What is x, if \(\displaystyle{1}!+{2}!+{3}!+\cdots+{\left({x}-{1}\right)}!+{x}!={k}^{{2}}\) and k

Karen Smith

Karen Smith

Answered question

2022-03-13

What is x, if 1!+2!+3!++(x1)!+x!=k2 and k is an integer?

Answer & Explanation

Veronica Riddle

Veronica Riddle

Beginner2022-03-14Added 9 answers

Step 1
It can be proved that for x4, no solutions exist. The expressions
1!+2!+3!+4!=33
1!+2!+3!+4!+5!=153
1!+2!+3!+4!+5!+6!=873
1!+2!+3!+4!+5!+6!+7!=5913
end with the digit 3. Now, for x4 the last digit of the sum 1!+2!+3!++x! is equal to 3 and therefore, this sum cannot be equal to a square of a whole number k (because the square of a whole number cannot end with 3).
Therefore, 1 and 3 are the only solutions for x.

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