High School Math Award -- MAA

MAA Math Award

The Mathematical Association of America offers annual math test to high school students called the AMC 12. The purpose of this test is to stimulate interest in mathematics and to develop talent through solving challenging problems. When I found out that I received the high score at my school, I asked my math teacher what kind of award I would be getting, and I thought he said a "pen," but it turned out to be a "pin".

Here are two sample problems. The solutions are provided below.

1. A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by 25% without altering the volume, by what percent must the height be decreased?

(A) 10 (B) 25 (C) 36 (D) 50 (E) 60

This problem harder than it looks. I had to think about it for awhile before I could get a definite answer. Hint: Imagine a cylinder with simple dimensions and see what happens when you increase the diameter by the specified amount, while keeping the volume the same. (Solution below)

2. The set of all real numbers x for which

log₂₀₀₄(log₂₀₀₃(log₂₀₀₂(log₂₀₀₁x )))

is defined is { x | x > c }. What is the value of c?

(A) 0 (B) 2001²⁰⁰² (C) 2002²⁰⁰³ (D) 2003²⁰⁰⁴ (E) (2001²⁰⁰²)²⁰⁰³

This problem is easier than it looks. Hint: Ignore the large logarithm bases initially. Just solve the problem for base 10 logarithms first, then think about the large bases.

Answer to problem 1: The volume of a cylinder is the area of the circular base multiplied by the height: V = pi*r*r*h. Suppose that the original jar has a radius of 1 and a height of 1. Then the volume of the jar is pi*1*1*1 = pi. The new jar has a radius 25% larger, or 1.25 = 5/4, a smaller height x, and the same total volume, pi. Thus, its volume V = pi = pi*(5/4)*(5/4)*x, so x = 1 / [(5/4)*(5/4)] = 1/(25/16) = 16/25 = 64/100 = .64. The height must be reduced from 1.00 to .64, or a reduction of 36 percent. Thus, the correct answer is (C).

Answer to problem 2: To start, let's ignore the large logarithm bases, and consider the expression using conventional base-10 logarithms:

log(log(log(logx )))

The problem is asking: What is the set of real numbers x for which the expression is defined (not undefined)? The value of x can be large, but there is a lower limit to its allowed range. What is the lower limit of x in this expression?

Let's review a few principles of logarithms:

1. The argument of the logarithm must be greater than zero.

2. The logarithm of 1 is 0.

3. The logarithm of 10 is 1.

4. The logarithm of 10¹⁰ is 10.

5. The logarithm function is monotonically increasing. In other words, if the argument gets bigger, its logarithm gets bigger as well.

Remember that a logarithm can be calculated only for a positive number, not for zero or for a negative number. Starting from the outside (leftmost) logarithm:

log(log(log(log x))) = log( a number greater than 0 )

Therefore log(log(log x)) must be a number greater than 0. In order for this expression to be greater than 0, its argument must be greater than 1:

log(log(log x) = log( a number greater than 1 )

Therefore log(log x) must be > 1. In order for this expression to be greater than 1, its argument must be greater than 10, the logarithm base:

log(log x) = log( a number greater than 10)

Therefore log x must be > 10. In order for this expression to be greater than 10, the argument x must be greater than 10¹⁰, because log (10¹⁰) is 10.

So we have solved the problem for the case of base-10 logarithms. So what about those crazy large bases?

log₂₀₀₄(log₂₀₀₃(log₂₀₀₂(log₂₀₀₁x )))

It's really the same principle:

The argument of log₂₀₀₄ must be greater than 0.

The argument of log₂₀₀₃ must be greater than 1, because log₂₀₀₃(1) = 0.

The argument of log₂₀₀₂ must be greater than 2002, the logarithm base, because log₂₀₀₂(2002) = 1.

The argument of log₂₀₀₁ must be greater than 2001²⁰⁰², because log₂₀₀₁(2001²⁰⁰²) = 2002.

Thus, the correct answer is (B).