Binomial theorem for positive integral Index

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$7^{103_{}}$ when divided by $25$ leaves the remainder

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Solution

We have,
$\begin{array}{l} 7^{103} = 7 (49)^{51} \\ \Rightarrow 7^{103} = 7 (50 - 1)^{51} \\ \Rightarrow 7^{103} = 7 \{^{51} C_{0} (50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} \dots - 1\} \\ \Rightarrow 7^{103} = 7 \{(50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} \dots\} - 7 \\ \Rightarrow 7^{103} = 7 \{(50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} \dots\} - 7 - 18 + 18 \\ \Rightarrow 7^{103} = 7 \{(50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} \dots\} - 25 + 18 \end{array}$
$7^{103_{}}$ = Multiple of $25 + 18$
Hence, remainder = 18.

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

7103 when divided by 25 leaves the remainder

We have, 7103=7(49)51⇒ 7103=7(50−1)51⇒ 7103=7 51C0(50)51−51C1(50)50+51C2(50)49…−1⇒ 7103=7(50)51−51C1(50)50+51C2(50)49…−7⇒ 7103=7(50)51−51C1(50)50+51C2(50)49…−7−18+18⇒ 7103=7(50)51−51C1(50)50+51C2(50)49…−25+187103 = Multiple of 25 + 18 Hence, remainder = 18.

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$7^{103_{}}$ when divided by $25$ leaves the remainder