400 Challenging Programming Problems in School Mathematics

If you are looking for programming problems to challenge yourself or to give to your students, then Project Euler is one good place to start.  According to the website, “Project Euler exists to encourage, challenge, and develop the skills and enjoyment of anyone with an interest in the fascinating world of mathematics.” True indeed, the site has collected more than 400 programming problems in mathematics.

Project Euler Programming Problems

Many of the problems are accessible even to Grade 8 students  but some are a bit challenging.  Non-computer science majors and non-mathematics majors who want to keep their “problem solving and mathematics on the edge” are also encouraged to try. To give you an idea of the difficulty of the problems, one problem is shown above.

If you want to take the challenge of Project Euler, you may want keep track of your progress by registering.

Would you have made it into Harvard in 1869?

If you think your tough in math and can pass the 1869 Harvard math exam, don’t rejoice so much because there are other subjects as well.

Click the picture to view full exam

Other subjects included are History and Geography, Latin Grammar, English to Latin Translation, Greek Grammar with accents, and Greek Composition :-D.

Surprisingly, according to the original article, 185 out of 210 examinees passed the exam.

Problem Set 2

PROBLEMS

1.) Find a linear function f(x) such that f(1) = 42 and f(2) = 47.

2.) Solve for x: 4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 170.

3.) Prove that the product of 3 consecutive numbers is always divisible by 6.

4.) Prove that if p is prime, a and b are integers, and a \equiv b\mod p, then a^p \equiv b^p \mod p.

SOLUTIONS AND PROOFS

Post Date: October 20, 2009

1. Solution: This is just the same as saying, find the equation of the line passing through (1,42) and (2,1337). So, by point slope formula, we have, y - y_1 = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\Rightarrow y - 42 = \displaystyle\frac{1337 - 42}{2 - 1}(x - 1). \Rightarrow y = f(x) = -18x + 92.

2.) Solution:4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 4^x(1 + 4 +4^2 +4^3) = 170 \Rightarrow 4^x(85) = 170\Rightarrow 4^x = 2 \Rightarrow 2^{2x}=2^1 \Rightarrow x = \displaystyle\frac{1}{2}.

3.) Proof: A number is divisible by 6 if it is divisible by 2 and 3. A product of 3 consecutive numbers is divisible by 2 because at least one of them is even, so it remains to show it is divisible by 3.

If a number is divided by 3, its possible remainders are 0, 1, and 2.  Assume n, n +1 and n+2 be the three consecutive numbers, and r be the remainder if n is divided by 3.

Case 1: If r=0, we are done.

Case 2: If r = 1, then n + 2 \Rightarrow r=0

Case 3: If r = 2, then n + 1 \Rightarrow r = 0.

Since the product of the three consecutive numbers is even, and for each case of r, one of the consecutive numbers is divisible by 3, the product of three consecutive numbers is divisible by 6. \blacksquare

4.) Proof: From definition, a^p \equiv b^p \mod p \Leftrightarrow b = a + kp for some k \in \mathbb{Z}.

Raising both sides of the equation to p, we have b^p = (a + kp)^p. By the binomial theorem,  b^p = (a + kp)^p = a^p + \displaystyle {p \choose 1}a^{p-1}kp + \displaystyle{p \choose 2}a^{p-2}k^2p^2 + \ldots + k^pp^p.

Notice that every term aside from a^p is divisible by p^2. (Why?). Therefore,  a^p \equiv 0 \mod p^2 .

Hence, then a^p \equiv b^p \mod p. \blacksquare

Problem Set 1

PROBLEMS

1.) The sum of two numbers is 18 and there difference is -4. What are the two numbers?

2.) Find the values of p, q and r if:

p + q   + r = 3 \frac{1}{2}
pq + qr + rp = - 2 \frac{1}{2}
pqr = -2

3.) Prove that \displaystyle\frac{x + y}{2} \geq \sqrt{xy}

4.) Define cos(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} and sin(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

a.) Prove that cos(-x) = cos(x)
b.) Prove that sin(x) = -sin(x)

SOLUTIONS AND PROOFS
Posted October 13, 2009

1.) Solution: Let x and y be the two numbers. Then, x + y = 18 and x - y = -4. Adding the equations, we have (x + y) + (x - y) = 18 + (-4) \Rightarrow 2x = 16 \Rightarrow x = 8. Substituting it to the first equation gives us y = 10. Therefore, the two numbers are 6 and 10.

2.) Solution: From the given, p, q and r are roots of of the cubic equation x^3 - \frac{7x^2}{2} - \frac{5x}{2} + 2. Factoring, we have (x+1)(x-4)(2x + 1)=0. Therefore, x = 1, 4 or -1/2

3.) Proof: We know that the square of the difference of any two numbers is always positive or 0. Let x, y be any two numbers. Then, (x - y)^2 \geq 0. Expanding, we have x^2 - 2xy + y^2 \geq 0. Adding 4xy to both sides of the equation yields x^2 + 2xy + y^2 \geq 4xy \Rightarrow (x + y)^2 \geq 4xy. Getting the square root of both, we have, x + y \geq 2 \sqrt{xy} \Rightarrow \frac{x + y}{2} \geq \sqrt{xy}. \blacksquare

4.) Proof (a): We want \cos (-x) so we will just replace x‘s with -x. Therefore, cos(-x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-x)^{2n}}{(2n)!} =  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1 \cdot x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^{2n} \cdot x^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}= cos(x). \blacksquare

Proof of 4b is left as an exercise. It’s very similar to the proof of 4a.

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