Olympiad problems

2016 AMC 10, 22

For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have? $\textbf{(A) }110 \qquad \textbf{(B) } 191 \qquad \textbf{(C) } 261 \qquad \textbf{(D) } 325 \qquad \textbf{(E) } 425$

2016 AMC 12/AHSME, 18

Tags: 2016 AMC 10A , Divisors , 2016 AMC 12A , AMC 10A , AMC 12A , AMC 12 , number theory

For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have? $\textbf{(A) }110 \qquad \textbf{(B) } 191 \qquad \textbf{(C) } 261 \qquad \textbf{(D) } 325 \qquad \textbf{(E) } 425$

2021 AMC 10 Spring, 14

Tags: AMC 12 , AMC 10 , amc10A 2021 , AMC 10A , probs , AUKAAT

All the roots of polynomial $z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16$ are positive integers. What is the value of $B$? $\textbf{(A)}\ -88 \qquad\textbf{(B)}\ -80 \qquad\textbf{(C)}\ -64\qquad\textbf{(D)}\ -41 \qquad\textbf{(E)}\ -40$

2021 AMC 12/AHSME Spring, 12

Tags: AMC 12 , AMC 10 , amc10A 2021 , AMC 10A , probs , AUKAAT

All the roots of polynomial $z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16$ are positive integers. What is the value of $B$? $\textbf{(A)}\ -88 \qquad\textbf{(B)}\ -80 \qquad\textbf{(C)}\ -64\qquad\textbf{(D)}\ -41 \qquad\textbf{(E)}\ -40$