The misterious digit

Let the given sum be $N$ . We need to find $N \bmod 10$ .

$\begin{aligned} N_9 & \equiv 123456789^{123456789} \equiv 9^{123456789} \equiv (10-1)^{123456789} \equiv -1 \text{ (mod 10)} \end{aligned}$

$\begin{aligned} N_8 & \equiv 12345678^{12345678} \equiv 8^{12345678} \\ & \equiv (10-2)^{12345678} \equiv 2^{4n+2} & \small \color{#3D99F6} \text{Since }2^{4n} \equiv 6 \text{ (mod 10)} \text{, where }n \text{ is an integer.} \\ & \equiv 6 \times 4 \equiv 4 \text{ (mod 10)} \end{aligned}$

$\begin{aligned} N_7 & \equiv 1234567^{1234567} \equiv 7^{1234567} & \small \color{#3D99F6} \text{Since } \gcd(7,10) = 1 \text{, Euler's theorem applies} \\ & \equiv 7^{1234567 \bmod \phi (10)} \equiv 7^{1234567 \bmod 4} & \small \color{#3D99F6} \text{Euler's totient function }\phi (10) = 4 \\ & \equiv 7^3 \equiv 3 \text{ (mod 10)} \end{aligned}$

$\begin{aligned} N_6 & \equiv 123456^{123456} \equiv 6^{123456} \equiv 6 \text{ (mod 10)} & \small \color{#3D99F6} \text{All powers of 6 end with 6.} \end{aligned}$

$\begin{aligned} N_5 & \equiv 12345^{12345} \equiv 5^{123456} \equiv 5 \text{ (mod 10)} & \small \color{#3D99F6} \text{All powers of 5 end with 5.} \end{aligned}$

$\begin{aligned} N_4 & \equiv 1234^{1234} \equiv 4^{1234} \equiv 4^{2n} \equiv 6 \text{ (mod 10)} & \small \color{#3D99F6} \text{Since }4^{2n} \equiv 6 \text{ (mod 10)} \text{, where }n \text{ is an integer.} \end{aligned}$

$\begin{aligned} N_3 & \equiv 123^{123} \equiv 3^{123} \equiv 3(10-1)^{61} \equiv -3 \text{ (mod 10)} \end{aligned}$

$\begin{aligned} N_2 & \equiv 12^{12} \equiv 2^{12} \equiv 2^{4n} \equiv 6 \text{ (mod 10)} \end{aligned}$

$\begin{aligned} N_1 & \equiv 1^1 \equiv 1 \text{ (mod 10)} \end{aligned}$

Therefore,

$\begin{aligned} N & \equiv N_9 + N_8 + N_7 + N_6 + N_5 + N_4 + N_3 + N_2 + N_1 \\ & \equiv -1 + 4 + 3 + 6 + 5 + 6 - 3 + 6 + 1 \equiv 27 \equiv \boxed 7 \text{ (mod 10)} \end{aligned}$

