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I cannot seem to solve this problem, although it seems extremely easy.

Let us say that there is an auditorium and we are to seat several boys and girls. There are exactly 14 boys seated in each row and exactly 10 girls seated in each column. If there are 3 empty chairs how do we show that the least number of chairs that can meet these conditions is 567 chairs?

Thanks!

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1 Answer 1

Let $r$ be the number of rows, and $c$ be the number of columns. Then there are $14r$ boys, and $10c$ girls. So we have the total number of seats $rc=14r+10c+3$ or $rc-14r-10c-3=0$. Completing the multiple, we get $(r-10)(c-14)=143=11*13$. So $r-10=11$ and $c-14=13$ or $r=21$ and $c=27$ giving 567 chairs. There is another solution with $r-10=13$ and $c-14=11$ giving 575 chairs.

An example seating 'b' is boy, 'g' is girl, and 'e' is empty.

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bbbbbbbbbgggggggggggggbbbbb
ggggggggbbbbbbbbbbbbbbggggg
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ggggggbbbbbbbbbbbbbbggggggg
bbbbbbgggggggggggggbbbbbbbb
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bbbbbgggggggggggggbbbbbbbbb
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It seems like it needs to be shown that the 567 solution is admissible by exhibiting an arrangement. –  Alex R. Apr 15 '12 at 4:35

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