maths summation problem n^2007, n=[1,2007]?

Question

1^2007 + 2^2007 + 3^2007 + ... + 2007^2007. Find the last two digits of the sum of the summation.

1^2007 + 2^2007 + 3^2007 +... + 2007^2007. Find the last two digits of the answer to the summation. Please show me how to do it as well. Thank you.

Answer 1

It was intriguing for me coming second to check scythian's answer and it turned out that it's correct. The required number is the remainder R of the given sum by modulo 100 - here is a possible way to find it.
We have the relationship
a^2007 + b^2007 = (a + b)(a^2006 - a^2005*b + a^2004*b^2 - . . . - a*b^2005 + b^2006),
according which if we combine
1^2007 + 1999^2007, 2^2007 + 1998^2007, . . , 999^2007 + 1001^2007, each of these sums along with 1000^2007 will be divisible by 2000, hence by 100 also, so the last 7 addends remain:
R ≡ 2000^2007 + 2001^2007 + . . + 2007^2007 ≡
≡ 0 + 1^2007 + 2^2007 + . . + 7^2007 ≡
≡ 1^7 + 2^7 + 3^7 + 4^7 + 5^7 + 6^7 + 7^7 ≡
≡ 1 + 28 + 87 + 84 + 25 +36 + 43 ≡ 4 (mod 100), so the answer is indeed 04. I used above that for a = 1,2,3,4,5,6,7 we have
a^(n + 20) ≡ a^n (mod 100), /n=1,2, . ./

EDIT: well I'm not second but already third in the company, typing all that above.
EDIT 2: now I see scythian also has supplied a correct explanation.

Answer 2

The last 2 digits of this sum are 04.
First, we are looking for the answer mod 100,
and since the base repeats every 100 times,
our sum is equal to
1^2007 + 2^2007 + ... 7^2007 +
20*(1^2007 + 2^2007 + ... + 99^2007).
Next, a^2007 = a^7(mod 100) for any a.
Here's an outline of the proof:
Again it suffices to prove this for a < 100.
Suppose (a,100) = 1.
We calculate φ(100) = 40
and, by Euler's theorem, a ^40 = 1(mod 100)
So a^2000 = 1(mod 100) and a^2007 = a^7(mod 100).
Also, 2^2000 = 76(mod 100) and 2^7 = 28(mod 100)
But 76 *28 = 28(mod 100), so the result holds
for 2. Also 5^2000 = 25(mod 100) and 5^7 = 25(mod 100)
so the result holds for 5. Also 10^2007 = 0(mod 100)
and the remaining cases are easily checked.
The upshot of all this is that our sum
reduces to evaluating.
1^7 + 2^7 + ... + 7^7+ 20(1^7 + 2^7 + ... + 99^7)
= 4 + 20(1^7 + 2^7 + 99^7) (mod 100).
Now, it suffices to find this last sum mod 5, for if we
call this sum s,
then our answer will be 4 + 20(s + 5m) = 4 + 20s(mod 100).
But this last sum is just 20(1^7 + 2^7 + 3^7 + 4^7) = 0(mod 5).
Consequently, the last 2 digits of the sum are 04.
Great problem! It really had me going for a while!

Answer 3

04

First, consider the sum 1^2007 + ... + 100^2007. Rearrange into pairs as follows: (1^2007 + 99^2007) + (2^2007 + 98^2007) + ....+ 50^2007 + 100^2007. Each pair can be factorized, with one factor being 100, while 50^2007 and 100^2007 obviously both end in 00. Thus any sum 1^2007 + .... + (100A)^2007 ends in 00, so that we need only look at the sum 1^2007 + ... + 7^2007.

Answer 4

At this moment, I don't seem to have the solution.

However, if you are asking the last two digits of the sum of the last digits of each term, then this can be done:

Observed that all the ten digits have last digit of their powers repeating cyclically at most a period of 4 (some only 2 or 1).
Since 2007 mod 4 = 3, thus the last digit of each terms can be determined to be:
0 1 6 1 6 5 6 1 6 1
Thus the sum of last digit of each term of the sum is
200 * 1 + 200 * 6 + ... + 200 *1 + 200 * 0 + (1+6+1+6+5+6+1)
= 200 * (1+6+1+6+5+6+1+6+1+0) + 26
= 200 * 33 + 26
= 6600 + 26
So, last 2 digits are 26

Now, looking at this, hmmm...
Let n1..n10 are the last two digits of numbers ending with 1..9,0
If we look at the sum of the last two digits of each term of the sum,
200 * n1 + 200 * n2 + ... + 200 * n9 + 200 * n10 + (n1+n2+n3+n4+n5+n6+n7)
= 200 * (n1+n2+n3+n4+n5+ n6+n7+n8+n9+n10) + (n1+n2+n3+n4+n5+n6+n7)
= 100m + (n1+n2+n3+n4+n5+n6+n7)
where m is some integer.
So, the last two digits depends on the sum of (n1+n2+n3+n4+n5+n6+n7)
Oooops...

Let see, what about we consider 1 to 7 separately:
Let x1..x7 be the last two digits of the terms 1^2007 to 7^2007
sum of the last two digits of each term of the sum,
(x1+x2+x3+x4+x5+x6+x7) + 200 * n8 + 200 * n9 + 200 * n10 + ... + 200 * n7
= (x1+x2+x3+x4+x5+x6+x7) + 200 * (n8+n9+n10+n1+n2+ n3+n4+n5+n6+n7)
= (x1+x2+x3+x4+x5+x6+x7) + 100k
where k is some integer.

By studying the pattern of the 2nd last digit for numbers 1 to 7, we observed that all repeat after every 20 increments in power. i.e. last 2 digits of 3^21 = 3^41 = 3^(20z+1) = 03
the last two digits for
1^2007 and 1^7 is 00
2^2007 and 2^7 is 28
3^2007 and 3^7 is 87
4^2007 and 4^7 is 84
5^2007 and 5^7 is 25
6^2007 and 6^7 is 36
7^2007 and 7^7 is 43

So, answer is 04 (the above sum is 304).

On hind thoughts, we can also consider the powers of 2001 to 2007, which are (200 + 1) to (200 + 7), and when raised to the power of 2007, only the last term in the expansion matters, since all the other terms have the factor 200 in them.

In fact, we also observed that 8, 9 and 0 also repeat for every 20 increments in power, and the last two digits of their 7th powers are 52, 69, and 00 respectively.

Now, still thinking why 20 increments? Multiple of 4 increments since the last digit repeat for every 4 increments. Then, why 5 * 4? (Note: 20 is not the least number increments before repetition for all. E.g. the obvious ones are 10, 5)

For info, the number of increments in power till repeat, for 1, 2, 3, ... 21, ... are:
1, 20, 30, 10, 1, 5, 4, 20, 10, 1, 10, 20, 10, 2, 6, 20, 4, 10, 1, 5, ...

Answer 5

incredibly that summation would not equivalent one different than while n=a million . As n strategies infinity it diverges, meaning the summation additionally is going to infinity (or -infinity summation (a million/n) from i=a million to n = a million/a million + a million/2 + a million/3 + a million/4........ notice we are over 2 with in simple terms 4 words. we can coach that it diverges simply by fact the necessary of a million/n is: ln(n) + C and ln(n) strategies infinity as as n strategies infinity. right that could be a sequence that strategies a million for you. a million/2 + a million/4 + a million/8 + a million/sixteen + a million/32 ....a million/(2^n) we would show this sequence as summation a million/2^n from i=a million to n or summation 2^(-n) from i=a million to n