Whoa, Triple Factorial?

Number Theory Level 2

Given that

( ( 3 ! ) ! ) ! 3 ! = k × n ! \large \dfrac{{}{((3!)!)!}}{{3!}}= {k \times n!}

where k k and n n are positive integers and n n is as large as possible. Find k + n k+n .


The answer is 839.

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Anish Harsha by Anish Harsha , 20, India

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2 solutions

Rishabh Jain
Jan 11, 2016

( ( 3 ! ) ! ) ! = ( 6 ! ) ! = 720 ! \color{#D61F06}{((3!)!)!=(6!)!=720!} Hence, 720 ! 3 ! = 120 × 719 ! \text{Hence, }\color{#EC7300}{\frac{720!}{3!}=120\times 719!} Clearly maximum possible value of n=719. ∴ n = 719 , k = 120 ∴n=719,k=120 n + k = 839 \Large n+k=\color{#20A900}{839}

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Well explained, same logic

Mahdi Raza - 1 year, 1 month ago

We first note that 3 ! = 6 3!=6 and ( ( 3 ! ) ! ) ! ⇒ ( 6 ! ) ! = 720 ! \left( \left( 3! \right) ! \right) !\Rightarrow\left( 6! \right) !=720! .

Solving the equation, 720 ! 3 ! = k × n ! 720 × 719 × 718 × ⋯ × 1 6 = k × n ! 120 × 719 × 718 × ⋯ × 1 = k × n ! 120 × 719 ! = k × n ! \begin{aligned} \frac{720!}{3!}&=k \times n!\\\frac{720\times719\times718\times\dots\times1}{6}&=k \times n!\\120\times719\times718\times\dots\times1&=k \times n!\\120\times719!&=k \times n! \end{aligned} Therefore, largest value of n n is 719 719 and k = 120 k=120 .

Hence, k + n ⇒ 120 + 719 = 839 k+n\Rightarrow 120+719=\boxed { 839 } .

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