Difference between revisions of "1999 AHSME Problems/Problem 14"
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| − | So, the four girls | + | So, the four girls sang <math>4, 5, 5</math> and <math>7</math> songs in trios. That means there were <math>\frac{4 + 5 + 5 + 7}{3} = 7</math> songs sung, and the answer is <math>\boxed{A}</math>. |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1999|num-b=13|num-a=15}} | {{AHSME box|year=1999|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:34, 5 July 2013
Problem
Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang 
songs, which was more than any other girl, and Mary sang 
songs, which was fewer than any other girl. How many songs did these trios sing?
Solution
Alina and Tina must sing more than , but less than , songs. Therefore, Alina sang 
or 
songs, and Tina sang or songs, with possible combinations.
However, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by 
. Thus, 
must be divisible by , and 
must be divisible by .
Since 
by the bounds of 
, we have 
. Because the middle number must be a multiple of , we set 
, leading to 
and 
.
So, the four girls sang 
and songs in trios. That means there were 
songs sung, and the answer is 
.
See Also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
