Question:
If \(793 \equiv x \pmod{29}\) and \(0 \le x < 29\), then the sum of distinct prime divisors of \(x\) is:
If \(793 \equiv x \pmod{29}\) and \(0 \le x < 29\), then the sum of distinct prime divisors of \(x\) is:
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Always find remainder using nearest multiple method.
Updated On: Jun 15, 2026
- \(2 \)
- \(5 \)
- \(7 \)
- \(10 \)
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The Correct Option is C
Solution and Explanation
Concept:
Find remainder using division method.
Step 1: Find remainder of 793 by 29.
\[ 29 \times 27 = 783 \] \[ 793 - 783 = 10 \]
Step 2: So value of x.
\[ x = 10 \]
Step 3: Find prime factors.
\[ 10 = 2 \times 5 \]
Step 4: Sum of primes.
\[ 2 + 5 = 7 \]
Step 1: Find remainder of 793 by 29.
\[ 29 \times 27 = 783 \] \[ 793 - 783 = 10 \]
Step 2: So value of x.
\[ x = 10 \]
Step 3: Find prime factors.
\[ 10 = 2 \times 5 \]
Step 4: Sum of primes.
\[ 2 + 5 = 7 \]
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