Find a positive integer less than 8 for x that satisfies x = 720 + 3 (mod 7).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem:**
Find a positive integer less than 8 for \( x \) that satisfies the following congruence:
\[ x \equiv 7^{20} + 3 \pmod{7} \]
**Solution Explanation:**
1. **Understanding the Congruence:**
\[
x \equiv 7^{20} + 3 \pmod{7}
\]
This implies we are looking for the remainder of \( 7^{20} + 3 \) when divided by 7.
2. **Simplifying \( 7^{20} \mod 7 \):**
Since \( 7 \equiv 0 \pmod{7} \), it follows that any power of 7 is also congruent to 0 modulo 7:
\[
7^{20} \equiv 0^{20} \equiv 0 \pmod{7}
\]
3. **Adding 3:**
Now substitute back into the equation:
\[
x \equiv 0 + 3 \equiv 3 \pmod{7}
\]
4. **Conclusion:**
The positive integer less than 8 that satisfies the original congruence is:
\[
x = 3
\]
Thus, the solution is \( x = 3 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F72f9ec78-21fa-4141-958a-3b1eb6c54d04%2F08df0037-2dd6-49bf-8f21-a431b899c1e7%2F0j9jzu0e_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find a positive integer less than 8 for \( x \) that satisfies the following congruence:
\[ x \equiv 7^{20} + 3 \pmod{7} \]
**Solution Explanation:**
1. **Understanding the Congruence:**
\[
x \equiv 7^{20} + 3 \pmod{7}
\]
This implies we are looking for the remainder of \( 7^{20} + 3 \) when divided by 7.
2. **Simplifying \( 7^{20} \mod 7 \):**
Since \( 7 \equiv 0 \pmod{7} \), it follows that any power of 7 is also congruent to 0 modulo 7:
\[
7^{20} \equiv 0^{20} \equiv 0 \pmod{7}
\]
3. **Adding 3:**
Now substitute back into the equation:
\[
x \equiv 0 + 3 \equiv 3 \pmod{7}
\]
4. **Conclusion:**
The positive integer less than 8 that satisfies the original congruence is:
\[
x = 3
\]
Thus, the solution is \( x = 3 \).
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