AMC 12A Problem and Solution

2017 AMC 12A

Problem20

How many ordered pairs between and

such that is a positive real number and is an integer

, inclusive, satisfy the equation

Solution

By the properties of logarithms, we can rearrange the equation to read with . If , we may divide by it and get which implies

. Hence, we have possible values , namely

Since exactly have

,

is equivalent to solutions

, each possible value yields

to each

. In total, we

, as we can assign solutions.

Problem21

A set is constructed as follows. To begin,

. Repeatedly, as long as possible,

if is an integer root of some polynomial for some , all of whose coefficients are elements of , then is put into . When no more elements can be added to , how many elements does have?

Solution 1

At first,

.

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2017 AMC 12A

At this point, no more elements can be added to . To see this, let

with each in . is a factor of , and is in , so has to be a factor of some element in . There are no such integers left, so there can be no more additional elements.

has elements

Problem22

A square is drawn in the Cartesian coordinate plane with vertices at

,

,

,

. A particle starts at

. Every second it moves with

equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is of

that the particle will move from

, , or

,

,

to each

,

,

,

. The particle will eventually hit the square for the first time,

either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is is

?

, where

and are relatively prime positive integers. What

Solution

We let

and

be the probability of reaching a corner before an edge when starting at an

), an \

), and the middle respectively.

\

Starting in the middle, there is a chance of moving to an inside edge and a chance of moving to an inside corner, so

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2017 AMC 12A

Starting at an inside edge, there is a chance of moving to another inside edge, a chance of moving to an inside corner, a chance of moving into the middle, and a chance of reaching an outside edge and stopping. Therefore,

Starting at an inside corner, there is a chance of moving to an inside edge, a chance of moving into the middle, a chance of moving to an outside edge and stopping, and finally a chance of reaching that elusive outside corner. This gives

Solving this system of equations gives

Since the particle starts at

it is

we are looking for, so the final answer is

Problem23

For certain real numbers , , and , the polynomialdistinct roots, and each root of

is also a root of the polynomialWhat is

?

has three

Solution

Let

and be the roots of

. Let be the additional root of and the cubic term of

. Then from Vieta's

formulas on the quadratic term of , we obtain the following:

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2017 AMC 12A

Thus

.

, the linear term of

, and the

Now applying Vieta's formulas on the constant term of linear term of

, we obtain:

Substituting for we obtain:

in the bottom equation and factoring the remainder of the expression,

It follows that Now we can factor

. But in terms of

so as

Then

and

Hence

.

Solution 2

Since all of the roots of

are distinct and are roots of

, and the degree of is one

more than the degree of , we have that

for some number . By comparing

coefficients, we see that

Expanding and equating coefficients we get that

The third equation yields

, and the first equation yields

. So we have that

. Thus,

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