Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1984 AIME Problems/Problem 7"

(→‎Solution 2)
(→‎Solution 4 (Not really a solution, DON'T DO THIS ON A REAL TEST))
(45 intermediate revisions by 7 users not shown)
Line 12: Line 12:
 
<cmath>\begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\
 
<cmath>\begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\
 
&=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*}</cmath>
 
&=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*}</cmath>
−
So this function reiterates with a period of 2 for <math>x</math>. It might be tempting at first to assume that <math>f(1004) = 999</math> is the answer; however, that is not true since the solution occurs slightly before that. Start at <math>f^3(1004)</math>:
+
So this function reiterates with a period of 2 for <math>x</math>. It might be tempting at first to assume that <math>f(1004) = 1001</math> is the answer; however, that is not true since the solution occurs slightly before that. Start at <math>f^3(1004)</math>:
 
<cmath>f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}</cmath>
 
<cmath>f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}</cmath>
  
 +
Note that we should also be suspicious if our answer is <math>1001</math>- it is a <math>4</math>-digit number, and we were not asked to, say, divide our number by <math>13</math>.
  
 
== Solution 2==
 
== Solution 2==
−
We start by finding values of the function right under 1000 since they require little repetition.
+
We start by finding values of the function right under <math>1000</math> since they require iteration of the function.
  
−
<math></math>\begin{align*}f(999)=f(f(1004))=f(1001)=998
+
<cmath>f(999)=f(f(1004))=f(1001)=998</cmath>
−
f(998)=f(f(1003))=f(1000)=997
+
<cmath>f(998)=f(f(1003))=f(1000)=997</cmath>
−
f(997)=f(f(1002))=f(999)=998
+
<cmath>f(997)=f(f(1002))=f(999)=998</cmath>
−
f(996)=f(f(1001))=f(998)=997\\
+
<cmath>f(996)=f(f(1001))=f(998)=997</cmath>
  
−
Soon we realize the <math>f(k)</math> for integers <math>k<1000</math> either equal <math>998</math> or <math>997</math> based on it parity. (If short on time, a guess of 998 or 997 can be taken now.)
+
Soon we realize the <math>f(k)</math> for integers <math>k<1000</math> either equal <math>998</math> or <math>997</math> based on its parity. (If short on time, a guess of <math>998</math> or <math>997</math> can be taken now.)
−
If <math>k</math> is even <math>f(k)=997</math> if <math>k</math> is odd <math>f(k)=998</math>. <math>84</math> has even parity, so <math>f(84)=997</math>.
+
If <math>k</math> is even, <math>f(k)=997</math>. If <math>k</math> is odd, <math>f(k)=998</math>. <math>84</math> has even parity, so <math>f(84)=997</math>.
 +
The result may be rigorously shown through induction.
 +
 
 +
== Solution 3 ==
 +
 
 +
Assume that <math>f(84)</math> is to be performed <math>n+1</math> times. Then we have
 +
<cmath>f(84)=f^{n+1}(84)=f(f^n(84+5))</cmath>
 +
In order to find <math>f(84)</math>, we want to know the smallest value of
 +
<cmath>f^n(84+5)\ge1000</cmath>
 +
Because then
 +
<cmath>f(84)=f(f^n(84+5))=(f^n(84+5))-3</cmath>
 +
From which we'll get a numerical value for <math>f(84)</math>.
 +
 
 +
Notice that the value of <math>n</math> we expect to find is basically the smallest <math>n</math> such that after <math>f(x)=f(f(x+5))</math> is performed <math>\frac{n}{2}</math> times and then <math>f(x)=x-3</math> is performed back <math>\frac{n}{2}</math> times, the result is greater than or equal to <math>1000</math>.
 +
 
 +
In this case, the value of <math>n</math> for <math>f(84)</math> is <math>916</math>, because
 +
<cmath>84+\frac{916}{2}\cdot5-\frac{916}{2}\cdot3=1000\Longrightarrow f^{916}(84+5))=1000</cmath>
 +
Thus
 +
<cmath>f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\boxed{997}</cmath>
 +
 
