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

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==Solution==
 
==Solution==
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Notice that <math>\frac{f(1)}{25} = 19</math>, <math>\frac{f(2)}{25} = 191</math>, <math>\frac{f(3)}{25} = 1911</math>, and <math>\frac{f(n)}{25}</math> ends in <math>111</math> for all <math>n \ge 4</math>. So, the last 3 digits of <math>\frac{f(1)+f(2)+ \cdots + f(100)}{25}</math> are the last <math>3</math> digits of <math>19 + 191 + 911 + 111 \cdot 97</math>, which are <math>\boxed{888}</math>.
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We can easily find that <math>\frac{f(1)}{25}=\frac{475}{25}=19,\frac{4775}{25}=191,\frac{47775}{25}=1911.</math> Thus, we claim<math>\text{}^1</math> that <math>\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1.</math> Now, we find we can easily find that <math>(\frac{f(1)+f(2)+ \cdots + f(100)}{25})\pmod{1000}\equiv(19+191+911+(111)(97))\pmod{1000}\equiv 11888 \pmod{1000}=\boxed{888}.</math>  
  
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~Mathdreams
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<math>\text{}^1</math> This will be a proof by induction.
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Base Case: <math>\frac{f(1)}{25}=\frac{475}{25}=19=19\underbrace{111 \cdots 1}_{(1-1=0)\text{one's}}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
  
 
==Solution 2 (More Algebraic)==
 
==Solution 2 (More Algebraic)==

Revision as of 16:35, 11 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 $\frac{f(1)}{25}=\frac{475}{25}=19,\frac{4775}{25}=191,\frac{47775}{25}=1911.$ Thus, we claim$\text{}^1$ that $\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1.$ Now, we find we can easily find that $(\frac{f(1)+f(2)+ \cdots + f(100)}{25})\pmod{1000}\equiv(19+191+911+(111)(97))\pmod{1000}\equiv 11888 \pmod{1000}=\boxed{888}.$

$\text{}^1$ This will be a proof by induction. Base Case: $\frac{f(1)}{25}=\frac{475}{25}=19=19\underbrace{111 \cdots 1}_{(1-1=0)\text{one's}}1$ I claim that $\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1.$ We can easily find that $f(n+1)=10f(n)+25.$ Thus, since $\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1,$ $\frac{f(n+1)}{25}=10(19\underbrace{111 \cdots 1}_{(n-1)\text{one's}}1)+1=19\underbrace{111 \cdots 1}_{(n)\text{one's}}1$ as desired.

~pinkpig

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

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