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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?

pop32
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    Note: Since the factor of $100$ only contributes $0's$ to the result, it can be discarded. – lulu Jun 06 '20 at 18:11
  • See this question, or this one. – Dietrich Burde Jun 06 '20 at 18:12
  • Yep. But what's next? – pop32 Jun 06 '20 at 18:13
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    Please edit to show your efforts. Once you discard the $0's$, it's a simple matter of computing the order of $7\pmod {100}$. – lulu Jun 06 '20 at 18:14
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    $700^{2^{88}} = 7^{2^88}100^{2^{88}}$. The $100^M$ just adds a bunch of zeros to the end. So we can ignore them. The last two non-zero digits will by the last two non zero digits of $7^{2^88}$. As $7$ and $10$ are relatively prime the last two digits will be non-zero. So calculate $7^{2^{88}}\pmod {100}$. ..... Oh, I suppose we may get that it is $0a$. Does that count as the last two non-zero digits? Well, we'll worry about that if* it happens which it probably won't. – fleablood Jun 06 '20 at 18:16
  • How much modular arithmetic do you know? Do you know 1) Euler's Theorem and 2) did you understand what lulu means by "computing the order of $7\pmod{100}$? – fleablood Jun 06 '20 at 18:19
  • @fleablood I think it does since $7^4 =2401$. – rogerl Jun 06 '20 at 18:21
  • @rogerl I believe you are addressing me, and you are quite right. And I really should have realized that as $\phi(100) = 40$ and $2^{88}$ is very "four-y" that it'd be quite likely that would happen.....So what exactly does last two non-zero digits mean if the second to last digit is $0$. I feel in good faith in must mean "the last two digits after we clear out all the leading zeros". In which case this will be easy to see is $01$. As you pointed out. – fleablood Jun 06 '20 at 18:25
  • @fleablood I agree; if it doesn't mean that then this is an unfairly worded question. – rogerl Jun 06 '20 at 18:27
  • @lulu $\mod 100$ is not enough because the second to last digit is 0, see my answer. – markvs Jun 06 '20 at 18:56
  • @fleablood yes so i think we need mod 1000 – pop32 Jun 06 '20 at 19:13
  • So, use $\pmod {1000}$. It's still easy. Though I would read the question as asking for the first two digits after the long block of $0's$. – lulu Jun 06 '20 at 19:21
  • I can't think of any practical reason anyone would ask for the last two digits that are not $0$. I can see if there's a long bloc $2*2^{88}$ zeros someone would want to know, well, whats after that, but I can see no reason anyone would care, after the one we have a zero so we go one digit further.... but... whatever.... – fleablood Jun 06 '20 at 19:36

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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 $6,1$.

markvs
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