# 1)what is the unit place of the remainder if 2²°¹³ is divided by 2013 ?2)how many natural numbers n are there such that n!+10 is a perfect square ?

1
by Souvik118

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

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

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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.