Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 15"
(Created page with "==Problem== For all positive integers <math>n,</math> define the function <math>f(n)</math> to output <math>4\underbrace{777 \cdots 7}_{n\ \text{sevens}}5.</math> For example,...") |
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==Solution== | ==Solution== | ||
| − | + | 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 
define the function 
to output 
For example, 
, 
, and 
Find the last three digits of ![\[\frac{f(1)+f(2)+ \cdots + f(100)}{25}.\]](http://latex.artofproblemsolving.com/2/4/4/244944bb29f9d9d942d159a2b7178e3b06f5728a.png)
Solution
Notice that 
, 
, 
, and 
ends in 
for all 
. So, the last 3 digits of 
are the last 
digits of 
, which are 
.
~Mathdreams
