Let n = 3¹⁰⁰, then for n the value of ten’s digits is ?

To find the ten's digit of the number \( n = 3^{100} \), we need to determine the last two digits of this large number. The ten's digit is the second last digit in the number, and we can achieve this by calculating \( 3^{100} \mod 100 \). This will give us the last two digits directly.

Using the Chinese Remainder Theorem

To simplify our calculations, we can use the Chinese Remainder Theorem (CRT). We will find \( 3^{100} \mod 4 \) and \( 3^{100} \mod 25 \), and then combine these results.

Step 1: Calculate \( 3^{100} \mod 4 \)

Since \( 3 \equiv -1 \mod 4 \), we can simplify:

• \( 3^{100} \equiv (-1)^{100} \mod 4 \)
• \( (-1)^{100} = 1 \)

Thus, \( 3^{100} \mod 4 = 1 \).

Step 2: Calculate \( 3^{100} \mod 25 \)

For this part, we can use Euler's theorem. First, we find \( \phi(25) \), where \( \phi \) is the Euler's totient function:

• \( \phi(25) = 25 \cdot (1 - \frac{1}{5}) = 20 \)

According to Euler's theorem, since \( \gcd(3, 25) = 1 \), we have:

• \( 3^{20} \equiv 1 \mod 25 \)

Now, we need to compute \( 3^{100} \mod 25 \):

• \( 100 \mod 20 = 0 \)
• Thus, \( 3^{100} \equiv (3^{20})^5 \equiv 1^5 \equiv 1 \mod 25 \)

Combining Results

Now we have:

• \( 3^{100} \equiv 1 \mod 4 \)
• \( 3^{100} \equiv 1 \mod 25 \)

Using the CRT, since both congruences give us the same result, we conclude:

• \( 3^{100} \equiv 1 \mod 100 \)

Final Result

The last two digits of \( 3^{100} \) are 01. Therefore, the ten's digit of \( n = 3^{100} \) is:

• **0**

In summary, the ten's digit of \( 3^{100} \) is 0, as we derived through modular arithmetic and the Chinese Remainder Theorem. This method is efficient for handling large exponents and helps us avoid direct computation of the enormous number.