To prove that $10^{25} - 7$ is divisible by 3, we can use the divisibility rule for 3, which states that a number is divisible by 3 if the sum of its digits is divisible by 3. However, instead of computing all digits of $10^{25} - 7$, we can use modular arithmetic for efficiency.

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Step-by-Step Proof

1. Modulo 3 Analysis:

   • Any integer $a$ is congruent to its remainder when divided by 3. This is written as: $a \equiv r \pmod{3},$ where $r$ is the remainder.

2. Modulo of $10$ raised to a power:

   • We first evaluate $10 \mod 3$: $10 \div 3 = 3 \cdot 3 + 1 \quad \text{(remainder 1)}.$ Therefore: $10 \equiv 1 \pmod{3}.$

   • Using the property of exponents in modular arithmetic: $10^n \equiv 1^n \equiv 1 \pmod{3}.$

   • Hence: $10^{25} \equiv 1 \pmod{3}.$

3. Subtracting 7:

   • Now consider $10^{25} - 7$: $10^{25} - 7 \equiv 1 - 7 \pmod{3}.$

   • Simplify the subtraction: $1 - 7 = -6.$

   • Reduce $-6 \mod 3$: $-6 \div 3 = -2 \quad \text{(remainder 0)}.$

   • Thus: $10^{25} - 7 \equiv 0 \pmod{3}.$

4. Conclusion:

   • Since $10^{25} - 7 \equiv 0 \pmod{3}$, it is divisible by 3 without a remainder.

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Final Answer:

The number $10^{25} - 7$ is divisible by 3.

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Would you like me to explain any step in detail?

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Related Questions:

1. What is the significance of modular arithmetic in divisibility tests?
2. Can you prove $10^{n} - 1$ is divisible by 3 for any positive integer $n$?
3. How does the divisibility rule for 3 apply to large numbers?
4. What is the remainder of $10^{50} - 7$ when divided by 3?
5. How can you generalize divisibility rules for other numbers using modular arithmetic?

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Tip:

Modular arithmetic simplifies large computations by reducing numbers to smaller residues, making proofs and calculations efficient.