One day, back in the time before computers, little John was scribing the answer his arithmetic homework - to compute the value of $2^{50}$. He had, in the days prior, painstakingly calculated every last digit of that value - and was, to be quite honest, very proud of his result.

Unfortunately, this was also before pencils. It was also before little John would fully develop his fine motor skills. Consequently, as he wrote the last digit, he tipped over his inkwell, and spilled the blackest India ink all over his work! He was able to decipher most of the number, but he still couldn't make out one digit $\blacksquare$ towards the middle which had been completely obscured.

$$112589990\blacksquare 842624.$$

He cried all day. All night too. Here was his magnum opus (he wasn't very ambitious) ruined! His parents weren't much help - they didn't want to redo the computation because, honestly, it was pretty tedious, and they were too busy doing whatever it was people did before computers.

So, they took him, and his ruined paper to the local doctor (of mathematics). The parents explained the situation and the doctor (of mathematics), seeing the pitiful boy, took a look at the paper, and, after making a few calculations (far too few to have calculated the value of $2^{50}$ anew), wrote the correct digit above the ink-splotch.

Examining the paper, the parents discovered that the doctor (of mathematics) had used every other digit on the child's homework as part of the computation and had never written a number greater than $16$ (or less than $0$).

How did the doctor (of mathematics) find the missing digit?

  • $\begingroup$ To make it more difficult, you can black out two (or three (or four)) digits. $\endgroup$ – Lopsy Feb 24 '15 at 0:52
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    $\begingroup$ @Lopsy True, though the solution I have in mind, though it's happy to generalize to more digits, seems more elegant when done with one blacked out digit. (If we wanted to make it really hard, we could black out every digit ;-) ) $\endgroup$ – Milo Brandt Feb 24 '15 at 0:57
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    $\begingroup$ This was before the days when anyone cared about... what? The suspense is killing me! $\endgroup$ – Michael Feb 24 '15 at 5:28
  • $\begingroup$ Did he try looking at the blotting paper? $\endgroup$ – KSmarts Feb 24 '15 at 17:09

He used the digital root property, that the (repeated) sum of the digits in a numbers equals its value modulo 9. Working $\bmod 9$,

$$2^{50} = 2^{2 + 3\times 16} = 2^2 \times (2^3)^{16} = 4 \times 8^{16} = 4 \times (-1)^{16} =4 $$

So, $112589990?842624 = 4 \bmod 9$. Adding up the digits gives $70$, which is $7 \bmod 9$, so the remaining digit must be $6$ to produce a sum of $4 \bmod 9$.

  • $\begingroup$ > 2^{50} = 4 \times 8^{16} = 4 \times (-1)^{16} =4 I'm interested in understanding what you're doing here to get the 4. I understand how you would do it using 50 mod 6 = 2 (3rd number in sequence - the digital roots for powers of 2 run in the sequence 1, 2, 4, 8, 7 and 5) but I don't get what you're doing and would like to know (I'm just someone that stumbled upon this site, I'm not a math person). $\endgroup$ – Dois Feb 24 '15 at 6:50
  • $\begingroup$ @Dois I added some more steps. Does it makes sense now? $\endgroup$ – xnor Feb 24 '15 at 6:52
  • $\begingroup$ @Dois the gist is that $-1 \bmod 9$ is the same as $8 \bmod 9$. $\endgroup$ – No. 7892142 Feb 24 '15 at 8:15
  • $\begingroup$ Ah, that's what I needed to get (that -1 mod 9 is the same as 8 mod 9), I didn't understand what was happening there. Thanks. $\endgroup$ – Dois Feb 24 '15 at 9:36

Alternatively, use the alternating sum of the digits and $\mod 11$.

Then, $$2^{50} \equiv \left( 2^{10} \right)^5 \equiv 1^5 =1 \mod 11$$ by Fermat's Little Theorem.

Then $-1+1-2+5-8+9-9+9-0\,+\,?-8+4-2+6-2+4 \equiv 1 \mod 11$.

This gives $?+6 \equiv 1 \mod 11$, and hence $? \equiv -5 \equiv 6 \mod 11$ and hence $?=6$.


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