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References

• Euler's theorem )
• Euler's totient function

First, we'll calculate the digit unit of each number, then we're going tu sum all of them. Let's start with ${123456789}^{123456789}$ . Our attention is focused on the base's digit unit, and in this case it's $9$ . There are two ways to calculate the digit unit of this enormous number. If you know that the sequence of power of $9$ , so $9$ , $81$ , $729$ ,... has also a sequence of unit digits, you're advantaged: the number will end with a $9$ if the exponent is odd, instead the number will end with $1$ if it's even. You can quickly understand that, since the exponent is odd, the number will end with $9$ . But there is also another way to solve it, and personally I prefer this one: $[123456789 \times 123456789 \times ... \times 123456789]_{10}$ = $[9 \times 9 \times 9 \times ... \times 9]_{10}$ . At this step we know that $9 \times 9 = 81$ so we can substitute each couple of $9$ with a $1$ . The $9$ is repeated $123456789$ times, and this is an odd number, so just one $9$ will be left. Here is the next step: $[1 \times 1 \times 1 \times ... \times 1 \times 9]_{10}$ = $[9]_{10}$ . We can conclude that if we divide ${123456789}^{123456789}$ by $10$ we obtain $9$ . For any of the numbers to sum the methods are all the same as shared above, so, I know, it's pretty boring. Let's calculate the digit number of ${12345678}^{12345678}$ . In this case it's much better to use the first way, so we have to know that the sequence of unit digits of the power of $8$ is: $8$ , $4$ , $2$ , $6$ , $8$ , $4$ , $2$ , $6$ , ... where the period $8 4 2 6$ is repeated endlessly. Now another problem approaches: in this sequence, what digit will be in the $12345678$ place? It's easier than it looks: since the period in made by 4 elements, we only have to discover what number multiple of $4$ is the nearest to $12345678$ . It can be also the same number because it's even, we only have to verify that you can divide it by 4. I'll verify it this way, by dividing it by 2 and verifying the result is still even. But if I calculate $\frac{12345678}{2}$ I obtain an odd number, so it means that $12345678$ is a multiple of $2$ , but not of $4$ . If any integer $n$ is even but not multiple of $4$ , then $n-2$ and $n+2$ are certainly multiples of $4$ . In this case, it means that $12345676$ is a multiple of 4: the number in the $12345676$ position will be the last number of the sequence, so it will be $6$ . The sequence will repeat, and we can conclude that the number in the $12345678$ position is the second of the sequence, so it's $4$ . This can be applied to all of the powers to sum: discover the unit digit sequence of the power of the number (and the number is the base's unit digit), and calculate wich number of the sequence will be in the position determined by the exponent. There are 3 particular powers in this expression that make we calculate the unit digit much more quickly: $123456^{123456}$ , $12345^{12345}$ and $1234^{1234}$ . The unit digit sequence of the powers of $6$ is $6$ , $6$ , $6$ , $6$ ,... so it will always be $6$ . The same happens with $5$ . Calculating the unit digits of $123456^{123456}$ and $12345^{12345}$ is, then, very easy: respectively, $6$ and $5$ . Also calculating the $1234^{1234}$ unit digit is easier than the other cases: the unit digit sequence is $4$ , $6$ , $4$ , $6$ , ... (so it's like the $9$ ) : the number will end with $4$ if the exponent is odd, instead it will end with $6$ . Since the exponent $1234$ is even the number will end with $6$ . At the end, once we have substituted each power with its unit digit, we obtain: $9+4+3+6+5+6+7+6+1=...7$ . We just wanted to know the unit digit that is $7$ .

Let $x = 123456789^{123456789} + 12345678^{12345678} + 1234567^{1234567} + 123456^{123456} + 12345^{12345} + 1234^{1234} + 123^{123} + 12^{12} + 1^1.$

As sum of five odd numbers and four even numbers, $x$ must be odd and thus $x \equiv 1 \pmod{2}.$

Now Fermat's Little Theorem tells us that $a^5 \equiv a \pmod{5}$ which implies that $a^{4n + m} \equiv a^m \pmod{5}$ . Also by the definition of conguerence $5k + a \equiv a \pmod{5}$ and thus also $\left(5k + a\right)^{4n + m} \equiv a^m \pmod{5}.$

Therefore $x \equiv 4^3 + 3^2 + 2^3+ 1^0 + 0^1 + 4^2 + 3^3 + 2^0 + 1^1 \equiv 64 + 9 + 8 + 1 + 0 + 16 + 27 + 1 + 1 \equiv 4 + 4 + 3 + 1 + 0 + 1 + 2 + 1 + 1 \equiv 2 \pmod{5}.$

Since $2$ and $5$ are coprime, by the Chinese Remainder Theorem , the solution to $\begin{cases} x \equiv 1 \pmod{2} \\ x \equiv 2 \pmod{5} \end{cases}$ is unique $\pmod{10}$ .

It is easy to see that $\boxed{7}$ is the solution.