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In mathematics an automorphic number (sometimes referred to as a circular number) is a number whose square "ends" in the same digits as number itself. Thus $5$ is automorphic since $25$ ends in $5$. Similarly, $625^2 = 390625$, so $625$ is also automorphic.

Given an integer $N$. How could we find the sum of all $N$ digit numbers which are automorphic? Obviously I am not looking for brute-forcing.

You may also like to take a look at what I call the inspiration problem. My solution for that is not much of mathematically rigorous, as the option were given and I knew that $25$ is automorphic. I simply subtracted from the options, I got $74,75,76$ and $77$. Again as only $1,5$ and $6$ are one digits automorphic numbers, so $74$ and $77$ could be discarded. Also any number ending with $5$ when squared and we take a modulo by $100$ gives $25$ so $75$ rejected. Hence $76$ is the other number.

However, I guess, this is not exactly a mathematical approach, since if the options are not there I don't think I could had solved it, also having more digits makes things more complicated. Hence,I was thinking if there is any other techniques for this?

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Your linked Wikipedia article has the answer: "For $k$ greater than $1$, there are at most two automorphic numbers with k digits" and "The sum of the two numbers is $10^k + 1$. The smaller of these two numbers may be less than $10^{k-1}$; for example with $k = 4$ the two numbers are $9376$ and $625$. In this case there is only one $k$ digit automorphic number; the smaller number could only form a $k$ digit automorphic number if a leading $0$ were added to its digits." –  Henry Oct 6 '11 at 10:56
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Given a multiple choice problem, I see nothing wrong with trying all the possible answers. As you say, that often doesn't give larger understanding, but it solves the task at hand. –  Ross Millikan Oct 6 '11 at 13:12

2 Answers 2

up vote 6 down vote accepted

$1,6,5$

$76, 25$

$376, 625$

$9376$

$90625$

$109376,890625$

For any $N+1$ digit automorphic number, the last $N$ digits must form an automorphic number. The leading digits of all the $N$-digit automorphic numbers sum to $9$. The sequence begins $11, 101, 1001, 10001-625, 100001-9376, 1000001 \cdots$

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(+1):Interesting observation! –  Quixotic Oct 6 '11 at 13:01

Your linked Wikipedia article has the answer:

For $k$ greater than $1$, there are at most two automorphic numbers with $k$ digits...

The sum of the two numbers is $10^{ k} +1$ . The smaller of these two numbers may be less than $10^{ k−1}$; for example with $k=4$ the two numbers are $9376$ and $625$. In this case there is only one $k$ digit automorphic number; the smaller number could only form a $k$ digit automorphic number if a leading $0$ were added to its digits.

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+1 from my side. –  Quixotic Oct 6 '11 at 18:47
    
There are no automorphic numbers other than 1 in bases 2,3,4 or 5. In any base other than 1, there cannot be any automorphic numbers of >1 digits ending with 1 because (a+b)^2=a^2+b^2+2ab and if b is set as the last digit, then the 2nd digit from the right, t, would have to be the last digit of 2t, which is >t if 2t has 1 digit and <t if t has 2 digits. –  Angela Richardson Oct 7 '11 at 7:53

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