We are excited to offer two challenge problems to students.  Level 1 is geared for students in grades K-2 and Level 2 is geared for students in grades 3-6.  However, students should work on the problem that is appropriately challenging for them.  We encourage all students to participate.  

There is a new RMC display in the 2nd floor hallway across from 3L where you can find the problems.  There is also a box where students can submit their answers.  

You can expect to see new problems twice a month.  Give it a try!

10-24-11 candles on the cake

10-24-11 not so magic squares

Last week’s solution

45. From the definition, the first and second digits of an upright integer automatically determine the third digit, which is the sum of the first two digits. Consider those upright integers beginning with 1: 101, 112, 123, 134, 145, 156, 167, 178, and 189; there is a total of 9 such numbers. (Note that the second digit may not be 9; otherwise, the last digit would be 1+9=10.) Beginning with 2, the upright integers are 202, 213, 224, 235, 246, 257, 268, and 279; there is a total of 8 such numbers. We may continue this pattern of analysis to show that the numbers of upright integers beginning with a digit of 3, 4, 5, 6, 7, 8, or 9 are 7, 6, 5, 4, 3, 2, and 1, respectively.

Therefore, there is a total of 9+8+7+6+5+4+3+2+1=45 three-digit upright integers.

From Eric Lass, 7th & 8th grade Math Teacher

Last week’s question and solution

Problem: Daniel is having dinner with a friend. He buys five dishes and his friend buys three dishes. At the last minute another friend comes to eat with them and pays $4 for his share of the meal. If all the dishes have the same value, how should the money be divided between Daniel and his first friend?

Solution: Since the second friend paid $4, the total cost of the meal must be $4 x 3 = $12. Eight dishes were eaten, so each one costs $1.50. Daniel brought 5 x $1.50 = $7.50 worth of food, and since his share was $4, he should receive $3.50. The first friend brought 3 x $1.50 = $4.50 worth of food and should receive 50 cents.

From Eric Lass, 7th & 8th grade Math Teacher

Last week’s question and solution

Problem

The average house price in Boomtown rose 30 percent each year for the last five years. If the average house price is currently $250,000, what was the average house price five years ago?

Solution

Approximately $67,332. If P was the average house price five years ago, then the current average price is 1.35P or 3.71293P. Thus, 250,000 = 3.71293P, so P=250,000/3.71293, which equals $67,332.27

From Eric Lass, 7th & 8th grade Math Teacher

Last week’s problem and solution

999 Coins

Starting with a single pile of 999 coins, a person does the following in a series of steps: In step one, he splits the pile into two nonempty piles. Thereafter, at each step, he chooses a pile with 3 or more coins and splits this pile into two piles. What is the largest number of steps that is possible?

Solution

997. The number of steps is one less than the number of piles, and 998 is the largest number of piles, 997 with 1 coin and 1 with two coins.

From Eric Lass, 7th & 8th grade Math Teacher

Last week’s problem and solution

A Quarter Pounder

A quarter-pound hamburger contains approximately 80 calories per ounce of meat, an average french fry contains about 14 calories, a cola contains about 10 calories per ounce, and a bun contains 200 calories. Suppose you have a quarter-pound hamburger with a bun and six ounces of cola. How many french fries can you eat and still keep your meal below 800 calories?

Solution
15. We have 80h + 14f + 10c + 200 < 800. Using the information in the problem we can write 80(4) + 14f + 10(6) + 200 < 800. This implies that 14f < 220 and f < 16. Therefore, you can eat 15 french fries.

From Eric Lass, 7th & 8th grade Math Teacher

How many french fries can you eat and still keep your meal below 800 calories?

Last week’s problem:

Perfect Squares and Cubes…NOT

How many numbers from 1 to 1 million, inclusive, are not perfect squares or perfect cubes?

Solution:

998,910. There are 1000 perfect squares between 1 and 1 million; these are the squares of the first 1000 integers. Similarly, there are 100 perfect cubesóthe cubes of the numbers from 1 to 100. Subtract the squares and the cubes from 1 million to get 998,900. However, every number that is a perfect sixth power has been subtracted twice (the largest of these is 106 = 1,000,000). Adding these back in gives 998,910.

Solution

The man wants to earn $500 in interest each year. Investing $4000 at 5 percent yields yearly interest of 0.05 x $4000, or $200; whereas $3500 invested at 4 percent yields 0.04 x $3500, or $140; so the remaining $2500 needs to be invested at a rate that yields interest of $160 per year (to have a total interest of $500 each year). If r is the annual percentage rate for the $2500, then

So the man needs to invest the remaining $2500 at 6.4 percent or higher to guarantee an interest income of at least $500 a year.

— Mr. Lass