A number theory problem by John Rex Alegria

Number Theory Level 2

2 ( x ! ) = ( ( 5 ! ) ! 5 ! ) 2 + ( 5 ! ) ! 5 ! 2(x!) = \left(\sqrt{\frac {(5!)!}{5!}}\right)^2 + \frac {(5!)!}{5!}

Find x x that satisfies the equation above.


The answer is 119.

John Rex Alegria by John Rex Alegria , 23, Philippines

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2 solutions

2 ( x ! ) = ( ( 5 ! ) ! 5 ! ) 2 + ( 5 ! ) ! 5 ! = ( 5 ! ) ! 5 ! + ( 5 ! ) ! 5 ! = 2 × ( 5 ! ) ! 5 ! ⟹ x ! = ( 5 ! ) ! 5 ! = 120 ! 120 = 119 ! ⟹ x = 119 \begin{aligned} 2(x!) & = \left(\sqrt{\frac {(5!)!}{5!}}\right)^2 + \frac {(5!)!}{5!} = \frac {(5!)!}{5!} + \frac {(5!)!}{5!} = 2 \times \frac {(5!)!}{5!} \\ \implies x! & = \frac {(5!)!}{5!} = \frac {120!}{120} = 119! \\ \implies x & = \boxed{119} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
John Rex Alegria
Sep 4, 2016

First we have to transform the equation into its simplest possible form.

If you'll notice, the 1st term's square root in the right side of the equation will be cancel out because it is squared.

2(x!)=[(5!)!]/5! + [(5!)!]/5!

By factoring we'll get,

2(x!)=[(5!)!]/5!(2)

Cancel out the 2 in both sides of the equation and we'll get,

x!=[(5!)!]/5!

Then,

x!=[(1×2×3×4×5)!]/1×2×3×4×5

x!=(120!)/120

x!=(1×2×3×4×…120)/120

x!=(1×2×3×4×…119)

x!=119!

Therefore,

x=119

Pretty tricky ehh...

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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