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

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~Mathdreams
 
~Mathdreams
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==Solution 2 (More Algebraic)==
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<cmath>\sum_{n=1}^{100} f(n) = 5(100)+70(\underbrace{1+11+111+1111+ \cdots}_{\text{100 1s}}) + 100(44444 \cdots )</cmath>  We only care about the last <math>3</math> digits, so we evaluate <math>1+11+111+1111+ \cdots</math>. Note the expression is simply <math>1(100)+10(99)+100(98)+1000(97)+ \cdots + 10^{100}</math>, so factoring a <math>10</math> we have <math>1(10)+99+10(98)+ \cdots + 10^{99}</math>. Now, we can divide by <math>25</math> to get <cmath>20+28(1(10)+99+10(98)+100(97) \cdots)+4(444444 \cdots )</cmath> Evaluate the last <math>3</math> digits to get <cmath>20+28(10+99+980+700)+4(444)=\boxed{888} \mod 1000</cmath>
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<math>\linebreak</math>
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~Geometry285

Revision as of 12:34, 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

Notice that $\frac{f(1)}{25} = 19$, $\frac{f(2)}{25} = 191$, $\frac{f(3)}{25} = 1911$, and $\frac{f(n)}{25}$ ends in $111$ for all $n \ge 4$. So, the last 3 digits of $\frac{f(1)+f(2)+ \cdots + f(100)}{25}$ are the last $3$ digits of $19 + 191 + 911 + 111 \cdot 97$, which are $\boxed{888}$.

~Mathdreams

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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