last three digits of $7^{100}-3^{100}$ How can I find the las three digits of $7^{100}-3^{100}$ ? I know one way is to use $7^{100}=(10-3)^{100}=\sum_{k=0}^n{100 \choose k}10^{100-k}(-3)^k$ but I'm totally stuck... 
Thanks
 A: Well, note that for $0\le k\le 97,$ we have that $1000$ is readily a factor of $\binom{100}{k}10^{100-k}(-3)^k,$ so the last three digits of all of those numbers will be $0$s. Hence, the last three digits of $7^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}+\binom{100}{100}10^{0}(-3)^{100},$$ that is, of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}(-3)^{99}+3^{100}.$$ Hence, the last three digits of $7^{100}-3^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}.$$
Since $\binom{100}{98}=50\cdot99$ and $\binom{100}{99}=100$, the last three digits of this number are easily found.
A: As $\displaystyle 7^2=49=50-1, 7^4=(50-1)^2=2500-2\cdot50\cdot1+1=2400+1$
$\displaystyle\implies 7^{100}=(2400+1)^{25}=1+25\cdot2400+\binom{25}2(2400)^2\equiv1\pmod{1000}$
and $3^{100}=9^{50}=(10-1)^{50}\equiv1-50\cdot10+\binom{50}2\cdot10^2\pmod{1000}\equiv1$
as $-50\cdot10+\binom{50}2\cdot10^2=-500+2500\cdot49=500(-1+5\cdot49)\equiv0\pmod{1000}$
A: Essentially we need to find  $\displaystyle7^{100}-3^{100}\pmod{1000}$
Using Carmichael function, $\displaystyle\lambda(1000)=\cdots=100$
As $\displaystyle(7,1000)=1\implies 7^{100}\equiv1\pmod{1000}$ so is $\displaystyle(3,1000)=1$
A: $7^1 = 7$
$7^2 = 49$
$7^3 = 343$
$7^4 = 2401$, drop the $2$. 
So $7^8 = \dots801$, $7^{16} = \dots601$, etc?
