Interesting Problems from the AMATYC Student Math League Exams 2003

Interesting problems from the AMATYC Student Math League Exams 2003

(November 2003, #1) If L has equation , M is its reflection across the y-axis, and N is its reflection across the x-axis, which of the following must be true about M and N for all nonzero choices of a, b, and c?

Suppose that L has an x-intercept of A and a y-intercept of B. Then its slope is . M has an x-intercept of -A and a y-intercept of B. So its slope is . N has an x-intercept of A and a y-intercept of -B,. So its slope is .

So the correct answer is C) the slopes are equal.

Or just notice that the addition of the dashed line must result in a parallelogram which forces M and N to have the same slope.

Or notice that the equation of M must be so its slope is , and the equation of N must be so its slope must be .[See the section on Graph Properties]

(November 2003, #2) A collection is made up of an equal number of pennies, nickels, dimes, and quarters. What is the largest possible value of the collection which is less than $2?

. So the largest possible value is .

So the correct answer is A)$1.64.

(November 2003, #3) When the polynomial is divided by , the remainder is . What is the remainder when is divided by ?

, so . But this means that the remainder when is divided by is

So the correct answer is B). [See the section on Polynomial Properties]

(November 2003, #4) If , find .

So the correct answer is B) 55.

(November 2003, #5) What is the remainder when is divided by ?

. So the answer is .

So the correct answer is A) .[See the section on Polynomial Properties]

(November 2003, #6) Let p be a prime number and k an integer such that has two positive integer solutions. What is the value of ?

must factor into where a and b are integers. But the only way this can happen is if one of them is p and the other is 1. This means that , so . This rearranges into .

So the correct answer is B) . [See the section on Polynomial Properties]

(November 2003, #7) What is the least number of prime numbers (not necessarily different) that 3185 must be multiplied by so that the product is a perfect cube?

. So to turn it into a perfect cube with the least number of multiplications by primes, we would need to multiply it by .

So the correct answer is E) 5. [See The Fundamental Theorem of Arithmetic]

(November 2003, #8) Two adjacent faces of a three-dimensional rectangular box have areas 24 and 36. If the length, width and height of the box are all integers, how many different volumes are possible for the box?

and , and , where H is a common factor of 36 and 24. So the number of different volumes is the same as the number of common factors of 36 and 24. The common factors are 1, 2, 3, 4, 6, and 12.

So the correct answer is E) 6.

(November 2003, #9)

So the correct answer is C) . [See the section on Trigonometric Formulas]

(November 2003, #10) The counting numbers are written in the pattern at the right. Find the middle number of the 40th row.

2 3 4

5 6 7 8 9

10 11 12 13 14 15 16

The middle numbers are generated by adding consecutive even numbers to 1.

Or another pattern is that the middle number of the nth row is given by , so .

So the correct answer is A) 1561. [See the section on Algebraic Formulas]

(November 2003, #11) The solution set of is a subset of the solution set of which of the following inequalities?

Yes

So the correct answer is C) .

(November 2003, #12) If and , find .

So the correct answer is A) . [See the section on Algebraic Formulas]

(November 2003, #13) Square ABCD is inscribed in circle O, and its area is a. Square EFGH is inscribed in a semicircle of O. What is the area of square EFGH?

The area of an inscribed square in a circle of radius r, is .

The area of an inscribed square in a semicircle of radius r, is .

So if , then the area of the square inscribed inside the semicircle is .

So the correct answer is B) . [See the section on Geometric Formulas]

(November 2003, #14) Consider all arrangements of the letters AMATYC with either the A’s together or the A’s on the ends. What fraction of all possible such arrangements satisfy these conditions?

A’s together:

A’s on the ends:

Total number of different arrangements:

So the correct answer is D) . [See the section on Sets and Counting]

(November 2003, #15) The year 2003 is prime, but its reversal, 3002, in not. In fact, 3002 is the product of exactly three different primes. Let N be the sum of these three primes. How many other positive integers are the products of exactly three different primes with this sum N?

, so . , since the sum of three distinct primes not equal to 2 must be an odd number, one of the primes must be 2.

Now let’s check the possible prime values for, and see which ones make a prime number as well.

So there are two other numbers: and

So the correct answer is C) 2.

(November 2003, #16) In a group of 30 students, 25 are taking math, 22 English, and 19 history. If the largest and smallest number who could be taking all three courses are M and m respectively, find .

So .

So the correct answer is E) 25. [See the section on Sets and Counting]

(November 2003, #17) A boat with an ill passenger is 7½ miles north of a straight coastline which runs east and west. A hospital on the coast is 60 miles from the point on shore south of the boat. If the boat starts toward shore at 15 mph at the same time an ambulance leaves the hospital at 60 mph and meets the ambulance, what is the total distance(to the nearest .5 mile) traveled by the boat and the ambulance?

So the total distance traveled is .

So the correct answer is E) 62.5.

(November 2003, #18) If each letter in the equation represents a different decimal digit, find T’s value.

and to produce a 6-digit number, MYM must be at least 323. So let’s start checking the squares of MYM values:

, so .

So the correct answer is E) 7.

(November 2003, #19) If a, b, c, and d are nonzero numbers such that c and d are solutions of and a and b are solutions of , find .

If c and d are solutions of , then

, so and .

If a and b are solutions of , then

, so and .

So we get the system

Solving the linear part

Leads to

With solutions of . Plugging this into the other two equations leads to and . So the solution of the system is

, so .

So the correct answer is A) .

(November 2003, #20) Al and Bob are at opposite ends of a diameter of a silo in the shape of a tall right circular cylinder with radius 150 feet. Al is due west of Bob. Al begins walking along the edge of the silo at 6 feet per second at the same moment that Bob begins to walk due east at the same speed. The value closest to the time in seconds when Al first can see Bob is

At a point on the circle where a tangent line will intersect the x-axis, , it will intersect the x-axis at the point .

Here’s why:

The slope of the tangent line at the point on the circle is , so an equation of the tangent line is . Setting y equal to zero and solving for x to find the x-intercept results in .

Al’s path can be parametrized by . So we need to solve the equation . An approximate solution of this equation is 48.00747736.

Here’s the graph of both sides of the equation:

So the correct answer is C) 48.