Math Problem Statement

can 3 to the power of 444 + 4 to the power of 333 be divide by 5

Solution

We are asked if 3444+43333^{444} + 4^{333} is divisible by 5. To check this, we can use modular arithmetic.

Step-by-Step Solution:

We want to calculate 3444+4333mod53^{444} + 4^{333} \mod 5, i.e., the remainder when 3444+43333^{444} + 4^{333} is divided by 5.

Step 1: Find 3444mod53^{444} \mod 5

Using Fermat's Little Theorem, we know that if aa is not divisible by pp, then: ap11modpa^{p-1} \equiv 1 \mod p For p=5p = 5, this tells us 341mod53^4 \equiv 1 \mod 5.

So, 3444=(34)111×301111×11mod53^{444} = (3^4)^{111} \times 3^0 \equiv 1^{111} \times 1 \equiv 1 \mod 5.

Step 2: Find 4333mod54^{333} \mod 5

Similarly, we can use Fermat’s Little Theorem for 44 as well. Since 441mod54^4 \equiv 1 \mod 5, we know: 4333=44×83+1=(44)83×41183×44mod54^{333} = 4^{4 \times 83 + 1} = (4^4)^{83} \times 4^1 \equiv 1^{83} \times 4 \equiv 4 \mod 5

Step 3: Add the results modulo 5

Now we compute the sum modulo 5: 3444+43331+450mod53^{444} + 4^{333} \equiv 1 + 4 \equiv 5 \equiv 0 \mod 5

Conclusion:

Yes, 3444+43333^{444} + 4^{333} is divisible by 5.


Would you like more details or have any questions?

Here are 5 related questions to explore further:

  1. How does Fermat's Little Theorem help in modular arithmetic?
  2. Can we generalize this method for any prime divisor?
  3. What happens if we try dividing by numbers other than 5?
  4. How would the solution change if the exponents were different?
  5. Can this method be used for larger bases and powers efficiently?

Tip: Modular arithmetic often simplifies large exponentiation problems by reducing the problem into manageable cycles or periodic patterns.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic
Exponentiation

Formulas

Fermat's Little Theorem: a^(p-1) ≡ 1 (mod p) for a prime p

Theorems

Fermat's Little Theorem

Suitable Grade Level

Grades 11-12