1995 AIME Problem 13

Attempt Problem 13 of the 1995 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AIME solutions, or check the answer key.

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13.

Let

f(n)

be the integer closest to

\sqrt[4]{n}.

Find

\sum_{k=1}^{1995}\frac1{f(k)}.

Answer

Answer: 400

Concepts:radical counting integers in a range sum of first n squares

Difficulty rating: 1940

Small hint:

Count the integers

n

for which

j-\tfrac12<\sqrt[4]n<j+\tfrac12

Big hint:

The number of occurrences of

f(n)=j

simplifies to

4j^3+j

Solution:

For

j\geq1,

the number of positive integers

n

for which

f(n)=j

is

\begin{aligned}\left(j+\frac12\right)^4&-\left(j-\frac12\right)^4\\&=4j^3+j.\end{aligned}

For

j=1,\ldots,6,

these account for

\sum_{j=1}^6(4j^3+j)=1785

values, and their contribution to the requested sum is

\begin{aligned}\sum_{j=1}^6\frac{4j^3+j}{j}&=4\sum_{j=1}^6j^2+6\\&=370.\end{aligned}

The remaining

1995-1785=210

values have

f(n)=7,

contributing

30.

The total is

400.