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Q.
If $6^n-5 n$ is divided by 25 , then remainder is
Binomial Theorem
Solution:
For two numbers a and $b$, if we can find numbers $q$ and $r$ such that $a=b q+r$, then we say that $b$ divides a with $q$ as quotient and $r$ as remainder. Thus, in order to show that $6^n-5 n$ leaves remainder 1 when divided by 25 , we prove that $6^n-5 n=25 k+1$, where $k$ is some natural number.
We have,
$(1+a)^n={ }^n C_0+{ }^n C_1 a+{ }^n C_2 a^2+\ldots+{ }^n C_n a^n$
For $a=5$, we get
$(1+5)^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+\ldots+{ }^n C_n 5^n$
i.e., $(6)^n=1+5 n+5^2 \cdot{ }^n C_2+5^3 \cdot{ }^n C_3+\ldots+5^n$
i.e., $6^n-5 n=1+5^2\left({ }^n C_2+{ }^n C_3 5+\ldots+5^{n-2}\right)$
or $6^n-5 n=1+25\left({ }^n C_2+5 \cdot{ }^n C_3+\ldots+5^{n-2}\right)$
or $6^n-5 n=25 k+1$, where $k={ }^n C_2+5 \cdot{ }^n C_3+\ldots+5^{n-2}$
This shows that when divided by $25,6^n-5 n$ leaves remainder 1.