 +
~ Nafer
 +
 
 +
== Solution 4 (Not really a solution, DON'T DO THIS ON A REAL TEST) ==
 +
 
 +
Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it <math>f</math>, you can call it whatever you want) with parameter <math>n</math> (or 84 in this case) and say if <math>n</math> is greater than 1000, then return <math>n-3</math>. Else, return <math>f(f(n + 5))</math>. If this doesn't make sense, then search 'Recursion' on YouTube and try to understand recursive functions. Python code:
 +
 
 +
    def f(n):
 +
        if n >= 1000:
 +
            return n - 3
 +
        else:
 +
            return f(f(n + 5))
 +
    print(f(84))
 +
 
 +
-NL008
 +
 
 +
 
 +
Or [python]def f(n):
 +
  if n < 1000:
 +
        return f(f(n + 5))
 +
  else:
 +
      return n-3
 +
print(f(84))[/python]
 +
 
 +
-Sernegeti22
  
 
== See also ==
 
== See also ==

Revision as of 16:14, 12 June 2024

Problem

The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$

Find $f(84)$.

Solution 1

Define $f^{h} = f(f(\cdots f(f(x))\cdots))$, where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \ldots f^{y}(1004)$. $1004 = 84 + 5(y - 1) \Longrightarrow y = 185$. So we now need to reduce $f^{185}(1004)$.

Let’s write out a couple more iterations of this function: \begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\ &=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*} So this function reiterates with a period of 2 for $x$. It might be tempting at first to assume that $f(1004) = 1001$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$: \[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}\]

Note that we should also be suspicious if our answer is $1001$- it is a $4$-digit number, and we were not asked to, say, divide our number by $13$.

Solution 2

We start by finding values of the function right under $1000$ since they require iteration of the function.

\[f(999)=f(f(1004))=f(1001)=998\] \[f(998)=f(f(1003))=f(1000)=997\] \[f(997)=f(f(1002))=f(999)=998\] \[f(996)=f(f(1001))=f(998)=997\]

Soon we realize the $f(k)$ for integers $k<1000$ either equal $998$ or $997$ based on its parity. (If short on time, a guess of $998$ or $997$ can be taken now.) If $k$ is even, $f(k)=997$. If $k$ is odd, $f(k)=998$. $84$ has even parity, so $f(84)=997$. The result may be rigorously shown through induction.

Solution 3

Assume that $f(84)$ is to be performed $n+1$ times. Then we have \[f(84)=f^{n+1}(84)=f(f^n(84+5))\] In order to find $f(84)$, we want to know the smallest value of \[f^n(84+5)\ge1000\] Because then \[f(84)=f(f^n(84+5))=(f^n(84+5))-3\] From which we'll get a numerical value for $f(84)$.

Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\frac{n}{2}$ times, the result is greater than or equal to $1000$.

In this case, the value of $n$ for $f(84)$ is $916$, because \[84+\frac{916}{2}\cdot5-\frac{916}{2}\cdot3=1000\Longrightarrow f^{916}(84+5))=1000\] Thus \[f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\boxed{997}\]

~ Nafer

Solution 4 (Not really a solution, DON'T DO THIS ON A REAL TEST)

Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it $f$, you can call it whatever you want) with parameter $n$ (or 84 in this case) and say if $n$ is greater than 1000, then return $n-3$. Else, return $f(f(n + 5))$. If this doesn't make sense, then search 'Recursion' on YouTube and try to understand recursive functions. Python code: 

   def f(n):
       if n >= 1000:
           return n - 3
       else:
           return f(f(n + 5))
   print(f(84))

-NL008


Or [python]def f(n): 

  if n < 1000:
       return f(f(n + 5))
  else:
      return n-3

print(f(84))[/python]

-Sernegeti22

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_7&oldid=220415"