Find the last 3 digits of $2015^{18}+2015^{19}+2015^{20}+...+2015^{2018}$.

Question

Obvious I would find the sum of the numbers mod 100, but how would I handle this when the powers are so large? Calculating one already seemds ridiculous.

Answer 1

Consider that since $2015$ to any power will have a final digit of $5$, since the last digit of $2015$ is $5$. (You can check the first few of these using a calculator)

Also consider that you have an even number of elements. Anything ending in $5$ multiplied by an even number will have a last digit of $0$. ($5 \times 2 = 10$, $15 \times 3 = 45$, $25 \times 4 = 100$). You can therefore conclude that the last digit will be a $0$.

You should be able to make conclusions about the next two digits using a similar method of generalizations.

Answer 2

Modulo $1000$, $15^3=15^5$; so, for $k\geq 3$, $$15^k=\begin{cases}15^3,&\ k\ \text{ odd}\\ \ \\ 15^4,&\ k\ \text{ even}\end{cases}$$ Then, still modulo $1000$, $$ S=\sum_{k=0}^{2000} 2015^{18+k}=\sum_{k=0}^{2000} 2015^{k+2} =15^{2}+1000\times15^3+1000\times 15^4=15^2=625. $$ So the last three digits are 625.

Answer 3

Note that the last 3 digit for $2015^2$ are the same for $15^2=225$ and then they repeat periodically

• $15^3=...375$
• $15^4=...625$
• $15^5=...375$
• ...

then

$$2015^{18}+2015^{19}+2015^{20}+…+2015^{2018}\equiv1001\cdot625+1000\cdot 375 \pmod{1000}\equiv \Large{625}$$

Answer 4

We need to work modulo $1000 = 2^3 5^3$. Since $2015$ is divisible by $5$, the sum is congruent to $0$ modulo $5^3.$ Reducing the sum modulo $2^3$ yields

$$1 -1 + 1 - 1 +\cdots -1+1 \equiv 1 \pmod{8},$$

since $2015 \equiv 15 \equiv -1 \pmod{8}$. So you need to use Chinese remainder theorem to solve the system

$$x\equiv 0 \pmod{125}$$ $$x\equiv 1 \pmod{8}$$

which yields $x \equiv 625 \pmod{1000}.$

Answer 5

As $2015\equiv15\pmod{1000},$ we need $\displaystyle\sum_{r=18}^{2018}15^r\pmod{1000}$

As $(15^n,1000)=5^3$ for $n\ge3,$

let us find $\displaystyle S=\sum_{r=18-3}^{2018-3}15^r\pmod{1000/5^3}$

As $\displaystyle15\equiv-1\pmod8,S\equiv\sum_{r=18-3}^{2018-3}(-1)^r\equiv-1\pmod8$

$\displaystyle15^3S\equiv15^3\cdot-1\pmod{15^3\cdot8}\equiv-15^3\pmod{8\cdot5^3}$ as $5^3|15^3$

Now $-15^3=-225\cdot15=-3375\equiv-375\equiv1000-375\pmod{1000}$ as $S>0$