Art of Problem Solving

Euc20205/Sub-Problem 1: Difference between revisions

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==Problem==
(a) For each positive real number x, define <math>f(x)</math> to be the number of prime numbers
(a) For each positive real number x, define <math>f(x)</math> to be the number of prime numbers
<math>p</math> that satisfy <math>x \le p \le x + 10</math>. What is the value of <math>f(f(20))</math>?
<math>p</math> that satisfy <math>x \le p \le x + 10</math>. What is the value of <math>f(f(20))</math>?
==Solution==
Because <math>20</math> is a relatively small number, we can just bash this out. We first need to calculate <math>f(20)</math>, so the numbers in this interval are <math>20</math> to <math>30</math>. The only prime numbers in this range are <math>23</math> and <math>29</math>, so <math>f(20)=2</math>. Then, we need to find <math>f(2)</math>. The range we have is <math>2</math> to <math>12</math>, and the prime numbers that are in this interval are <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>11</math>, so <math>f(2)=\boxed{5}</math>.
~Baihly2024

Latest revision as of 17:05, 12 October 2025

Problem

(a) For each positive real number x, define $f(x)$ to be the number of prime numbers $p$ that satisfy $x \le p \le x + 10$. What is the value of $f(f(20))$?

Solution

Because $20$ is a relatively small number, we can just bash this out. We first need to calculate $f(20)$, so the numbers in this interval are $20$ to $30$. The only prime numbers in this range are $23$ and $29$, so $f(20)=2$. Then, we need to find $f(2)$. The range we have is $2$ to $12$, and the prime numbers that are in this interval are $2$, $3$, $5$, $7$, and $11$, so $f(2)=\boxed{5}$.

~Baihly2024