Finding the Remainder for \(111^{222} + 222^{333} + 333^{444}\) divided by 5

This solution explains how to find the remainder when the expression \(111^{222} + 222^{333} + 333^{444}\) is divided by 5. We will use the principles of modular arithmetic to simplify this calculation.

Understanding Modular Arithmetic and Remainders

Modular arithmetic helps us find remainders. The notation '\(a \equiv b \pmod{n}\)' signifies that \(a\) and \(b\) leave the same remainder when divided by \(n\). A crucial property for solving this problem is that if \(a \equiv b \pmod{n}\), then \(a^k \equiv b^k \pmod{n}\) for any positive integer \(k\). This property allows us to reduce large exponents significantly.

Our goal is to determine the remainder of the sum \(111^{222} + 222^{333} + 333^{444}\) when divided by 5. We can achieve this by calculating the remainder of each term (\(111^{222}\), \(222^{333}\), and \(333^{444}\)) individually when divided by 5, and then summing these remainders (also modulo 5).

Calculating the Remainder for Each Term

Remainder of \(111^{222}\) divided by 5

• First, let's find the remainder of the base, 111, when divided by 5. Using division, \(111 = 5 \times 22 + 1\). The remainder is 1. In modular arithmetic notation: \(111 \equiv 1 \pmod{5}\).
• Now, we apply this congruence to the exponentiation: \(111^{222} \equiv 1^{222} \pmod{5}\).
• Since 1 raised to any power equals 1, the calculation simplifies to: \(111^{222} \equiv 1 \pmod{5}\).

Remainder of \(222^{333}\) divided by 5

• First, find the remainder of the base, 222, when divided by 5. \(222 = 5 \times 44 + 2\). The remainder is 2. So, \(222 \equiv 2 \pmod{5}\).
• Next, we need to find the remainder of \(2^{333}\) when divided by 5. To do this, we examine the pattern of remainders for powers of 2 modulo 5:
  • \(2^1 \equiv 2 \pmod{5}\)
  • \(2^2 \equiv 4 \pmod{5}\)
  • \(2^3 \equiv 8 \equiv 3 \pmod{5}\)
  • \(2^4 \equiv 16 \equiv 1 \pmod{5}\)
  • \(2^5 \equiv 32 \equiv 2 \pmod{5}\)
  The remainders (2, 4, 3, 1) repeat every 4 powers. The cycle length is 4.
• To determine the remainder of \(2^{333}\), we find the position in this cycle by calculating the remainder of the exponent, 333, when divided by the cycle length, 4. \(333 = 4 \times 83 + 1\). The remainder is 1.
• Since the remainder of the exponent is 1, \(2^{333}\) will have the same remainder as the first term in the cycle, which is \(2^1\). \(2^{333} \equiv 2^1 \pmod{5}\). \(2^{333} \equiv 2 \pmod{5}\).
• Therefore, the remainder for this term is: \(222^{333} \equiv 2 \pmod{5}\).

Remainder of \(333^{444}\) divided by 5

• Find the remainder of the base, 333, when divided by 5. \(333 = 5 \times 66 + 3\). The remainder is 3. So, \(333 \equiv 3 \pmod{5}\).
• Now, we find the remainder of \(3^{444}\) when divided by 5. Let's look at the pattern of remainders for powers of 3 modulo 5:
  • \(3^1 \equiv 3 \pmod{5}\)
  • \(3^2 \equiv 9 \equiv 4 \pmod{5}\)
  • \(3^3 \equiv 27 \equiv 2 \pmod{5}\)
  • \(3^4 \equiv 81 \equiv 1 \pmod{5}\)
  • \(3^5 \equiv 243 \equiv 3 \pmod{5}\)
  The remainders (3, 4, 2, 1) also repeat every 4 powers. The cycle length is 4.
• To find the position in this cycle for \(3^{444}\), we calculate the remainder of the exponent, 444, when divided by the cycle length, 4. \(444 = 4 \times 111 + 0\). The remainder is 0.
• When the remainder of the exponent is 0, it corresponds to the last element in the cycle (equivalent to the exponent being the cycle length itself). Therefore, we use the remainder of \(3^4\). \(3^{444} \equiv 3^4 \pmod{5}\). \(3^{444} \equiv 1 \pmod{5}\).
• Hence, the remainder for this term is: \(333^{444} \equiv 1 \pmod{5}\).

Combining the Remainders

We have successfully calculated the remainders for each part of the expression:

• \(111^{222} \equiv 1 \pmod{5}\)
• \(222^{333} \equiv 2 \pmod{5}\)
• \(333^{444} \equiv 1 \pmod{5}\)

Now, we add these individual remainders together:

Sum of remainders = \(1 + 2 + 1 = 4\).

Finding the Final Remainder

The final step is to find the remainder of this sum (4) when divided by 5.

Since \(4 < 5\), the remainder when 4 is divided by 5 is simply 4.

Using modular arithmetic notation: \(4 \equiv 4 \pmod{5}\).

Therefore, the remainder of the original expression \(111^{222} + 222^{333} + 333^{444}\) when divided by 5 is 4.