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Expert: - 7/1/2010


Question
Hi Scott,

We had a test in my intermediate algebra class today.  On it was this question:

1.)A marching band has 52 members and there are 24 in the pom-pom squad.  They wish to form several hexagons and squares like those diagrammed below.  Can it be done with no people left over? If so, how many hexagons and squares should they form?

  B     B              P     P
B     P      B             B
  B     B              P     P

The answer I gave was: Not possible. The closest one can get it 7 hexagons and 4 squares, which would still leave 1 Pom Pom person out.  

I took the grueling "trial-and-error" approach and calculated possible combinations.  Was my answer technically correct and is there an easier, more algebraic way to obtain that answer?  

Thanks in advance,
Josh

Answer
The people must be grouped into groups of two groups.
The 1st group has n(6B+P).  The 2nd group has m(B+4P).

If there are 52 people in the band, with 24 on the pom-pom squad, that leaves 28 players.
We need 6n+m=28 and n+4m=24, and the answers need to be in integers for n and m.
Since there are two equations and two variables, there is only one solution.

The 1st equation says that m=28-6n.  Putting this in the 2nd equation gives n+4(28-6n)=24.
This multiplies out to n + 112 - 24n = 24.  That goes to -23n = 88, so the answer is not
an integer.

If there was 1 more player, so that there were 29 players and 53 in the band,
there would be 4 hexagons and 5 squares.  This would give 25+5=29 band members
and 4+20=24 pop-pom girls.

If there were 43 pom-pom girls, that would give 3 hexagons and 10 squares.
This would give 18+10=28 band members and 3+40=43 pom-pom girls.

If there were 4 less pom-pom girls, that would give 4 hexagons and 4 squares.
That would be 24+4=28 players and 4+16=20 pom=pom girls.

If there were only 6 band players (that's 22 less), there would be 6 squares.
That's 6 players and 24 pom-pom girls, which is what we have.

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