If you’ve seen the math club flyer or read our article in the September Cheetah Chatter (page 13), you’ve seen this year’s set of sample problems.  I hope you’re enjoying them!

There are many, many interesting problems that kids learn to solve in math club.  This is only a small subset.  Don’t know what a “subset” is?  Don’t worry….  We cover that in math club too. 🙂

If you’ve given all the problems your best effort, you can check what you did against the answers here.  Note:  Even the most advanced students should find it very challenging to get all four correct!

1. Add all the odd numbers from 1 to 25, including 1 and 25. What do you get?

Answer:  169.  One thing we learn in math club is that there are many ways to solve the same problem.  Here are three different ways to solve this problem.  Each method is faster than the last one.

• Make a table of running totals:
• 1  3  5  7  9  11  13  15  17   19   21   23   25
• 1  4  9 16 25 36 49 64 81 100 121 144 169

The answer is 169.  (By the way, do you see the pattern in the bottom row?)

• We are adding numbers in an arithmetic series.
• arithmetic series count = (last-first)/jump + 1 = (25-1)/2+1=13
• arithmetic series mean = (first+last)/2 = (1+25)/2 = 13
• sum = count x mean = 13 x 13 = 169
• The sum of the first n odd numbers is n^2.  13^2 = 13 x 13 = 169.

2. Angela is 3 years older than Betty, who is 7 years younger than Charlie, who is 8 years older than Deron, who is 4 years younger than Evelynn, who is 11. How old is Angela?

Answer:  11.  Two ways to solve:

• Work backwards, one at a time:  Evelyn is 11 => Deron is 11-4=7 => Charlie is 7+8=15 => Betty is 15-7=8 => Angela is 8+3=11.
• Work backwards, all at once:  A = 11-4+8-7+3 = 11.

3. Trina is driving her car at 30 miles per hour. How many minutes will she take to go 4 miles?

Answer:  8.  Two ways to solve:

• Proportional reasoning:  60 miles per hour is 1 mile per minute.  So half of that, 30 miles per hour, is 1 mile per 2 minutes — that is, each mile takes her 2 minutes.  4 miles would take 4 x 2 = 8 minutes.
• Unit factors:  4 miles x (1 hour / 30 miles) x (60 minutes / 1 hour) = (4 x 1 x 60) / (30 x 1) minutes = 8 minutes.  [miles and hours cancel out]

4. (extra challenge)  Harry lives in Gry ndor, Cedric lives in Hu epu , Cho lives in Ravenclaw, and Draco lives in Slytherin. One day, the Un-Sorter Hat assigns them to new houses. Every student goes to one of the four houses, no student stays in their original house, and no two students go to the same house. How many ways are there to assign the four students to the four houses, using these rules?

Answer:  9.  We are looking for the number of “derangements” (arrangements where everything is out of place) for 4 items.  Two ways to solve:

• Logical reasoning:
• Suppose Harry goes to Hufflepuff (which we can write as Harry => H), Cedric’s house.  Cedric has 3 choices of where to go, and each one determines the arrangement of all students:
• If Cedric goes to Gryffindor (Cedric => G), Draco needs to leave Slytherin and go to Ravenclaw (Draco => R), which puts Cho in Slytherin (Cho => S).
• If Cedric goes to Ravenclaw (Cedric => R), it forces Draco => G and Cho => S.
• If Cedric goes to Slytherin (Cedric => S), it forces Cho => G and Draco => R.
• Similarly, Harry => R and Harry => S each give 3 possible arrangements.  3 + 3 + 3 = 9.
• Derangement Shortcut:  To find the number of derangements of 4 items, calculate the alternating sum of permutations 4!/0! – 4!/1! + 4!/2! -4!/3! + 4!/4! = 24/1 – 24/1 + 24/2 – 24/6 + 24/24 = 24 – 24 + 12 – 4 + 1 = 9.

Advanced note:  There is an even faster shortcut to calculate derangements, if you have a scientific calculator.  Calculate 4!/exp(1) and round to the nearest integer.  4!/exp(1) = 8.829…, which rounds to 9.  [On your calculator the exp() function might be labeled as the inverse of ln(), or it might be called e^x.]