AMC 10A 2019

Evaluate the parenthesised expression first:

$2^0 - 1 + 5^2 - 0 = 1 - 1 + 25 - 0 = 25.$

Then $(25)^{-1} \times 5 = \dfrac{1}{25} \times 5 = \dfrac{1}{5}$.

Answer: C.

We only need the last four digits of $20! - 15!$.

$20! \pmod{10^4}$: $20!$ contains the factors $5, 10, 15, 20$ (four multiples of $5$) and at least four factors of $2$, so $20!$ is divisible by $10^4$. Thus $20! \equiv 0 \pmod{10^4}$.

$15! \pmod{10^4}$: $15!$ has three factors of $5$ (from $5, 10, 15$) and many factors of $2$, so it ends in exactly three zeros. Computing $15! = 1{,}307{,}674{,}368{,}000$, the last four digits are $8000$.

So $20! - 15! \equiv 0 - 8000 \equiv -8000 \equiv 2000 \pmod{10^4}$.

The last four digits of $20! - 15!$ are $\ldots 2000$. The hundreds digit (third from the right) is $\boxed{0}$.

Answer: A.

Let $A$ and $B$ be Ana's and Bonita's ages this year.

From "this year Ana's age is the square of Bonita's age": $A = B^2$.

From "last year Ana was $5$ times as old as Bonita": $A - 1 = 5(B - 1)$.

Substitute $A = B^2$:

$B^2 - 1 = 5B - 5 \implies B^2 - 5B + 4 = 0 \implies (B - 1)(B - 4) = 0.$

So $B = 1$ or $B = 4$. If $B = 1$, then $A = 1$ and they'd be the same age, contradicting "different years." So $B = 4$ and $A = 16$.

Thus $n = A - B = 12$.

Answer: D.