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Use the binomial theorem to prove that remainder is always $1$ when $6^n-5n$ is divided by $25$

I haven't tried anything yet. I know only mathematical induction theorem to prove this type of question, but I can solve it with mathematical induction.

  • 2
    Perhaps writing $6=1+5$ and applying the binomial theorem to $6^n=(5+1)^n$ might help. – 2017-01-24

3 Answers 3

5

$$6^n-5n=(5+1)^n-5n=1+\sum_{k=2}^n\binom nk5^k=1+25\sum_{k=0}^{n-2}\binom n{k+2}5^k$$

1

Hint: Try expand $$1=6-5=(6-5)^n$$

-1

$$\begin{align} 6^n-5n &= (1+5)^n -5n \\ &=((n\mathop{\mathsf{C}}0)+(n\mathop{\mathsf{C}}1) \cdot 5+(n\mathop{\mathsf{C}}2) \cdot 5^2+ \cdots +(n\mathop{\mathsf{C}}n) \cdot 5^n) -5n \\ &= 1+5n-5n+((n\mathop{\mathsf{C}}2) \cdot 5^2+(n\mathop{\mathsf{C}}3) \cdot 5^3 + \cdots + (n\mathop{\mathsf{C}}n) \cdot 5^n) \\ &= 1+5^2((n\mathop{\mathsf{C}}2)+(n\mathop{\mathsf{C}}3)\cdot 5 +\cdots+(n\mathop{\mathsf{C}}n)5^{n-2}) \end{align}$$ Hence $1$ will always be the remainder when divided by $25.$