math

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AMC 10-2007 (B25)

How many pairs of positive integers (a, b) are there such that a and b have no common factors greater than 1 and a/b + 14b/a is an integer?

(A) 4    (B) 6    (C) 9    (D) 12    (E) infinitely many

We can rewrite the fractions as (9a^2 + 14b^2) / 9ab. Therefore, 3|b. Since a and b are coprime, a|9a^2 + 14b^2, so a|14. Similarly, b|9. However, b cannot be 9 as 81a|9a^2 + 81*14 only has solutions when 3|a, which can’t happen when a|14. So, b=3, and a can be {1, 2, 7, 14}. Checking shows us that all 4 work, so the answer is A (4).

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Suppose triangle ABC has 3 acute angles A, B, and C. Which quadrant(s) is the point (cos B - sin A, sin B - cos A) in?

Since ∠C < π/2 (defn of acute), we know that ∠A + ∠B > π/2.

A > π/2 - B so sin A > sin π/2 - B so sin A > cos B. Therefore, cos B - sin A < 0.

With the same logic, we can find that sin B > cos A, so sin B - cos A > 0.

A point with the coordinates (less than 0, greater than 0) is in teh second quadrant.

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AMC 10-2019 (A12)

Melanie computes the mean µ, the median M, and the modes of the 365 values that are the dates in the months of 2019. Thus her data consists of 12 1s, 12 2s, ... , 12 28s, 11 29s, 11 30s, and 7 31s. Let d be the median of the modes. Which of the following statements is true?

(A) µ < d < M    (B) M < d< µ     (C) d = M = µ     (D) d < M < µ     (E) d < µ < M

d obviously must be less than M, so we can rule out (B) and (C). Since there are 365 entries, M is the 183rd number, which lies between 15*12 + 1 (180) and 16*12 (192). Therefore, M=16. We can also see that d = 14.5. If we ignored the 11 29s, 11 30s, and 7 31s, µ would also be =14.5. Because there are larger elements given, though, we can see that µ is greater than d. On the other hand, since there are fewer 29's, 30's, and 31's than the rest of the numbers, the mean has to be lower than the median. So, the answer is E (d < µ < M).

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AMC 10-2019 (A4)

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

(A) 75    (B) 76    (C) 79    (D) 84    (E) 91

The maximum number of balls we can draw without drawing 15 of the same color comes from this scenario: we draw 14 red, 14 green, 14 yellow, 13 blue, 11 white, and 9 black. This gives us 75 balls. Drawing one more ball will guarantee us 15 of the same color, so the answer is B (76).