We use Euler's totient theorem, which states that if a and n are coprime, then a^( (n)) 1 n. Here, a=19 and n=100. First, we calculate (100). Since 100 = 2² · 5², (100) = 100(1 - 1/2)(1 - 1/5) = 40.

We use Euler's totient theorem, which states that if a and n are coprime, then a^( (n)) 1 n. Here, a=19 and n=100. First, we calculate (100). Since 100 = 2² · 5², (100) = 100(1 - 1/2)(1 - 1/5) = 40.

Now we need to reduce the exponent 1919 modulo (100) = 40. Dividing 1919 by 40, we get 1919 = 40 × 47 + 39. So, 1919 39 40. However, using 39 directly can be cumbersome. We can use 39 -1 40 for easier calculation.

Now we need to reduce the exponent 1919 modulo (100) = 40. Dividing 1919 by 40, we get 1919 = 40 × 47 + 39. So, 1919 39 40. However, using 39 directly can be cumbersome. We can use 39 -1 40 for easier calculation.

According to Euler's totient theorem, 19^(1919) 19^(39) 100. Since 39 -1 40, we can write 19^(39) 19^(-1) 100. This means we need to find the multiplicative inverse of 19 modulo 100.

The product of the last two digits of $(1919)^{1919}$ is _____

Answer: 63

Step-by-step solution

Step 1: Find the last two digits of the base

To find the last two digits of $(1919)^{1919}$ , we only need to consider the last two digits of the base, which is $19$ . So, we need to find the last two digits of $(19)^{1919}$ .

Step 2: Find the Euler's totient function for 100

We use Euler's totient theorem, which states that if $a$ and $n$ are coprime, then $a^{\phi(n)} \equiv 1 \pmod{n}$ . Here, $a=19$ and $n=100$ . First, we calculate $\phi(100)$ . Since $100 = 2^2 \cdot 5^2$ , $\phi(100) = 100(1 - 1/2)(1 - 1/5) = 40$ .

Step 3: Reduce the exponent modulo $\phi(100)$

Now we need to reduce the exponent $1919$ modulo $\phi(100) = 40$ . Dividing $1919$ by $40$ , we get $1919 = 40 \times 47 + 39$ . So, $1919 \equiv 39 \pmod{40}$ . However, using $39$ directly can be cumbersome. We can use $39 \equiv -1 \pmod{40}$ for easier calculation.

Step 4: Apply Euler's totient theorem

According to Euler's totient theorem, $19^{1919} \equiv 19^{39} \pmod{100}$ . Since $39 \equiv -1 \pmod{40}$ , we can write $19^{39} \equiv 19^{-1} \pmod{100}$ . This means we need to find the multiplicative inverse of $19$ modulo $100$ .

Step 5: Find the multiplicative inverse of 19 modulo 100

We need to find an integer $x$ such that $19x \equiv 1 \pmod{100}$ . We can test values or use the extended Euclidean algorithm. By testing, we find that $19 \times 79 = 1501$ , which leaves a remainder of $1$ when divided by $100$ . Thus, $19^{-1} \equiv 79 \pmod{100}$ .

Step 6: Determine the last two digits and their product

The last two digits of $(1919)^{1919}$ are $79$ . The last digit is $9$ and the second to last digit is $7$ . The product of these digits is $7 \times 9 = 63$ .

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The product of the last two digits of (1919)1919(1919)^{1919}(1919)1919 is _____