What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?

Question

What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?

My work:

1^5 ends with 1. 2^5 ends with 2. 3^5 ends with 3. And so on. Do I simply add the ending digits to get my answer?

Answer 1

Hint:

$r^{2n+1}+(m-r)^{2n+1}\equiv0\pmod m$ for integers $r,n,m$

Set $m=100, 2n+1=5$

$r=1,2,3\cdots,49$

So, we are left with $50^5\pmod{100}\equiv?$

Answer 2

Write the sum as $$S=(1^5+99^5)+(2^5+98^5)+(3^5+97^5)+.....+ (49^5+51^5)+50^n$$ Then $100=1+99,=2+98= 3+97,...$ is the common factor of all the baraketed terms. Because $(a^5+b^5)=(a+b)M(a,b)$. Next, $50^5$ is also divisible by 100. So the last two digits in $S$ are $00$.

Answer 3

$1^5+99^5=(1+99)(1-99+99^2-99^3+99^4)=100(\text{positive number})$ what can you see?? Also $2^5+98^5=(100)(2^4-2^3(98)+2^2(98)^2-2(98)^3+98^4)$

My solution is a special case of lab's

Answer 4

Using Faulhaber's formula: $$S_{99}=\frac{4a^3-a^2}{4}=a^2\cdot \frac{4a-1}{4}=\frac{9900^2}{4}\cdot \frac{2\cdot 9900-1}{3}=33\cdot 25\cdot 9900\cdot 19799,$$ where: $$a=\frac{n(n+1)}{2}.$$ Hence, the answer is $00$.