Math Is Fun Forum

Radhika

Member

Registered: 2006-11-01

Posts: 2

I need help

5) If n  ≥ 4, show that 1! + 2! + 3! +.....+n!  Cannot  be a perfect square.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: I need help

When n=4, f(4) = 1! + 2! + 3! + 4! = 1+2+6+24 = 33, which is not a perfect square.

Because 5! and all higher factorials have a 2 and a 5 multiplied together, they are all multiples of 10 and so all end in 0.

This means that adding 5!, 6!, 7!, etc. to 33 will not change its last digit. So when n ≥4, the last digit of the function is 3.

However, perfect squares can only have last digits of 0, 1, 4, 5, 6 and 9. This means that there are no perfect squares that end in 3 and so there cannot be any factorial sums (that contain 4!) that are also perfect squares.

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Why did the vector cross the road?
It wanted to be normal.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: I need help

It is supposed to be concerning some famous theorem :lol

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X'(y-Xβ)=0

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: I need help

Can't seem to find such a theorem.  I found two equivalent ways to achieve the same sequence, but both involve summations themselves.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."