I dropped in on Ernie a few days ago. As I pushed open the door to his workshop I could hear a clanking, hissing noise and then saw Ernie emerging from a great clouds of steam, his face blackened with what looked like soot.

"Just in time to help me debug my calculator," he called out as he saw me.

"The clock-work calculator?" I asked.

"No. Clockwork is just so 20th century," Ernie retorted. "In my opinion, the future is in steam! Just let me stoke it up a little more and I can show you how it works." Ernie added a couple of shovels of coal to a small boiler and the hissing sound rose in intensity.

The calculator was a labyrinth of oily tubes, valves, gear-boxes, and brass widgets, like some sort of mechanically-augmented steam-punk tuba.

There were three long rows of clock-dials (each with a single pointer hand), each face numbered $0$ to $9$, and a small keyboard. The 14 keys were labelled with the digits 0 to 9 and the letters A, B, and D.

"If you press a series of digit buttons then the A or B button, you can load a number into the $A$ or $B$ register," he explained.

"What does the D button do?" I asked.

"That," Ernie replied, "is to test the Division and Decrement functions. The machine checks if $A$ divides perfectly into $B$ and, if it does, it completes the division and outputs $B \div A$ on the $C$ register. Then it decrements $A$ twice."

"What happens if $A$ doesn't divide into $B$?" I asked.

"Then it doesn't do anything. I haven't built in fractions or decimal points yet, and I hate approximate answers!" said Ernie.

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Ernie asked me to type my 'lucky' 3-digit number into the $A$ register, but unlike Ernie I don't believe in lucky numbers, so I typed in the first number I could think of - which happened to be my auto-bank PIN $a$. When I pressed the A button, the indicator arms on the first three dials dial of the $A$ register swung round to point at the digits I had entered.

"Now enter a big number for the $B$ register," Ernie commanded. I typed in $n$ digits, but when I hit the B button, the first $n$ indicator arms all swung round to the digit $1$. Ernie muttered something about the steam pressure being too low to actuate the internal digital transfer mechanism (or maybe a sticky bypass needle actuator) but then pointed out that $b = 111111...11111$ was still a valid integer. "Try the D button now," he said.

I pressed D. The hissing and clanking rose in intensity, and the indicator arms on the $C$ register swung round wildly. After some seconds there was a mechanical "ding" and the arms froze in place, displaying a very long number in the $C$-register, and $a$ was decremented by $2$.

"Well at least Division and Decrementation work properly," said Ernie, after checking the results. "Lucky that your $a$ was a factor of that $b$. Try a few more divisions and I will see if I can sort out the problems with data-entry," he said.

So I left $a-2$ in the $A$ register and entered another number $b_2$ (of different length) into the $B$ register. Once again $b_2$ turned to all $1$s, and when D was pressed, an answer was output to register $C$ and $a-2$ was decremented to $a-4$. I repeated the process twice more without changing the $A$ register, and the same sequence of events occurred each time.

But after my fourth calculation Ernie waved me to a stop. "I think I have found part of the problem," he said. He made a small adjustment to a set of valves with a mallet and asked me to try again.

His adjustments had reset all the registers back to $0$, so I re-entered my 3-digit PIN into $A$, and entered a new $b$. My fingers were getting a bit tired from all the number pressing, so this time I just typed in a 6-digit number. But when I pressed B, there was still obviously a problem - the indicator arms all pointed at $0$s and $1$s.

"Not perfect, but better than before", Ernie said, then pointed out that I still had a valid 6-digit integer in the $B$ register and I should try the D button again.

As on all the previous occasions, $a$ divided perfectly into $b$, and after displaying an answer in the $C$ register, $a$ was replaced by $a-2$.

I repeated the process thrice more without altering the $A$ register, but typing a 7-digit, then an 8-digit, then a 9-digit number into the $B$ register (each constructed from $0$s and $1$s), and the same sequence of events occurred each time.

Eventually Ernie lamented that "...it might take a bit more thought to work out what is going wrong with data entry and decrementation, but at least it is doing the actual division perfectly." So we stopped for lunch.

"I suppose you will be telling this story on the web," he commented after we had eaten. I was surprised and a little embarrassed because I hadn't realized Ernie had even known about my stories. But Ernie reassured me that he didn't mind his 'web-presence'. "But if I can make a suggestion," he continued, "it might be safest to change your PIN if you are going to tell the story exactly as it happened."

Well - I did change my PIN even though I couldn't see how it could possibly be determined from my story. I'm guessing that Ernie was just playing a practical joke on me.

Can anybody tell me if it really is possible to calculate my old PIN from what I have told you?

  • $\begingroup$ "but typing an 7-digit, then a 8-digit" I think you got your an/a uses swapped around here. $\endgroup$
    – warspyking
    Commented Jul 29, 2015 at 4:17
  • 4
    $\begingroup$ @warspyking Problem was, I originally started writing this in Kzijekistanese (for a very different web-site), and in that language the word for '7' starts with a vowel and the word for '8' starts with a consonant - hence the error. My sincerest apologies. $\endgroup$
    – Penguino
    Commented Jul 29, 2015 at 4:26
  • $\begingroup$ Nothing to be sorry for, we all make mistakes. Excellent work on yet another Ernie puzzle btw. I can never solve these :P $\endgroup$
    – warspyking
    Commented Jul 29, 2015 at 4:57
  • $\begingroup$ Got a question about the dividing. Is the machine using the number with 0s and 1s or the originally entered number (that wasn't displayed correctly)? $\endgroup$
    – B1indfire
    Commented Jul 29, 2015 at 13:35
  • $\begingroup$ The calculator is doing the correct division for the number displayed in the B register. $\endgroup$
    – Penguino
    Commented Jul 29, 2015 at 20:57

