N = (98^13 + 99^13+ 101^13 + 102^13). Find the remainder when N is divided by 25.

To find the remainder when \( N = 98^{13} + 99^{13} + 101^{13} + 102^{13} \) is divided by 25, we can calculate the remainders of each term modulo 25 and then add them up.

### Step 1: Calculate the remainders of the bases modulo 25.
\[
98 \mod 25 = 98 - 4 \times 25 = 98 - 100 = -2 \equiv 23 \pmod{25}
\]
\[
99 \mod 25 = 99 - 4 \times 25 = 99 - 100 = -1 \equiv 24 \pmod{25}
\]
\[
101 \mod 25 = 101 - 4 \times 25 = 101 - 100 = 1 \pmod{25}
\]
\[
102 \mod 25 = 102 - 4 \times 25 = 102 - 100 = 2 \pmod{25}
\]

### Step 2: Compute the powers modulo 25.
Since we are interested in the powers modulo 25, we can use properties of modular arithmetic. Notice that if \( a \equiv b \pmod{25} \), then \( a^{13} \equiv b^{13} \pmod{25} \).

\[
23^{13} \equiv (-2)^{13} \equiv -2^{13} \pmod{25}
\]
\[
24^{13} \equiv (-1)^{13} \equiv -1 \pmod{25}
\]
\[
1^{13} \equiv 1 \pmod{25}
\]
\[
2^{13} \pmod{25}
\]

Now, we compute these powers modulo 25:
- \( 2^{13} \) modulo 25:
\[
2^2 = 4,\quad 2^4 = 16,\quad 2^8 \equiv 16 \times 16 = 256 \equiv 6 \pmod{25}
\]
\[
2^{12} = 6 \times 16 = 96 \equiv 21 \pmod{25},\quad 2^{13} = 21 \times 2 = 42 \equiv 17 \pmod{25}
\]

So:
\[
23^{13} \equiv -17 \equiv 8 \pmod{25}
\]
\[
24^{13} \equiv -1 \equiv 24 \pmod{25}
\]
\[
1^{13} \equiv 1 \pmod{25}
\]
\[
2^{13} \equiv 17 \pmod{25}
\]

### Step 3: Sum the results modulo 25.
\[
N \equiv 8 + 24 + 1 + 17 \pmod{25}
\]
\[
N \equiv 50 \pmod{25}
\]

Since 50 is divisible by 25:
\[
N \equiv 0 \pmod{25}
\]

### Final Answer:
The remainder when \( N \) is divided by 25 is **0**.

Would you like a more detailed breakdown of any part of this solution, or do you have any further questions?

Here are some additional questions related to this concept:
1. How would the calculation change if one of the numbers was raised to a different power?
2. What are other methods for calculating large powers modulo a number?
3. How does the result change if instead of dividing by 25, we divide by 24?
4. Can you find the remainder when the sum \( 98^{14} + 99^{14} + 101^{14} + 102^{14} \) is divided by 25?
5. How can you generalize this problem to \( N = (x_1^k + x_2^k + x_3^k + x_4^k) \mod m \)?

**Tip**: When computing large powers modulo a number, using properties like Fermat's Little Theorem can simplify the process significantly, especially with prime moduli.

Learn how to calculate the remainder when the sum N = (98^13 + 99^13 + 101^13 + 102^13) is divided by 25. Step-by-step solution using modular arithmetic and powers modulo 25. Includes detailed calculations and application of Fermat's Little Theorem.

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