Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 1"

 
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==Problem==
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Let <math>T</math> be the sum of all positive integers of the form <math>2^r\cdot3^s</math>, where <math>r</math> and <math>s</math> are nonnegative integers that do not exceed <math>4</math>.  Find the remainder when <math>T</math> is divided by <math>1000</math>.
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==Solution==
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We note that the required sum is equal to <math>(2^0+2^1+2^2+2^3+2^4)(3^0+3^1+3^2+3^3+3^4)=(31)(121)=3751</math>,
 
We note that the required sum is equal to <math>(2^0+2^1+2^2+2^3+2^4)(3^0+3^1+3^2+3^3+3^4)=(31)(121)=3751</math>,
 
which is <math>\framebox{751}</math> mod 1000.
 
which is <math>\framebox{751}</math> mod 1000.
 
~AbbyWong
 
~AbbyWong

Latest revision as of 09:17, 23 November 2023

Problem

Let $T$ be the sum of all positive integers of the form $2^r\cdot3^s$, where $r$ and $s$ are nonnegative integers that do not exceed $4$. Find the remainder when $T$ is divided by $1000$.

Solution

We note that the required sum is equal to $(2^0+2^1+2^2+2^3+2^4)(3^0+3^1+3^2+3^3+3^4)=(31)(121)=3751$, which is $\framebox{751}$ mod 1000. ~AbbyWong