Question

Find the number of zeros at the end of $(5!)^{5!}+(10!)^{10!}+(50!)^{50!}+(100!)^{100!}$.

• A. 11
• B. 12
• C. 110
• D. 120

Hint:

For sums of huge factorial powers, the trailing zeros are typically the minimum $v_5$ among terms; check the smallest term and confirm it doesn’t have more.

Answer 1

The Correct Option is D

Trailing zeros are governed by the power of $10$, i.e.\ $\min(v_2,\,v_5)$. For $n!\,$, \[ v_5(n!)=\sum_{k\ge 1}\left\lfloor\frac{n}{5^k}\right\rfloor. \] For $(n!)^{n!}$ we have $v_5\big((n!)^{n!}\big)=n!\cdot v_5(n!)$.
[2mm] Compute the smallest among the four terms: \[ v_5\big((5!)^{5!}\big)=5!\cdot v_5(5!)=120\cdot 1=120. \] For $n=10,50,100$ the values are much larger ($v_5(10!)=2$, etc.), hence every term is divisible by $10^{120}$ and the \emph{minimum} $v_5$ across the sum is $120$. The sum cannot gain an extra factor of $10$ because the first term has exactly $120$ factors of $5$. Therefore the number of trailing zeros is $120$. \[ \boxed{120} \]