We are tasked with proving the following statement by contrapositive:

Original Statement: If $10a \equiv 10b \pmod{n}$, then $n \mid 10$.

By the contrapositive method, we first rewrite the original statement in contrapositive form:

Contrapositive:

If $n \nmid 10$, then $10a \not\equiv 10b \pmod{n}$.

This is what we need to prove.

Proof:

1. Assume $n \nmid 10$:

   • This means $n$ does not divide 10, so 10 is not a multiple of $n$. Consequently, $n$ and 10 are relatively prime, i.e., $\gcd(n, 10) = 1$.

2. Consider $10a \equiv 10b \pmod{n}$:

   • If $10a \equiv 10b \pmod{n}$, this implies $10a - 10b = 10(a - b)$ is divisible by $n$, i.e., $n \mid 10(a - b)$.

3. Apply the fact that $n$ does not divide 10:

   • Since $n \nmid 10$, we know that $n$ cannot divide 10 directly. For $n \mid 10(a - b)$ to hold, it must be the case that $n$ divides $a - b$, i.e., $n \mid (a - b)$.
   • Therefore, if $n \nmid 10$, then the only way $n$ divides $10(a - b)$ is if $n$ divides $a - b$. In other words, $a \equiv b \pmod{n}$.

4. Conclusion:

   • If $n \nmid 10$, then it is not possible for $10a \equiv 10b \pmod{n}$, unless $a \equiv b \pmod{n}$. Hence, $10a \not\equiv 10b \pmod{n}$ when $n \nmid 10$.

Thus, we have proven the contrapositive: If $n \nmid 10$, then $10a \not\equiv 10b \pmod{n}$, completing the proof.

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Would you like more details or have any questions?

Here are 5 related questions:

1. What is the significance of using the contrapositive in proofs?
2. How does the concept of greatest common divisors apply to modular arithmetic?
3. Can you prove a similar result if we replaced 10 with another integer, like 12?
4. How does one determine if two numbers are relatively prime?
5. What is the importance of dividing both sides of a congruence relation in proofs?

Tip: Always check if two numbers are relatively prime when dealing with divisibility and congruence relations. This can simplify the proof or provide useful insights!