quad\textcolor[rgb]{1,0,0}{ x^{a*b} \mod d = (x^a\mod d)^b \mod d }\\
= ( ( 1 )^4 )^{175} \mod 100 \quad | \quad 1^4 = 1 \\
= ( 1 )^{175} \mod 100 \quad | \quad 1^{175} = 1 \\
= 1 \mod 100 \\\\
7^{2800} \mod 100 = 1 \mod 100 \quad | \quad 1 \mod 100 = 01$$
the last 2 digits of 7^2800 = 01
The last digit of 7^x rotates among 1, 7, 9, and 3 because:
7^0 = 1 7^4 = 7^4 x 7^0 = 2401 7^8 = 7^4 x 7^4 x 7^0 = 5764801 ...
7^1 = 7 7^5 = 7^4 x 7^1 = 16707 7^9 = 7^4 x 7^4 x 7^1 = 40353607 ...
7^2 = 49 7^6 = 7^4 x 7^2 = 117649 ...
7^3 = 343 7^7 = 7^4 x 7^3 = 823543 ...
What I'm trying to show is extra factors of 7^4 doesn't change the last digit of the answer.
So divide the exponent by 4,
if the remainder is 0, the last digit will be a 1
if the remainder is 1, the last digit will be a 7
if the remainder is 2, the last digit will be a 9
if the remainder is 3, the last digit will be a 3
Dividing the exponent of 2800 by 4, gives a remainder of 0, so the last digit will be a 1.
Thank you for the response. What you answered is something i already know... my question is to find the last 2 digits and not just the last digit. So essentially, when 7^2800 is divided by 100, it will give the required answer.
what are the last 2 digits of 7^2800 or what is $$7^{2800} \mod 100$$ ?
$$7^{2800} \mod 100 \quad | \quad 2800 = 2^4*175\\
= 7^{ 16 * 175 } \mod 100 \\
= ( 7^{16} )^{175} \mod 100 \quad | \quad 7^{16}= (7^4)^4\\
= ( (7^4)^4 )^{175} \mod 100 \quad | \quad 7^{4}= (7^2)^2\\
= ( ( (7^2)^2 )^4 )^{175} \mod 100 \quad | \quad 7^2 = 49\\
= ( ( (49)^2 )^4 )^{175} \mod 100 \quad | \quad 49^2 = 2401\\
= ( ( 2401 )^4 )^{175} \mod 100 \quad | \quad 2401 \mod 100 = 1 \quad\textcolor[rgb]{1,0,0}{ x^{a*b} \mod d = (x^a\mod d)^b \mod d }\\
= ( ( 1 )^4 )^{175} \mod 100 \quad | \quad 1^4 = 1 \\
= ( 1 )^{175} \mod 100 \quad | \quad 1^{175} = 1 \\
= 1 \mod 100 \\\\
7^{2800} \mod 100 = 1 \mod 100 \quad | \quad 1 \mod 100 = 01$$
the last 2 digits of 7^2800 = 01