April 10, 2017

State competitions have officially wrapped up and we know have our 224 national competitors determined! Let’s get excited for another round of competition by trying some of this year’s state competition problems that the competitors tackled!

Eleven boastful bees are all lined up in a row. Each bee, after the first one, brags that it collected one more than twice as many grains of pollen as the bee in front of it. If the first bee has 100 grains of pollen, how many grains did the last bee collect?

[Sprint #11]   

Let bi be the number of grains collected by bee number i. Then b2 = 2b1 + 1, b3 = 2b2 + 1 = 4b1 + 3, b4 = 2b3 + 1 = 8b1 + 7, and so on.  We see that the coefficient of b1 goes up by a factor of 2 each time and the added constant always stays 1 less than the coefficient of b1. We can generalize as the formula bi = (2i ­– 1)b1 + 2i – 1 – 1. Therefore, b11 = (210)b1 + 210 – 1. Therefore, the number of grains collected by the last be is 1024 × 100 + 1023 = 102,400 + 1023 = 103,423 grains.

 

A positive integer q is the product of a prime number and a perfect square. Additionally, q is the product of a different prime number and a perfect cube. What is the least possible value of q?

 

[Sprint #17]   

The two least primes are 2 and 3, so q has the form 2a3b for some positive integers a and b. The specification that q is the product of a prime and a power in two different ways indicates: q = 2 × 2a – 1 × 3b and q = 3 × 2a × 3b – 1, where one of 2a – 1 × 3b and 2a × 3b – 1 is a perfect square and the other is a perfect cube. To be a perfect square, both exponents must be even (divisible by 2); to be a perfect cube, both exponents must be divisible by 3. If 2a – 1 × 3b is to be the perfect square and 2a × 3b – 1 the perfect cube, then a needs to be odd and divisible by 3, and b needs to be even and leave a remainder upon division by 3: The least positive integers satisfying these properties are a = 3 and b = 4. Similarly, if we reverse the role of the perfect square and perfect cube, we get a = 4 and b = 3. To obtain the least q, we want the lesser exponent on the greater base, so a = 4 and b = 3, which yields q = 24 × 33 = 2 × 63 = 2 × 216 = 432.

 

A pole of radius 1 inch and height h inches is perpendicular to the ground. A snake is coiled around the pole in a spiral pattern, making a 60-degree angle with the ground. If the snake is 9π inches long and reaches from the ground to the top of the pole, then h can be written in the form aπ√b, where a is a rational number and b is a positive integer that is not divisible by the square of any prime. What is the value of a + b? Express your answer as a common fraction.

[Sprint #19]   

Imagining the pole to be like a rolled-up sheet of paper, make a vertical slice down the poll and unroll it into a flat sheet. The width of the sheet is the circumference of the original pole, which is 2π inches, because the radius is 1 inch. The layout of the snake on this sheet forms a 30‑60‑90 right triangle, with the short leg being the circumference of the pole. For one wind around the pole, the snake goes up vertically 3 times as far and “uses up” 2 times as much snake. Therefore, 4π inches of snake reaches a height of 23 × π inches. Since the snake is 9π inches long in total, we can set up the ratio 9π/4π = h/(23 × π), where h represents the height of the pole. Therefore, h = (9/4) × 23 × π inches = (9/2) × √3 × π inches, so a = 9/2, b = 3 and a + b = 15/2.

 

Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2
contains ONLY PROBLEMS.