Question

Find the unit's place digit of \((1!)^{1! + (2!)^{2!} + (3!)^{3!} + \cdots + (100!)^{100!}\).}

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

For unit-digit problems, reduce each term modulo 10 and ignore terms that become zero early (like \(n!\) for \(n\ge 5\)).

Answer 1

The Correct Option is D

For \(n\ge 5\), \(n!\) ends with 0, hence \((n!)^{n!}\) contributes unit digit \(0\). Only \(n=1,2,3,4\) matter:
\((1!)^{1!}=1,\ (2!)^{2!}=2^2=4,\ (3!)^{3!}=6^6\) ends with \(6\),
\((4!)^{4!}=24^{24}\) has the same unit pattern as \(4^{24}\) (even power) \(=6\).
Sum of unit digits \(=1+4+6+6=17\)
so the final unit digit is \(7\).