Find the reminders when
(a) 2^14 divisible by 17
(b) 11^35 divisible by 13
(c) 1! + 2! + 3! + · · · + 100! divisible by 16.
(d) Calculate 3385(mod 479).      detail explaination

Let's solve each part systematically.

### Part (a) \( 2^{14} \mod 17 \)

We want to find the remainder when \( 2^{14} \) is divided by 17, i.e., \( 2^{14} \mod 17 \).

This can be solved using modular exponentiation.

**Step 1: Apply Fermat's Little Theorem.**

Since 17 is prime, Fermat’s Little Theorem tells us that:
\[
a^{p-1} \equiv 1 \mod p \quad \text{for any } a \text{ such that } \gcd(a, p) = 1
\]
Here, \( p = 17 \), and Fermat’s Theorem implies:
\[
2^{16} \equiv 1 \mod 17
\]
Thus, \( 2^{14} \mod 17 \) can be computed by breaking it down using successive squaring.

**Step 2: Compute \( 2^{14} \mod 17 \) directly using successive squaring.**
\[
2^2 = 4, \quad 2^4 = 16 \mod 17 = -1, \quad 2^8 = (-1)^2 = 1 \mod 17
\]
Thus:
\[
2^{14} = 2^{8} \cdot 2^4 \cdot 2^2 = 1 \cdot (-1) \cdot 4 = -4 \mod 17
\]
\[
\text{Therefore, } 2^{14} \mod 17 = 13.
\]

### Part (b) \( 11^{35} \mod 13 \)

We want to find the remainder when \( 11^{35} \) is divided by 13, i.e., \( 11^{35} \mod 13 \).

**Step 1: Apply Fermat's Little Theorem.**

Since 13 is prime, Fermat’s Little Theorem tells us that:
\[
11^{12} \equiv 1 \mod 13
\]
Thus, we can reduce the exponent 35 mod 12:
\[
35 \mod 12 = 11
\]
Thus:
\[
11^{35} \mod 13 = 11^{11} \mod 13
\]

**Step 2: Compute \( 11^{11} \mod 13 \) using successive squaring.**
\[
11^2 = 121 \mod 13 = 4
\]
\[
11^4 = (11^2)^2 = 4^2 = 16 \mod 13 = 3
\]
\[
11^8 = (11^4)^2 = 3^2 = 9 \mod 13
\]
Thus:
\[
11^{11} = 11^8 \cdot 11^2 \cdot 11 = 9 \cdot 4 \cdot 11 = 396 \mod 13 = 6.
\]
\[
\text{Thus, } 11^{35} \mod 13 = 6.
\]

### Part (c) \( 1! + 2! + 3! + \cdots + 100! \mod 16 \)

We are tasked with finding the remainder when the sum of factorials from 1 to 100 is divided by 16.

**Step 1: Recognize that \( n! \mod 16 = 0 \) for \( n \geq 16 \).**

This is because \( n! \) for \( n \geq 16 \) contains 16 as a factor, so for \( n \geq 16 \), \( n! \mod 16 = 0 \).

Thus, we only need to compute the sum of the factorials from \( 1! \) to \( 15! \mod 16 \).

**Step 2: Compute factorials modulo 16 for \( 1! \) through \( 15! \):**
\[
1! = 1 \mod 16 = 1
\]
\[
2! = 2 \mod 16 = 2
\]
\[
3! = 6 \mod 16 = 6
\]
\[
4! = 24 \mod 16 = 8
\]
\[
5! = 120 \mod 16 = 8
\]
\[
6! = 720 \mod 16 = 0
\]
Since \( 6! \) and beyond are divisible by 16, we only need to sum \( 1! \) through \( 5! \).

**Step 3: Sum the values:**
\[
1 + 2 + 6 + 8 + 8 = 25 \mod 16 = 9.
\]
Thus:
\[
1! + 2! + 3! + \cdots + 100! \mod 16 = 9.
\]

### Part (d) Calculate \( 3385 \mod 479 \)

We need to find the remainder when 3385 is divided by 479, i.e., \( 3385 \mod 479 \).

**Step 1: Perform the division:**
\[
3385 \div 479 = 7 \quad \text{(integer part of the division)}
\]
\[
3385 - (7 \times 479) = 3385 - 3353 = 32.
\]
Thus:
\[
3385 \mod 479 = 32.
\]

---

### Summary of Results:
(a) \( 2^{14} \mod 17 = 13 \)  
(b) \( 11^{35} \mod 13 = 6 \)  
(c) \( 1! + 2! + \cdots + 100! \mod 16 = 9 \)  
(d) \( 3385 \mod 479 = 32 \)

Would you like further details or have any questions?

---

### Related Questions:
1. How does Fermat's Little Theorem simplify modular exponentiation?
2. What is modular exponentiation, and why is it useful?
3. Can successive squaring be used for any large exponentiation problem?
4. How do factorials behave under modulo arithmetic?
5. What are other techniques for solving modulus problems quickly?

### Tip:  
For large exponent problems, always look for ways to reduce the exponent using properties like Fermat’s Little Theorem or Euler’s Theorem to simplify calculations!

Explore solutions to modular arithmetic problems and factorials using Fermat's Little Theorem and modular exponentiation. Learn how to compute remainders efficiently for large powers and factorials under modulus constraints.

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