Question

find the values when n is equal or greater than 1 for which 1!+2!+3!+...+n! is a perfect square

Answer 1

\[1!+2!+3!\] works ,you get 9

Answer 2

yes but i am looking for all the values

Answer 3

n=1 :)

Answer 4

i think, Only 1 and 3.
If n =>4 the unit digit of the expression is always 3 . So u can conclude that there are no perfect square of the form 1!+2!+...+n! If n=4,5,6,....

Answer 5

\[
n! + (n+1)! = n!+(n+1)n! = (n+2)n!
\]

Answer 6

\[
(n+2)n! + (n+2)! = (n+2)n!+(n+2)(n+1)n! = (n+2)(n+2)^2n!
\]Interesting pattern.

Answer 7

Should be\[
(n+2)n! + (n+2)! = (n+2)n!+(n+2)(n+1)n! = (n+2)^2n!
\]

Answer 8

so, ?

Answer 9

I actually don't understand your proof.

Answer 10

You're saying last digit must be 3?

Answer 11

1! = 1 = 1^2 (true it is a perfect square)
1! + 2! = 3 is not a perfect square
1! + 2! + 3! = 9 = 3^2 (true)
1! + 2! + 3! + 4! = 23 is not a perfect square
1! + 2! + 3! + 4! + 5! = 143 is not a perfect square
1! + 2! + 3! + 4! + 5! + 6! = 863 is not a perfect square
..... and so on, will has the unit always 3
while, no square numbers has a unit "3"

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = ...6
5^2 = ...5
6^2 = ...6
7^2 = ...9
8^2 = ...4
9^2 = ...1
10^2 = ... 0

Answer 12

1 and 3

Answer 13

agree with @RadEn :) there is no positive integer m such that\[m^2 \equiv 3 \ \ \ (\text{mod}
\ 10) \]on the other hand for \(n\ge 5\)\[1!+2!+3!+...+n! \equiv \ 1!+2!+3!+4!\equiv33 \equiv 3 \ \ (\text{mod} 10)\]trying \(n=1,2,3,4\) gives \(n=1,3\) as answer.