Almost The Hyperfactorial

Note that $(n+20)^{n+20}\equiv n^{(n+20)}\equiv n^n \pmod{10}$ for all positive $n$ ; we can reduce the exponent modulo $4=\phi(10),$ by Euler's Theorem. Thus the sequence $n^n \pmod{10}$ has a period of 20. Let $S=\sum_{n=1}^{20}n^n$ . Since our sum runs through $\frac{1000}{20}=50$ periods, with one term $n=1001$ left over, we have $\sum_{n=1}^{1001}n^n\equiv50S+1001^{1001}\equiv1^1=\boxed{1}\pmod{10}$

Consider the first forty terms and further split them into 4 partitioned sums:

$\begin{aligned} S_1 &=& 1^1 + 2^2 + 3^3 + \ldots + 10^{10} \\ S_2 &=& 11^{11} + 12^{12} + 13^{13} + \ldots + 20^{20} \\ S_3 &=& 21^{21} + 22^{22} + 23^{23} + \ldots + 30^{30} \\ S_4 &=& 31^{31} + 32^{32} + 33^{33} + \ldots + 40^{40} \\ \end{aligned}$

We further define $S_n = (10n-9)^{10n-9} + (10n-8)^{10n-8} + \ldots + (10n)^{10n}$ .

Consider modulo $10$ , the last digit of $a^b$ is simply $( a \bmod {10} )^{b \ \bmod 4 }$ and with a little bit of modular arithmetic, we can easily see that

$\begin{aligned} S_1 & \equiv & S_5 \equiv S_9 \equiv \ldots \equiv S_{97} \equiv 7\\ S_2 & \equiv & S_6 \equiv S_{10} \equiv \ldots \equiv S_{98} \equiv 7\\ S_3 & \equiv & S_7 \equiv S_{11} \equiv \ldots \equiv S_{99} \equiv 7\\ S_4 & \equiv & S_8 \equiv S_{12} \equiv \ldots \equiv S_{100} \equiv 7 \\ \end{aligned}$

Hence, the summation in question is congruent to

$\begin{aligned} \displaystyle \sum_{n=1}^{1001} n^n & = & (S_1 + S_2 + S_3 + \ldots + S_{100} ) + 1001^{1001} \\ & \equiv & 100(7) + 1 \equiv \boxed 1 \end{aligned}$

Consider if the number on top of sigma is 1000. The sum of n=1 to n=10 ends with a 7. So does the sum of n=11 to 20 and...... and the sum of 991 to 1000.

100*7=700, which ends with 0. So the last digit of the sum of when n=1 to n=1000 is 0

Now 1001^1001 obviously ends with a 1.

1+0=1.

As it ask for the last digit I focused on that. We know that 1^2, always ends in 1, 2^2 in 4, so the sequence is: 1,4,9,6(16), 5(25), 6(36) ,9(49), 4(64), 1(81), 0 (10). If we sum the last digits it will end at 5 (45). If we do it 1000 times it will end in 0, plus the last 1 (1001), that will end in 1.

I just looked at the first 10 values of n and focused on the last digits. I manually worked out the last digits would be 1,4,7,4,5,6,3,6,9,0, which add to 5. (I noticed the symmetry and number bonds so feel there was a way of doing this work quicker) but then 11^11 until 20^20 would have the same last digits so wold also equal 5. This would repeat until 1000, so I wold have 200 sets of numbers with last digits of 5. These added make last digit of 0 then +1 for 1001^1001 term. similar to other methods but interesting that most people went for sets of 20 and I went sets of 10.

Each last digit besides one will be summed 100 times. Out of these: 9, 5, 6, 0 have repeating last digit in there power sequences. So the sum of these powers will be: 100 9,5,6,0 which will have no effect on our summnations last digit since 100 n’s last digit will be zero for all integer ns. 4, 7, 3, 2, 8 have repeating last digit power sequence some even number apart , ex: 3, 9, 27, 81, 243... 4, 16, 64, 256... 7, 49, xx3, xxx1, xxxx7... 2, 4, 8, 16, 64, 32... 8, 64, 512, 4096, xxxx8

For n^n a number with the samelast digit will occur ever 10 away, ex: 3^3, 13^13, 23^23... So the next number with the same last digit will be two reapeats away + 2 for 3,7,2,8 since these repeat their last digit every 4 away (2 4)+2. Since the third number with the same last digit will be 2 ((2 4)+2) away we get (4 4)+4 so the pattern will repeat every two this is a multiple of 4. So from 1-1001 we will have 2 last digits appear 50 times each for 3,7,2,8: 50*n will always have a last digit of 0 for all integers so these numbers do no effect our summnations last digit.

4 will have the same last digit of 6 in n^n for all numbers since it’s power last digit pattern repeats every 2 and every n with a last digit of 4 is 10 away from each other, and 10 is a multiple of 2: 2 5. So all n with a last digit of 4 will have the same last digit of n^n and theses numbers will appear 100 times between 1-1001 so the amount 4 effects the last digit of our summnation is 100 n which will have a last digit of 0 for all integers n. So it wont effect our summnations last digit.

We have shown that all numbers with last digit 2-9 will not effect the above summnations fibal digit. So, only numbers with a last digit of 1 will effect out summnations last digit. Numbers with the last digit 1 will appear 101 times between 1-1001. So it will effect our summnations last digit by the last digit of 101*1 which is equal to 1. So our summnations last digit must be 1

Python 2.7:

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print str(sum([n**n for n in xrange(1,1002)]))[-1]