KCET · Maths · Mathematical Reasoning
The least positive integer \(\mathrm{x}\) satisfying \(2^{2010} \equiv 3 x(\bmod 5)\) is
- A 3
- B 4
- C 1
- D 2
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
Since, we know that
\(2^{2} \equiv-1(\bmod 5) \quad\left(\because 2^{2}+1\right.\) divisible by 5\()\)
Now, \(\quad 2^{2010}=\left(2^{2}\right)^{1005}\)
\(\equiv(-1)^{1005}(\bmod 5)\)
\(\Rightarrow \quad 2^{2010} \equiv-1(\bmod 5)\)
\(\Rightarrow \quad-1 \equiv 2^{2010}(\bmod 5)\)
But \(\quad 2^{2010} \equiv 3 x(\bmod 5)\)
\(\Rightarrow \quad-1 \equiv 3 x(\bmod 5)\) (by transitive relation)
\(\Rightarrow \quad x=3\)
\(2^{2} \equiv-1(\bmod 5) \quad\left(\because 2^{2}+1\right.\) divisible by 5\()\)
Now, \(\quad 2^{2010}=\left(2^{2}\right)^{1005}\)
\(\equiv(-1)^{1005}(\bmod 5)\)
\(\Rightarrow \quad 2^{2010} \equiv-1(\bmod 5)\)
\(\Rightarrow \quad-1 \equiv 2^{2010}(\bmod 5)\)
But \(\quad 2^{2010} \equiv 3 x(\bmod 5)\)
\(\Rightarrow \quad-1 \equiv 3 x(\bmod 5)\) (by transitive relation)
\(\Rightarrow \quad x=3\)
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