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Number Theory Level 3

$\huge 5,\quad 55,\quad 555,\quad 5555,\quad ...$ If general term of this sequence is $\dfrac{A}{B}(C^{n}-D)$ where A,B,C,D are positive integers, $n=1,2,3,4,...$ , and $\dfrac{A}{B}$ is an irreducible fraction. Find the value of $\dfrac{C!-B!}{(A+D)!}$

The answer is 4536.

1 solution

Chew-Seong Cheong
Apr 28, 2015

$\begin{array} {} n = 1 & \Rightarrow a_1 & = 5 & = 5(1) \\ n = 2 & \Rightarrow a_2 & = 55 & = 5(1+10) \\ n = 3 & \Rightarrow a_3 & = 555 & = 5(1+10+10^2) \\ n = 4 & \Rightarrow a_4 & = 5555 & = 5(1+10+10^2+10^3) \\ ... & \Rightarrow a_n & = 555...5 & = 5(1+10+10^2+...+10^{n-1}) \\ & & & = 5 \left(\dfrac{10^n - 1}{10-1} \right) \\ & & & = \dfrac{5}{9} \left(10^n - 1 \right)\end{array} $

$\Rightarrow \dfrac {C!-B!}{(A+D)!} = \dfrac {10!-9!}{(5+1)!} = \dfrac {3265920 - 362880}{720} = \dfrac{3628800} {720} = \boxed {4536}$

$\frac { 10!-9! }{ 6! } =\frac { 9!\times 9 }{ 6! } =9\times 8\times 7\times 9$ ._Easy for hand calculations.Otherwise same approach as yours.

shivamani patil - 5 years, 11 months ago

@Chew-Seong Cheong Great solution!

Typo: $10!=3628800\\ 10!-9!=3265920$

Sualeh Asif - 5 years, 11 months ago

Thanks for liking my solution and the corrections. I have corrected them.

Chew-Seong Cheong - 5 years, 11 months ago

Ha HaHa HaHaHa!

The answer is 4536.

1 solution

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