This problem is meant to introduce users to the concept of function problems where the function is not given.
Let f:\\mathbb{Z}^+\\rightarrow\\mathbb{Z}^+ (that notation means that the function accepts and returns positive integers) be a function such that f(1)=f(0)=1, and f(x) = xf(x-1). Find \\frac{f(100)}{f(99)}.
", "hints_md": "Doesn't this look like the factorial function?", "hints_html": "Doesn’t this look like the factorial function?
", "editorial_md": "We are given a function $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$defined by the following conditions:\n\n- $f(0) = 1$\n- $f(1) = 1$\n- For $x \\geq 2,f(x) = x \\cdot f(x-1)$\n\nOur task is to find the value of the expression: $\\frac{f(100)}{f(99)}$\n\n#### Understanding the Function\n\nThe given function $f(x)$ resembles a recursive definition often seen in factorial calculations. Let's explore this by evaluating the first few values:\n- $f(0) = 1$\n- $f(1) = 1$\n- $f(2) = 2 \\cdot f(1) = 2 \\cdot 1 = 2$\n- $f(3) = 3 \\cdot f(2) = 3 \\cdot 2 = 6$\n- $f(4) = 4 \\cdot f(3) = 4 \\cdot 6 = 24$\n\nThis pattern matches the definition of factorials:\n\n$f(n) = n!$\n\nThus, we can conjecture that $f(n) = n!$, the factorial of $n$.\n\n#### Proving the Conjecture\n\nTo formally prove that f(n)=n!f(n) = n!f(n)=n!, we proceed by mathematical induction.\n\n**Base Cases:**\n\n- $f(0) = 1$ matches $0! = 1$\n- $f(1) = 1$ matches $1! = 1$\n\n**Inductive Step:**\n\nAssume $f(k) = k!$ holds for some $k \\geq 1$. We need to show that $f(k+1) = (k+1)!$:\n\n$f(k+1) = (k+1) \\cdot f(k)$\n\nBy the induction hypothesis, $f(k) = k!$, so:\n\n$f(k+1) = (k+1) \\cdot k! = (k+1)!$\n\nThus, by the principle of induction, $f(n)=n!$ for all $n \\geq 0$.\n\n#### Calculating the Expression\n\nNow that we have $f(n)=n!$, the expression simplifies to:\n\n$\\frac{f(100)}{f(99)} = \\frac{100!}{99!}$\n\nBy the property of factorials, this reduces to:\n\n$\\frac{100 \\times 99!}{99!} = 100$\n\nThus, the answer is $100$.\n\n- botman", "editorial_html": "We are given a function f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+defined by the following conditions:
Our task is to find the value of the expression: \\frac{f(100)}{f(99)}
The given function f(x) resembles a recursive definition often seen in factorial calculations. Let’s explore this by evaluating the first few values:
This pattern matches the definition of factorials:
f(n) = n!
Thus, we can conjecture that f(n) = n!, the factorial of n.
To formally prove that f(n)=n!f(n) = n!f(n)=n!, we proceed by mathematical induction.
Base Cases:
Inductive Step:
Assume f(k) = k! holds for some k \\geq 1. We need to show that f(k+1) = (k+1)!:
f(k+1) = (k+1) \\cdot f(k)
By the induction hypothesis, f(k) = k!, so:
f(k+1) = (k+1) \\cdot k! = (k+1)!
Thus, by the principle of induction, f(n)=n! for all n \\geq 0.
Now that we have f(n)=n!, the expression simplifies to:
\\frac{f(100)}{f(99)} = \\frac{100!}{99!}
By the property of factorials, this reduces to:
\\frac{100 \\times 99!}{99!} = 100
Thus, the answer is 100.