2020 AMC 12A Problem 25

Question

The number

a = \dfrac{p}{q},

where

p

and

q

are relatively prime positive integers, has the property that the sum of all real numbers

x

satisfying

\lfloor x \rfloor \cdot \{x\} = a \cdot x^2

is

420,

where

\lfloor x \rfloor

denotes the greatest integer less than or equal to

x

and

\{x\} = x - \lfloor x \rfloor

denotes the fractional part of

x.

What is

p + q?

245

593

929

1331

1332

Answer: C

Concepts:floor and ceiling functions quadratic summation

Difficulty rating: 2520

Small hint:

On

x \in [n, n+1)

write

\lfloor x \rfloor = n

and

\{x\} = x - n,

turning the equation into a quadratic

a x^2 - n x + n^2 = 0

Big hint:

Parameterize the two roots after dividing by

n

as

1+\dfrac1u

and

1+u;

determine which positive integers

n

put a root in

[n,n+1)

Solution:

There are no negative solutions, while

x=0

is always a solution. For

n\ge1

and

x\in[n,n+1),

put

y=\frac{x}{n}.

The equation becomes

ay^2-y+1=0.

Its roots must be real, so

0\lt a\le\tfrac14.

If the two roots are

\alpha\le\beta,

their sum and product are both

\frac{1}{a},

so

(\alpha-1)(\beta-1)=1.

Write

\alpha=1+\dfrac1u,\qquad \beta=1+u

with

u\ge1.

Then

a=\dfrac{u}{(u+1)^2}.

The root

x=n\alpha

lies in

[n,n+1)

exactly when

n\lt u,

while

n\beta

never lies there for a positive integer

n.

The required positive total ensures

u\gt1.

Let

N

be the largest positive integer less than

u,

so

N\lt u\le N+1.

The sum of all solutions is therefore

\dfrac{u+1}{u}\cdot\dfrac{N(N+1)}2=420.

Because

\dfrac{u+1}{u}

decreases with

u,

these inequalities imply

\dfrac{N(N+2)}2\le420\lt\dfrac{(N+1)^2}{2},

which forces

N=28.

Substitution gives

406\cdot\dfrac{u+1}{u}=420,

so

u=29.

Hence

a=\dfrac{29}{30^2}=\dfrac{29}{900}.

Indeed the positive solutions are

x=\dfrac{30n}{29}

for

1\le n\le28,

and their sum is

420.

Therefore

p+q=29+900=929.

Thus, C is the correct answer.