Difference between revisions of "2002 AMC 12P Problems/Problem 22"

Revision as of 20:30, 10 March 2024

Problem

Under the new AMC $10, 12$ scoring method, $6$ points are given for each correct answer, $2.5$ points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between $0$ and $150$ can be obtained in only one way, for example, a score of $104.5$ can be obtained with $17$ correct answers, $1$ unanswered question, and $7$ incorrect answers, and also with $12$ correct answers and $13$ unanswered questions. There are scores that can be obtained in exactly three ways. What is their sum?

$\text{(A) }175 \qquad \text{(B) }179.5 \qquad \text{(C) }182 \qquad \text{(D) }188.5 \qquad \text{(E) }201$

Solution

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == Solution ==
-If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=21|num-a=23}}
 {{MAA Notice}}

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Difference between revisions of "2002 AMC 12P Problems/Problem 22"

Revision as of 20:30, 10 March 2024

Problem

Solution

See also