Question 753531
{{{7^2=49=50-1}}}
{{{7^4=(7^2)^2=49^2=(50-1)^2=2500-100+1=2401=2400+1}}}
For an integer {{{n}}},
{{{7^(4n)=2401^n=(2400+1)^n=2400^n+n*2400^(n-1)+(n(n-1)/2)*2400^(n-2)}}}+ ... +{{{(n(n-1)/2)*2400^2+n*2400+1}}}
All the terms rxcept the last one are multiples of 100, so {{{7^4n=100K+1}}} for some integer K, and ends in 01
{{{7^(4n+2)=(7^4n)(7^2)=(7^4n)*49=(100K+1)*49=4900K+49=49K*100+49}}} ends in {{{49}}}
{{{1994=1992+2=4*498+2}}} so
{{{7^1994}}}=.......{{{highlight(49)}}}
 
{{{3^1994=3*2*997=(3^2)^997=9^997=(10-1)^997=10^997-997*10^996+(997*996/2)*10^995}}}+....+{{{-(997*996/2)*10^2+997*10-1=100}}}({{{10^995-997*10^994+(997*996/2)*10^993}}}+....+{{{-997*996/2}}})+{{{997*10-1=100P+997*10-1=100P+9969=100(P+99)+69}}}
with {{{K=10^995-997*10^994+(997*996/2)*10^993}}}+....+{{{-997*996/2}}}
So {{{3^1994}}}=.......{{{highlight(69)}}}
 
{{{3^1994 + 7^1994=(100(P+99)+69)+(49K*100+49)=100(P+99)+49K*100+69+49=100(P+99)+49K*100+118=100(P+99)+49K*100+100+18=100(P+99+49K+1)+18=(49K+P+100)100+18}}}
So {{{3^1994 + 7^1994}}}=.......{{{highlight(18)}}}
 
{{{7^1994-3^1994=(49K*100+49)-(100(P+99)+69)=(49K-1)100+100+49-100(P+99)-69= (49K-1)100-100(P+99)+149-69=100(49K-1-P-99)+80=(49K-P-100)+80}}}
So {{{7^1994-3^1994}}}=.......{{{highlight(80)}}}