Countdown to 2014: Problem #2

Since we are to find the last digit we need to take $\sum_{i=1}^{2014} i^{2014}\pmod{10}.$ Which is equal to $:1^{2014}\pmod{10}+2^2014\pmod{10} +3^{2014}\pmod{10}+...............2013^{2014}\pmod{10}+2014^{2014}\pmod{10}.$ But we don't have to calculate them all as $1\equiv11\equiv21\equiv31\equiv41...............2011\pmod{10}$ and $2\equiv12\equiv22\equiv32\equiv42........2012\pmod{10}$ and $3\equiv13\equiv23\equiv33\equiv43............2013\pmod{10}$ and so on.Thus,we will just calculate $\sum_{i=1}^{10} i^{2014}\pmod{10}$ and the others will follow.Now,let us calculate $\sum_{i=1}^{10} i^{2014}\pmod{10}.$ $1^{2014}\equiv1\mod{10},2^{2014}\equiv4\pmod{10}.$ I found this by building a pattern of the last digits of powers of $2.$ $3^{2}\equiv-1\pmod{10}\Longrightarrow3^{2014}(-1)\equiv9\pmod{10}$ $4^{2014}\equiv6\pmod{10}$ I found this too by building a pattern. $5^{2014}\equiv5\pmod{5}.$ $6\equiv-4\pmod{10}\Longrightarrow6^{2014}\equiv4^{2014}\equiv6\pmod{10}.$ $7^{2014}\equiv(-3)^{2014}\equiv3^{2014}\equiv9\pmod{10}.$ Just go on doing this till $9$ and you will get $\sum_{i=1}^{10} i^{2014}\pmod{10}=5.$ Now we multiply this by $210$ because this was just from $1-10$ and if we multiply it by $210$ we will get from $1-2010.$ Now,just calculate the remaining $4$ and add and you will get the answer.

We need to check the last digits when 1, 2, 3,...,8, 9, 0 are respectively raised to the power 2014. Since these digits reappear in periods of 10, we have to simply multiply the last digit of the sum of 1^2014 + 2^2014 + ... + 3^2014 by [1024/10] to get the last digit when we work out the 2010th period. Then we have to add 1^2014 + 2^2014 + 3^2014 + 4^2014 (their last digits, of course!). And this exercise gives 5.

Sorry I don't have LATEX software.

These are all "brilliant" solutions. I did it a bit differently using rules of summation notation.

Recall that $\sum_{i=1}^n{i} = \frac{n*(n+1)}{2}$ and $\sum_{i=1}^n{i^3} = \frac{n^2*(n+1)^2}{4}$

So I intended to get rid of the summation notation in order to make it easier to interpret and look for a pattern.

$\sum_{i=1}^{2014}{i^{2014}} = \sum_{i=1}^{2014}((i^3)^{671}*i^1)$

Becomes

$(\frac{n^2*(n+1)^2}{2^2})^{671}*\frac{n*(n+1)}{2}$ where n = 2014

Simplified it becomes

$(\frac{n^2+n}{2})^{1343}$

After subbing in n = 2014 we get

$(2029105)^{1343}$

Since the last digit is 5. It won't change given any exponent so I didn't need to use mod.

I started by raising the integers 0-9 to different powers and finding the pattern in the ones digit. 1 -> 1 ... 2 -> 4 -> 8 -> 6 -> 2 ... 3 -> 9 -> 7 -> 1 -> 3 ... 4 -> 6 -> 4 ... 5 -> 5 ... 6 -> 6 ... 7 -> 9 -> 3 -> 1 -> 7 ... 8 -> 4 -> 2 -> 6 -> 8 ... 9 -> 1 -> 9 ... 0 -> 0 ... Now I need to figure out which number from the sequences to pick, so I look at the power (2014). The powers go 1,2,3,4...2014, which means there are 2014 powers. Label the first number in the sequences at index 0. So subtract 1 from the number of powers to get 2013, and get the remainder of 2013 when divided by the period length of each sequence. This gives the index of the number in each sequence: 2013%1=0 2013%2=1 2013%4=1 So I pick the first number in the sequences of period 1 (obviously) and the second number in sequences of periods 2 and 4. This gives the mapping of the integers 0-9 when raised to power 2014 in the ones digit as: 1 -> 1 2 -> 4 3 -> 9 4 -> 6 5 -> 5 6 -> 6 7 -> 9 8 -> 4 9 -> 1 0 -> 0 Incidentally, note these are the ones digit of the square of the integers. The numbers are also palindromic between 1 and 9. Then I add these to get the ones digit of the sum of every 10 terms in the summation: 1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0 = 45; ones digit of 45 = 5. So for every 10 terms in the summation I add 5. I note that 10 goes into 2014 201 times. So I add 5 201 times and get the ones digit only: 201 * 5 -> 5. This is the answer for up to 2010 terms in the summation. There are still 4 more terms left, and their ones digits are: 1 + 4 + 9 + 6 -> 0. Adding this to the previous result I get the final answer 5.

the way I did it, We can see if from the powers of two that, it follows a cycle of 4, after every 4 power the unit digit reappears, 2^1=2, 2^2=4, 2^3=8, 2^4=(1)6, 2^5=(3)2.. so the unit digit 2 reappears after 4 powers, if we see, all the numbers have this cycle of 4 after which the same unit digit reappears.. only exception is 5,6 which follows the cycle of 1 and 4,9 which follows the cycle of 2, now if 2 is raised to the power of 2014, so up to 2012(divisible by 4 which is the cycle number) power the unit digit will be 2 as unit digit after 4 cycle (2^4) is 2 so the unit digit of 2^2014 is simply, 2^2=4

3^1=3 3^2=9 3^3=(2)7 3^4=(8)1 3^5 is going to give 3 as unit digit. for 3^2014, 3^2012 will give 1 as unit digit, 3^2014 is actually 3^2 which give 9 as unit digit

for 4 and 9 it follows cycle of 2, after every 2 power 6 appears as unit digit for 4 and 1 appears as unit digit for 9, 4^2014 will give 6 as unit digit, 9^2014 will give 1 as unit digit,

5,6 follows the cycle of 1, so 5^2014 will give 5 and 6^2014 will give 6, for 7, 8 will also follow the same rule as 2,3, so 7^2014 will give 9 as unit digit, (7^2=>9) 8^2014 will give 4 as unit digit,

so adding up all the unit digits from 1^2014 to 9^2014 (1+4+9+6+5+6+9+4+1) we get 45, for every 10 terms we get the sum of unit digit 45, as 10^2014 will give 0 as unit digit, so does not count, so till 2010 terms we will find total 181 times the sum of 45 which is going to give 5 as unit digit and from 2011 to 2014 the sum of unit digit is 20 . so the final unit digit it 5.