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When trying to calculate $2^{100}$ with a malfunctioning calculator, the screen displayed a number with missing dgits, that is:

$$1x676506002282294014967032053yz$$

where $x,y$ and $z$ represent missing digits. Can you find all of the missing digits?

What I have so far is that $z=6$, since $2^{100}$ has the form $2^{4k}$. Then, since $2^{100}$ is divisible by 4, its last two digits must be a number divisible by 4, this means $y$ can be either $1, 3, 5, 7$ or $9$. Then I took into consideration that $2^{100}$ is also divisible by 8, so its last 3 digits should be a number divisible by 8. That means $y$ is reduced to the two posibilities $3$ or $7$. Then I thought I could try with the divisibility by 16, but I realized this process is far too slow and I was wondering if there's a faster way to fing what the value for $y$ is.

Then for $x$ I thought that, since $2^{100}$ is only made by powers of 2, I should find the posibilities for $x$ that result in a number that cannot be divided by any other number than powers of 2. But that also seems like a slow and bad a pproach.

Robert Shore
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NotAMathematician
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    You can compute $2^{100} \bmod 100$ (which gets you $y$ and $z$) by computing it $\bmod 4$ and $\bmod 25$ (using the totient theorem), then using the Chinese remainder theorem. To get $x$ you can compute $2^{100} \bmod 9$ (again using the totient theorem, although there's a shortcut here since $2^3 = 8 \equiv -1 \bmod 9$), which uniquely determines $x$ via the fact that the sum of the digits of a number is congruent to it $\bmod 9$. – Qiaochu Yuan Oct 01 '20 at 01:21
  • Consider also using other relatively convenient primes... you should be able to figure out what $2^{100}$ is equivalent to modulo $9$ as well as what it is equivalent to modulo $11$. Now, recall divisibility rules for modulo $9$ and for modulo $11$ to get additional relations between $x,y,z$ – JMoravitz Oct 01 '20 at 01:22
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    Closely related: https://math.stackexchange.com/questions/3059994/determining-the-missing-digits-of-15-equiv-1-square0767436-square000-without?rq=1 – JMoravitz Oct 01 '20 at 01:23
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    @QiaochuYuan Unless $x=0$ or $x=9$ ;) – N. S. Oct 01 '20 at 01:23
  • @N.S.: oh, yes, good point! – Qiaochu Yuan Oct 01 '20 at 01:25
  • Determine the last two digts as in the first dupe, then determine $x$ by casting out nines as in the 2nd dupe. We have hundreds of prior questions on both topics, which you can find by searching. – Bill Dubuque Oct 01 '20 at 01:49
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    $2^3 \equiv -1 \pmod 9$ is probably a big tool. $2^{100} \equiv (-1)^33*2 \equiv -2 \equiv 7\pmod 9$. Oh.... and $2^5 \equiv 1 \pmod {11}$ so $2^{100}\equiv 1 \pmod {11}$ if $x = 0$ or $9$ then this should get you $x$ will no ambiguity. – fleablood Oct 01 '20 at 01:52
  • Any possible way of doing so without using the totient theorem? – NotAMathematician Oct 01 '20 at 17:11

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Hint

Calculate $2^{100} \pmod{100}$. This gives you the last 2 digits. Last two digits always means modulo 100 arithmetic.

Then calculate $2^{100} \pmod{9}$. If this doesn't work, try $\pmod{11}$. If you are missing a digit, modulo 9 arithmetic tells you the digits, excepting when it is 0 or 9. Modulo 11 arithmetic always tells you the missing digit.

N. S.
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