2 Answers 2


If a number is even, it's never going to divide a number that is all 1s. Similarly, a number that is divisible by 5 also isn't going to divide all 1s. So a, a-2, a-4, and a-6 are all odd and not divisible by 5. The only way that this is possible is if a ends in 3. This section gives us no further information, since every number that is relatively prime to 10 will divide a repunit of some length.1

I couldn't come up with a neat way to do the second part, so I ran a brute force search of all 3-digit factors ending in 3 of 6-digit numbers consisting of 1s and 0s. This was the result:

143 divides 100100, 101101, 110110, 111111
183 divides 100101
193 divides 110010
213 divides 100110
273 divides 101010, 111111
393 divides 101001
653 divides 111010
803 divides 110011

Then, I tested the corresponding $a$-2 values with 7-digit numbers:

141 divides 1001100, 1011111
181 divides 1001111
271 divides 1011101, 1111100

And then the 8-digit numbers:

179 divides 11010111

Just to check:

177 divides 100101111, 101110011

The PIN must be 183.

1Proof from Wikipedia (https://en.wikipedia.org/wiki/Repunit#Properties):

Using the pigeon-hole principle it can be easily shown that for each $n$ and $b$ such that $n$ and $b$ are relatively prime, there exists a repunit in base $b$ that is a multiple of $n$. To see this consider repunits $R_1^{(b)},...,R_n^{(b)}$. Assume none of the $R_k^{(b)}$ is divisible by $n$. Because there are $n$ repunits but only $n-1$ non-zero residues modulo $n$ there exist two repunits $R_i^{(b)}$ and $R_j^{(b)}$ with $1\le i<j\le n$ such that $R_i^{(b)}$ and $R_j^{(b)}$ have the same residue modulo $n$. It follows that $R_j^{(b)} - R_i^{(b)}$ has residue $0$ modulo $n$, i.e. is divisible by $n$. $R_j^{(b)} - R_i^{(b)}$ consists of $j - i$ ones followed by $i$ zeroes. Thus, $R_j^{(b)} - R_i^{(b)} = R_{j-i}^{(b)}\times b^i$. Since $n$ divides the left-hand side it also divides the right-hand side, and since $n$ and $b$ are relatively prime, $n$ must divide $R_{j-i}^{(b)}$, contradicting the original assumption.

  • 1
    $\begingroup$ Glad I changed that PIN before publishing. $\endgroup$
    – Penguino
    Commented Jul 29, 2015 at 21:22
  • 1
    $\begingroup$ @Penguino You, sir, are an amazing actor, especially since you have to create a personality completely over text. $\endgroup$
    – warspyking
    Commented Jul 29, 2015 at 21:51

Building off of f" answer:

The machine converted Ernie's big numbers into giant strings of 1's, but these 1's were in fact divisible by his secret code. Thus we need to confirm the following 4 things to in fact show that 181 does work:

183 * some 1st giant number = some giant amount of 1's
181 * some 2nd giant number = some giant amount of 1's
179 * some 3rd giant number = some giant amount of 1's
177 * some 4th giant number = some giant amount of 1's

Also I should note that these are not unique. Instead the numbers below are the "smallest" which work.

I wrote a macro in excel to find these giant numbers after figuring out the 183 only to find that f" beat me to it.

Our input number for 183 is only:
which can be confirmed using Wolfram Alpha here.

Our input number for 181 is:
which can be confirmed using Wolfram Alpha here.

Our input number for 179 is:
which can be confirmed using Wolfram Alpha here.

Finally, our input number for 177 is:
which can be confirmed using Wolfram Alpha here.

Since I had written the script, I wanted to look at all numbers 1000 digits or less that could be divided by 183, 181, 179, and 177. (The smallest of which is listed above).

177: 171, 345, 519, 693, 867 digits (a repeating cycle of 174 digits)
179: 175, 353, 531, 709, 887 digits (a repeating cycle of 178 digits)
181: 177, 357, 537, 717, 897 digits (a repeating cycle of 180 digits)
183:  57, 117, 177, 237, 297 digits (a repeating cycle of  60 digits)

I was wondering if anyone knew how to predict the length of the number/cycle?

  • $\begingroup$ I stated that any number relatively prime to 10 divides a bunch of 1s, but I forgot to give the proof. I have added the proof from Wikipedia to my answer. $\endgroup$
    – f''
    Commented Jul 29, 2015 at 15:54
  • $\begingroup$ Anyway, no wonder the narrator's "fingers were getting a bit tired from all the number pressing"! $\endgroup$
    – f''
    Commented Jul 29, 2015 at 15:55
  • $\begingroup$ @f'' but big numbers are cooler than squiggly symbols $\endgroup$
    – qwertylpc
    Commented Jul 29, 2015 at 15:58
  • $\begingroup$ Consider the number of 1s in the product (which is 3 more than the number of digits in the other factor). If x 1s are divisible by n, then x 9s must be as well. Then we add 1, so 10^x must be 1 mod n. This shows that the pattern is that the number of 1s is a multiple of the order of 10 mod n. In turn, the order is always a factor of phi(n). $\endgroup$
    – f''
    Commented Jul 29, 2015 at 16:57
  • $\begingroup$ Phi(177) is 116, and in fact the order of 10 mod 177 is 58. 58 1s, 116 1s, 174 1s, etc. are all divisible by 177. $\endgroup$
    – f''
    Commented Jul 29, 2015 at 16:59

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