Question

The remainder when $2017^{2018} + 2018^{2019} + 2019^{2020}$ is divided by 5 is

Answer 1

2

Answer 2

To find the remainder when $2017^{2018} + 2018^{2019} + 2019^{2020}$ is divided by 5, we can use modular arithmetic. We need to find the remainder of each term when divided by 5 and then sum these remainders.

Step 1: Simplify the base of each term modulo 5.

• For the first term, $2017 \pmod{5}$: The last digit of 2017 is 7. $7 \div 5$ gives a remainder of 2. So, $2017 \equiv 2 \pmod{5}$.
• For the second term, $2018 \pmod{5}$: The last digit of 2018 is 8. $8 \div 5$ gives a remainder of 3. So, $2018 \equiv 3 \pmod{5}$.
• For the third term, $2019 \pmod{5}$: The last digit of 2019 is 9. $9 \div 5$ gives a remainder of 4. So, $2019 \equiv 4 \pmod{5}$.

Now, the expression becomes finding the remainder of $2^{2018} + 3^{2019} + 4^{2020}$ when divided by 5.

Step 2: Calculate the remainder for each power term.

• For $2^{2018} \pmod{5}$: Let's look at the cycle of powers of 2 modulo 5: $2^1 \equiv 2 \pmod{5}$

  $2^2 \equiv 4 \pmod{5}$

  $2^3 \equiv 8 \equiv 3 \pmod{5}$

  $2^4 \equiv 16 \equiv 1 \pmod{5}$

  The cycle length is 4. We need to find the remainder of the exponent $2018$ when divided by 4: $2018 \div 4 = 504$ with a remainder of 2. So, $2^{2018} \equiv 2^2 \equiv 4 \pmod{5}$.

• For $3^{2019} \pmod{5}$: Let's look at the cycle of powers of 3 modulo 5: $3^1 \equiv 3 \pmod{5}$

  $3^2 \equiv 9 \equiv 4 \pmod{5}$

  $3^3 \equiv 27 \equiv 2 \pmod{5}$

  $3^4 \equiv 81 \equiv 1 \pmod{5}$

  The cycle length is 4. We need to find the remainder of the exponent $2019$ when divided by 4: $2019 \div 4 = 504$ with a remainder of 3. So, $3^{2019} \equiv 3^3 \equiv 2 \pmod{5}$.

• For $4^{2020} \pmod{5}$: We can notice that $4 \equiv -1 \pmod{5}$. So, $4^{2020} \equiv (-1)^{2020} \pmod{5}$. Since 2020 is an even number, $(-1)^{2020} = 1$. Therefore, $4^{2020} \equiv 1 \pmod{5}$.

Step 3: Sum the individual remainders.

Now, we add the remainders found in Step 2: $2017^{2018} + 2018^{2019} + 2019^{2020} \equiv 4 + 2 + 1 \pmod{5}$

$4 + 2 + 1 = 7$

Step 4: Find the remainder of the sum modulo 5.

$7 \pmod{5}$

$7 \div 5 = 1$ with a remainder of 2.

So, $7 \equiv 2 \pmod{5}$.

The remainder when $2017^{2018} + 2018^{2019} + 2019^{2020}$ is divided by 5 is 2.