Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 15"
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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
Solution
Notice that , , , and ends in for all . So, the last 3 digits of are the last digits of , which are .
~Mathdreams