Find the last two nonzero digits of $700^{2^{88}}$ [duplicate]
Find the last two nonzero digits of $700^{2^{88}}$. I tried approach this problem with modular arithmetic but haven't got much development. Any insights?
Solution 1:
You need to find the last two nonzero digits of $a=7^{2^{88}}$. Note that $7^4 \mod 100 =1$, so the next to the last digit of $a$ is $0$, so we need $a \pmod{1000}$. Note that $7^{60}=1 \pmod{1000}$, so we need $2^{88}\mod 60$. Now $2^{88} =4^{14}\cdot 256= (4^3)^4\cdot 16\cdot 16 =16\cdot 16\cdot 16\equiv 16 \pmod{60}$. So $a=7^{16} \pmod{1000}=601$. So the last $2$ non-zero digits of $a$ are $6,1$.