- #1
PrudensOptimus
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Problem:
[tex](n!)^3[/tex] n = {1-99}
How many digit is the resulting [tex](n!)^3 [/tex]?
Attempted solution:
[tex]\Sigma^{99}_{n=1} (n!)^3 = (\Sigma^{99}_{n=1} (n!))^3 = 99(\frac{1!^3 - 99!^3}{2}) =~ 470[/tex] digits. But they say it's wrong. Please help.
[tex](n!)^3[/tex] n = {1-99}
How many digit is the resulting [tex](n!)^3 [/tex]?
Attempted solution:
[tex]\Sigma^{99}_{n=1} (n!)^3 = (\Sigma^{99}_{n=1} (n!))^3 = 99(\frac{1!^3 - 99!^3}{2}) =~ 470[/tex] digits. But they say it's wrong. Please help.