Difference between revisions of "2008 iTest Problems/Problem 51"
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Alexis imagines a <math>2008 \times 2008</math> grid of integers arranged sequentially in the following way: | Alexis imagines a <math>2008 \times 2008</math> grid of integers arranged sequentially in the following way: | ||
| − | <cmath>\begin{array}{r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r}1,&2,&3,&\ldots,&2008\\2009,&2010,&2011,&\ldots,& | + | <cmath>\begin{array}{r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r}1,&2,&3,&\ldots,&2008\\2009,&2010,&2011,&\ldots,&4016\\4017,&4018,&4019,&\ldots,&6024\\\vdots&&&&\vdots\\2008^2-2008+1,&2008^2-2008+2,&2008^2-2008+3,&\ldots,&2008^2\end{array}</cmath> |
| − | She picks one number from each row so that no two numbers she picks are in the same column. She them proceeds to add them together and finds that <math>S</math> is the sum. Next, she picks <math>2008</math> of the numbers that are distinct from the <math>2008</math> she picked the first time. Again she picks exactly one number from each row and column, and again the sum of all <math>2008</math> numbers is <math>S</math>. Find the remainder when <math>S</math> is divided by <math>2008</math>. | + | She picks one number from each row so that no two numbers she picks are in the same column. She them proceeds to add them together and finds that <math>S</math> is the sum. Next, she picks <math>2008</math> of the numbers that are distinct from the <math>2008</math> she picked the first time. Again she picks exactly one number from each row and column, and again the sum of all <math>2008</math> numbers is <math>S</math>. Find the remainder when <math>S</math> is divided by <math>2008</math>. |
==Solution== | ==Solution== | ||
Latest revision as of 23:07, 18 January 2024
Problem
Alexis imagines a 
grid of integers arranged sequentially in the following way:
![\[\begin{array}{r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r}1,&2,&3,&\ldots,&2008\\2009,&2010,&2011,&\ldots,&4016\\4017,&4018,&4019,&\ldots,&6024\\\vdots&&&&\vdots\\2008^2-2008+1,&2008^2-2008+2,&2008^2-2008+3,&\ldots,&2008^2\end{array}\]](http://latex.artofproblemsolving.com/1/a/4/1a4b1464ada14b3bb6ae040bcf9651336c1f891f.png)
She picks one number from each row so that no two numbers she picks are in the same column. She them proceeds to add them together and finds that 
is the sum. Next, she picks 
of the numbers that are distinct from the she picked the first time. Again she picks exactly one number from each row and column, and again the sum of all numbers is . Find the remainder when is divided by .
Solution
Notice that all the numbers of the first column are congruent to 
modulo , all of the numbers in the second column are congruent to 
modulo , and so on. That means the sum of Alexis's numbers is congruent to 
modulo . After calculating the sum and dividing by , we find that 
, so the remainder when is divided by is 
.
See Also
| 2008 iTest (Problems) | ||
| Preceded by: Problem 50 |
Followed by: Problem 52 | |
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