Difference between revisions of "1999 AHSME Problems/Problem 14"
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==Problem== | ==Problem== | ||
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==Solution== | ==Solution== | ||
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+ | Alina and Tina must sing more than <math>4</math>, but less than <math>7</math>, songs. Therefore, Alina sang <math>5</math> or <math>6</math> songs, and Tina sang <math>5</math> or <math>6</math> songs, with <math>4</math> possible combinations. | ||
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+ | 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 <math>3</math>. Thus, <math>7 + 4 + a + t</math> must be divisible by <math>3</math>, and <math>a+t - 1</math> must be divisible by <math>3</math>. | ||
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+ | Since <math>10 \le a+t \le 12</math> by the bounds of <math>5 \le a, t \le 6</math>, we have <math>9 \le a + t - 1 \le 11</math>. Because the middle number must be a multiple of <math>3</math>, we set <math>a + t - 1 = 9</math>, leading to <math>a + t = 10</math> and <math>a = t = 5</math>. | ||
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+ | 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>. | ||
==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 14:34, 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 |
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