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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~Mathdreams

Revision as of 10:34, 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