Question

The product of the last 2 digits of $ (1919)^{1919} $ is:

• A. 56
• B. 63
• C. 45
• D. 54

Hint:

For problems involving powers and last digits, use modular arithmetic to simplify the calculations.

Answer 1

The Correct Option is B

Step 1: Simplify the Problem
The objective is to determine the final two digits of \(1919^{1919}\), which is mathematically equivalent to calculating \( 1919^{1919} \mod 100 \). Since \(1919 \mod 100 = 19\), the problem simplifies to finding \( 19^{1919} \mod 100 \).
Step 2: Apply Euler's Theorem
Euler's theorem states that for coprime integers \(a\) and \(n\), \( a^{\phi(n)} \equiv 1 \mod n \), where \(\phi(n)\) is Euler's totient function. For \(n = 100\), \( \phi(100) = 100 \times \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{5}\right) = 40 \). As 19 and 100 are coprime, \( 19^{40} \equiv 1 \mod 100 \).
Step 3: Reduce the Exponent
The exponent \(1919\) can be expressed in terms of \(\phi(100) = 40\) as \( 1919 = 40 \times 47 + 39 \). Therefore, \( 19^{1919} = 19^{40 \times 47 + 39} = (19^{40})^{47} \times 19^{39} \equiv 1^{47} \times 19^{39} \equiv 19^{39} \mod 100 \).
Step 4: Calculate \(19^{39} \mod 100\)
Using exponentiation by squaring: \[ \begin{aligned} 19^1 &\equiv 19 \mod 100 \\ 19^2 &\equiv 361 \equiv 61 \mod 100 \\ 19^4 &\equiv (19^2)^2 \equiv 61^2 \equiv 3721 \equiv 21 \mod 100 \\ 19^8 &\equiv (19^4)^2 \equiv 21^2 \equiv 441 \equiv 41 \mod 100 \\ 19^{16} &\equiv (19^8)^2 \equiv 41^2 \equiv 1681 \equiv 81 \mod 100 \\ 19^{32} &\equiv (19^{16})^2 \equiv 81^2 \equiv 6561 \equiv 61 \mod 100 \\ \end{aligned} \] Expressing 39 as a sum of powers of 2: \( 39 = 32 + 4 + 2 + 1 \). Thus, \( 19^{39} \equiv 19^{32} \times 19^4 \times 19^2 \times 19^1 \equiv 61 \times 21 \times 61 \times 19 \mod 100 \). Performing the multiplications sequentially: \[ \begin{aligned} 61 \times 21 &\equiv 1281 \equiv 81 \mod 100 \\ 81 \times 61 &\equiv 4941 \equiv 41 \mod 100 \\ 41 \times 19 &\equiv 779 \equiv 79 \mod 100 \\ \end{aligned} \] The final two digits are \(\boxed{79}\).
Step 5: Compute the Product
The digits are 7 and 9; their product is \( 7 \times 9 = \boxed{63} \).
Final Answer
The correct option is \(\boxed{(2) 63}\).