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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 15"

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<math>\text{}^*</math> We proceed by induction. Our base case is <math>\tfrac{f(1)}{25}=\tfrac{475}{25}=19=19\underbrace{111 \cdots 1}_{(1-1=0)\text{one's}}1</math>
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<math>\text{}^*</math> We proceed by induction. Our base case is <math>\tfrac{f(1)}{25}=\tfrac{475}{25}=19.</math> Our inductive assumption is <cmath>\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{n-1 \text{ ones}},</cmath> and we wish to prove that this pattern holds for <math>f(n+1).</math>
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I claim that <math>\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1.</math> We can easily find that <math>f(n+1)=10f(n)+25.</math> Thus, since <math>\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1,</math> <math>\frac{f(n+1)}{25}=10(19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1)+1=19\underbrace{111 \cdots 1}_{(n)\text{one's}}1</math> as desired.
 
  
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~pinkpig
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We can easily find that <math>f(n+1)=10f(n)+25.</math> Using our inductive assumption, we obtain <cmath>\frac{f(n+1)}{25}=10 \cdot (19\underbrace{111 \cdots 1}_{n-1 \text{ ones}})+1=19 \cdot \underbrace{111 \cdots 1}_{n-1 \text{ ones}},</cmath> as desired. <math>\mathbb{Q.E.D.}</math>
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~Solution by pinkpig, <math>\LaTeX</math>/wording fixes by samrocksnature
  
 
==Solution 2 (More Algebraic)==
 
==Solution 2 (More Algebraic)==
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<math>\linebreak</math>
 
<math>\linebreak</math>
 
~Geometry285
 
~Geometry285
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==See also==
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
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#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 00:19, 12 July 2021

Problem

For all positive integers $n,$ define the function $f(n)$ to output $4\underbrace{777 \cdots 7}_{n\ \text{sevens}}5.$ For example, $f(1)=475$, $f(2)=4775$, and $f(3)=47775.$ Find the last three digits of \[\frac{f(1)+f(2)+ \cdots + f(100)}{25}.\]

Solution

We can easily find that $\tfrac{f(1)}{25}=19,\tfrac{f(2)}{25}=191,\tfrac{f(3)}{25}=1911,$ and so on. Thus, we claim$\text{}^*$ that \[\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{n-1 \text{ ones}}.\] Now, we find we can easily find that \[\frac{f(1)+f(2)+ \cdots + f(100)}{25}\equiv19+191+911+111 \cdot 97 \equiv 11888 \pmod{1000} \rightarrow \boxed{888}.\]



$\text{}^*$ We proceed by induction. Our base case is $\tfrac{f(1)}{25}=\tfrac{475}{25}=19.$ Our inductive assumption is \[\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{n-1 \text{ ones}},\] and we wish to prove that this pattern holds for $f(n+1).$

We can easily find that $f(n+1)=10f(n)+25.$ Using our inductive assumption, we obtain \[\frac{f(n+1)}{25}=10 \cdot (19\underbrace{111 \cdots 1}_{n-1 \text{ ones}})+1=19 \cdot \underbrace{111 \cdots 1}_{n-1 \text{ ones}},\] as desired. $\mathbb{Q.E.D.}$

~Solution by pinkpig, $\LaTeX$/wording fixes by samrocksnature

Solution 2 (More Algebraic)

\[\sum_{n=1}^{100} f(n) = 5(100)+70(\underbrace{1+11+111+1111+ \cdots}_{\text{100 1s}}) + 100(44444 \cdots )\] We only care about the last $3$ digits, so we evaluate $1+11+111+1111+ \cdots$. Note the expression is simply $1(100)+10(99)+100(98)+1000(97)+ \cdots + 10^{100}$, so factoring a $10$ we have $1(10)+99+10(98)+ \cdots + 10^{99}$. Now, we can divide by $25$ to get \[20+28(1(10)+99+10(98)+100(97) \cdots)+4(444444 \cdots )\] Evaluate the last $3$ digits to get \[20+28(10+99+980+700)+4(444)=\boxed{888} \mod 1000\] $\linebreak$ ~Geometry285


See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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