KCET · Maths · Binomial Theorem
The remainder when \(3^{100} \times 2^{50}\) is divided by 5 is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
Now, \(\quad 3^{2} \equiv 4(\bmod 5)\)
\(\Rightarrow \quad\left(3^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)\)
\(\Rightarrow \quad\left(3^{4}\right)^{25} \equiv(1)^{25}(\bmod 5)\)
\(\Rightarrow \quad 3^{100} \equiv 1(\bmod 5)\)
and \(\quad 2^{2} \equiv 4(\bmod 5)\)
\(\Rightarrow \quad\left(2^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)\)
\(\Rightarrow \quad\left(2^{4}\right)^{12} \equiv 1^{12}(\bmod 5) \equiv 1(\bmod 5)\)
\(\Rightarrow \quad 2^{48} \cdot 2^{2} \equiv 4(\bmod 5)\)
\(\therefore \quad 3^{100} \times 2^{50}=1 \times 4(\bmod 5)\)
Hence, remainder is 4 .
\(\Rightarrow \quad\left(3^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)\)
\(\Rightarrow \quad\left(3^{4}\right)^{25} \equiv(1)^{25}(\bmod 5)\)
\(\Rightarrow \quad 3^{100} \equiv 1(\bmod 5)\)
and \(\quad 2^{2} \equiv 4(\bmod 5)\)
\(\Rightarrow \quad\left(2^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)\)
\(\Rightarrow \quad\left(2^{4}\right)^{12} \equiv 1^{12}(\bmod 5) \equiv 1(\bmod 5)\)
\(\Rightarrow \quad 2^{48} \cdot 2^{2} \equiv 4(\bmod 5)\)
\(\therefore \quad 3^{100} \times 2^{50}=1 \times 4(\bmod 5)\)
Hence, remainder is 4 .
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