2001 AMC 10 Problems

Problem 1

The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is 10. What is the mean?

$\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 11$

Problem 2

A number $x$ is $2$ more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

$\mathrm{(A)}\ -4\leq x\leq -2 \qquad\mathrm{(B)}\ -2<x\leq 0 \qquad\mathrm{(C)}\ 0<x\leq 2$

$\mathrm{(D)}\ 2<x\leq 4 \qquad\mathrm{(E)}\ 4<x\leq 6$

Problem 3

The sum of two numbers is $S$ . Suppose $3$ is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$\mathrm{(A)}\ 2S+3 \qquad\mathrm{(B)}\ 3S+2 \qquad\mathrm{(C)}\ 3S+6 \qquad\mathrm{(D)}\ 2S+6 \qquad\mathrm{(E)}\ 2S+12$

Problem 4

What is the maximum number for the possible points of intersection of a circle and a triangle?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 3 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 5 \qquad\mathrm{(E)}\ 6$

Problem 5

How many of the twelve pentominoes pictured below have at least one line of symmetry?

$\mathrm{(A)}\ 3 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$

Problem 6

Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$ ?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 3 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 8 \qquad\mathrm{(E)}\ 9$

Solutions

1. The median is $n+6$ , therefore $n=4$ . Computation shows that the sum of all numbers is $99$ and thus the mean is $99/9=11$ .

2. The reciprocal of $x$ is $\frac 1x$ and the additive inverse is $-x$ . (Note that $x$ must be non-zero to have a reciprocal.) The product of these two is $\frac 1x \cdot (-x) = -1$ . Thus $x$ is $2$ more than $-1$ . Therefore $x=1$ .

3. The original two numbers are $x$ and $y$ , with $x+y=S$ . The new two numbers are $2(x+3)$ and $2(y+3)$ . Their sum is $2(x+3)+2(y+3)=2x+2y+12=2(x+y)+12 = 2S+12$ .

4. Each side of the triangle can only intersect the circle twice, so the maximum is at most 6. This can be achieved: $[asy] unitsize(0.3cm); draw( circle((0,-0.2),2.2) ); draw( (-2,-2)--(2,-2)--(0,3)--cycle ); [/asy]$

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