document.write( "Question 237643: k= 1!+ 2! +3!.......+n!\r
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Algebra.Com's Answer #174782 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "n=1                                        k = 1! =   1, a perfect square\r\n" );
document.write( "n=2                           k = 1! + 2! = 1 + 2 =   3\r\n" );
document.write( "n=3                  k = 1! + 2! + 3! = 1 + 2 + 6 =   9, a perfect square\r\n" );
document.write( "n=4        k = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 =  33\r\n" );
document.write( "n=5  k = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 + 120 = 153\r\n" );
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document.write( "...     ...     ...     ...     ...     ...     ... \r\n" );
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document.write( "Since 5!, which is 120, ends with a 0, the factorial of every \r\n" );
document.write( "integer 5 or higher will also end in a 0 because it will be a \r\n" );
document.write( "multiple of 120.  Since 1! + 2! + 3! + 4! = 33 ends with 3,\r\n" );
document.write( "the last digit of every sum after that will also end with a 3,\r\n" );
document.write( "That's because we will only be adding to it integers that end\r\n" );
document.write( "in 0.  Since no perfect square can end with 3,  1! + 2! + 3! = 9 \r\n" );
document.write( "is the last sum above that will be a perfect square.\r\n" );
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document.write( "So only the values n=1 and n=3 produce a sum for which k is\r\n" );
document.write( "a perfect square.  So the answer is 2 values of n.\r\n" );
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document.write( "Edwin
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