Consecutive Triangle

Number Theory Level 1

1 8 ! + 1 9 ! = x 10 ! \dfrac{1}{{8!}} + \dfrac{1}{{9!}} = \dfrac{x}{{10!}}

Find the value of x x satisfying the equation above.


Notation: ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 100.

Yash Jain by Yash Jain , 21, India

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7 solutions

⇒ 1 8 ! + 1 9 ! = x 10 ! \Rightarrow \dfrac{1}{\color{#D61F06}{8!}}+\dfrac{1}{\color{#3D99F6}{9!}}=\dfrac{x}{\color{#20A900}{10!}}

10 ! 8 ! + 10 ! 9 ! = x \dfrac{\color{#20A900}{10!}}{\color{#D61F06}{8!}}+\dfrac{\color{#20A900}{10!}}{\color{#3D99F6}{9!}}=x

10 × 9 × 8 ! 8 ! + 10 × 9 ! 9 ! = x \dfrac{10×9×\color{#D61F06}{8!}}{\color{#D61F06}{8!}}+\dfrac{10×\color{#3D99F6}{9!}}{\color{#3D99F6}{9!}}=x

x = ( 10 × 9 ) + 10 = 100 x=(10×9)+10=\boxed{100}

41 Helpful 1 Interesting 2 Brilliant 0 Confused
Akash Patalwanshi
Apr 29, 2016

1 8 ! + 1 9 ! = 9 ! + 8 ! ( 8 ! ) ( 9 ! ) = 8 ! ( 9 + 1 ) ( 8 ! ) ( 9 ! ) = 10 9 ! \frac{1}{8!} + \frac{1}{9!}= \frac{9! + 8!}{(8!) ( 9!)} =\frac{8!(9+1)}{(8!) (9!)} =\frac{10}{9!}

So 10 9 ! = x 10 ! \frac{10}{9!} = \frac{x}{10!}

→ x = ( 10 ) ( 10 ! ) 9 ! →x = \frac{(10)(10!)}{9!}

→ x = 100 →x = \boxed{100}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Correction: It should be 1 8 ! + 1 9 ! \frac{1}{8!} + \frac{1}{9!} in the first step.

Yash Jain - 5 years, 1 month ago

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Thanks ! Edited!!

akash patalwanshi - 5 years, 1 month ago
Sam Bealing
Apr 27, 2016

Relevant wiki: Factorials

1 8 ! + 1 9 ! = 10 × 9 10 × 9 × 8 ! + 10 10 × 9 ! = 90 10 ! + 10 10 ! = 100 10 ! \dfrac{1}{8!}+\dfrac{1}{9!}=\dfrac{10 \times 9}{10 \times 9 \times 8!}+\dfrac{10}{10 \times 9!}=\dfrac{90}{10!}+\dfrac{10}{10!}=\dfrac{100}{10!}

x = 100 x=\boxed{100}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

We multiply both the sides of the equation by 10 ! 10! . Then, we have:

⟹ 10 ! 8 ! + 10 ! 9 ! = x \implies \frac{10!}{8!} + \frac{10!}{9!} = x

⟹ 8 ! × 9 × 10 8 ! + 9 ! × 10 9 ! = x \implies \frac{8! \times 9 \times 10}{8!} + \frac{9! \times 10}{9!} = x

⟹ x = 90 + 10 \implies x=90+10

⟹ x = 100 \implies \boxed{x=100}

Couldn't get any better :3

3 Helpful 0 Interesting 0 Brilliant 0 Confused

There is one shortcut which is useful in competitive exams,

In general if a , b a,b and c c are three consecutive positive integers, then for

1 a ! + 1 b ! = x c ! \large \dfrac{1}{a!} + \dfrac{1}{b!} = \dfrac{x}{c!}

value of x x will be c 2 c^2 .

Yash Jain - 5 years, 1 month ago

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Yeah, we can do that, by making x x the subject of this formula.

Still, thanks. It would really benefit me in competitive exams.

Arkajyoti Banerjee - 5 years, 1 month ago

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Nice formula...I tried to prove and I succeed to prove.

Dharmendra Singh Hindustani - 4 years, 3 months ago

@Yash Jain Can you explain to me why the value of x x 1 a ! + 1 b ! = x c ! \frac { 1 } { a! } + \frac { 1 } { b! } = \frac { x } { c! } equals to c 2 c ^ { 2 } ?

. . - 3 months, 3 weeks ago

1 8 ! + 1 9 ! \frac {1}{8!} + \frac{1}{9!}

= = 9 9 ! + 1 9 ! \frac {9}{9!} + \frac{1}{9!}

= = 9 + 1 9 ! \frac {9 + 1}{9!}

= = 10 9 ! \frac {10}{9!}

= = 10 × 10 9 ! × 10 \frac {10 \times 10}{9! \times 10}

= = 100 10 ! \frac {100}{10!} = = x 10 ! \frac {x}{10!}

So x = 100 x =\boxed{100}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Vũ Nhật Huy
Dec 24, 2018

1 8 ! \frac{1}{8!} + 1 9 ! \frac{1}{9!} = x 10 ! \frac{x}{10!}

1 8 ! \frac{1}{8!} + 1 9 ! \frac{1}{9!} = x : 10!

x = ( 1 8 ! \frac{1}{8!} + 1 9 ! \frac{1}{9!} )*10!

x = 1 36288 \frac{1}{36288} *10!

x= 100

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Fahim Saikat
Jul 9, 2017

1 8 ! + 1 9 ! = 9 + 1 9 ! = 10 9 ! = 100 10 ! = x 10 ! \frac{1}{8!}+\frac{1}{9!}=\frac{9+1}{9!}=\frac{10}{9!}=\frac{100}{10!}=\frac{x}{10!}

therefore , x = 100 \boxed{x=100}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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