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2014-10-27T07:50:14+05:30

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Perhaps there is a simpler method than what  I find here.

2^{2013}=2^{11*3*61}=2048^{183}=(2013+35)^{183}\\\\Remainder=35^{183}=(35^3)^{61}=(21*2013+602)^{61}\\\\R=602^{61}=602*(602^2)^{30} =602(180*2013+64)^{30}\\\\=>R=602*64^{30}=602*(64^2)^{15}=602*(2013*2+70)^{15}\\

R=602*70^{15}=602*(70^3)^5=602*(170*2013+790)^5=>\\\\R=602*790^5=602*(790^2)^2*790\\\\=602*(310*2013+70)^2*790=602*70^2*790\\\\R=602*(2*2013+874)*790=602*874*790\\

R=874*(2013*236+512)=874*512\\\\R=(222*2013+602)=>602\\\\Units\ digit=2\\

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n! + 10 =  x²,  where x = integer

n = 1, 2 , 4, we don t  find a solution.

For n = 3 ,   x = 4    there is at least one solution.

Suppose n >= 6  then

n! + 10 = 10 + 120 * 6 * 7 *... (n-1) n 
           = 10 [ 1 + 12 * 6 * 7 * ... (n-1) n  ]  = x²

Let N = 
 1 + 12 * 6 * 7 * ... (n-1) n 

Since 10 N = x
²,    N is a multiple of 10.   Hence N-1 has a units digit 9.

M = N - 1 = 12 * 6 *  7 * 8 * .. (n-1) n     ends with a  units  digit  9

if n >= 10, then   M always has a units digit = 0.

For n = 5, 6, 7, 8 , and 9  the  Units digit of M is  : 

n = 5 =>  units digit  2  
n = 6  =>  units  digit  2
n = 7   =>      4
n = 8 =>     2
n = 9   =>    8           =>  no perfect squares for n = 5 to 9.

Hence there is only one solution  for n! + 10 = a perfect square.

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