SOLUTION: What are the last two digits of: 3^1994 7^1994 3^1994 + 7^1994 7^1994 - 3^1994

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: What are the last two digits of: 3^1994 7^1994 3^1994 + 7^1994 7^1994 - 3^1994      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 753531: What are the last two digits of:
3^1994
7^1994
3^1994 + 7^1994
7^1994 - 3^1994
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
7%5E2=49=50-1
7%5E4=%287%5E2%29%5E2=49%5E2=%2850-1%29%5E2=2500-100%2B1=2401=2400%2B1
For an integer n,
+ ... +%28n%28n-1%29%2F2%29%2A2400%5E2%2Bn%2A2400%2B1
All the terms rxcept the last one are multiples of 100, so 7%5E4n=100K%2B1 for some integer K, and ends in 01
ends in 49
1994=1992%2B2=4%2A498%2B2 so
7%5E1994=.......highlight%2849%29

+....+-%28997%2A996%2F2%29%2A10%5E2%2B997%2A10-1=100(10%5E995-997%2A10%5E994%2B%28997%2A996%2F2%29%2A10%5E993+....+-997%2A996%2F2)+997%2A10-1=100P%2B997%2A10-1=100P%2B9969=100%28P%2B99%29%2B69
with K=10%5E995-997%2A10%5E994%2B%28997%2A996%2F2%29%2A10%5E993+....+-997%2A996%2F2
So 3%5E1994=.......highlight%2869%29


So 3%5E1994+%2B+7%5E1994=.......highlight%2818%29


So 7%5E1994-3%5E1994=.......highlight%2880%29