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February's Problem
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The perimeter of triangle ABC is 60 units, ,

and the length of each side is a whole number. How many non-congruent triangles satisfy this description?

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A solution to this problem will appear along with next month’s problem.
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January's Problem
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The total value of exactly 10 coins is 27¢.

How many of those coins are nickels?

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METHOD 1: Solve a simpler problem.
Since all coins other than pennies must end in a value of 0¢ or 5¢, let's determine how many of the coins must be pennies. The number of pennies must end in 2 or 7.
There can't be 2 pennies because the value of 8 other coins is more than 25¢ and there can't be 12 or more pennies since that's already more than 10 coins. So there are 7 pennies.
The other three coins, worth 20¢, must be 1 dime and 2 nickels.

METHOD 2: Use algebra.
1 quarter is already 25¢ and there must be 9 other coins, so there are no quarters. Represent the number of pennies by p, the number of nickels by n, and the number of dimes by d.
We know there are 10 coins, so p + n + d = 10. The value of the coins is 27¢, so p + 5n +10d = 27.
Since p appears in both equations, subtract the first equation from the second to eliminate p. This gives 4n + 9d = 17.
Since 4n can't equal the odd number 17, d must be at least 1. Since 10¢ x 2 is greater than 17¢, d must be smaller than 2. Thus d = 1, leaving 4n + 9(1) = 17.
This simplifies to 4n = 8, and n = 2. There are 2 nickels.

METHOD 3: Make a table.
Construct a table with each row equaling a combination of coins totaling 27¢.

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December's Problem
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If a + b = 10 and a - b = 8, what is the value of 3a + b?

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METHOD 1: Find a combination that produces the desired expression.
Since a + b = 10, then 2a + 2b = 20. The sum of 2a + 2b and a - b is 3a + b, so the value is 20 + 8 = 28.

METHOD 2: Solve for each variable.
Because a + b = 10 and a - b = 8, a = 9; leading to b = 1. Thus 3a + b = 3(9) + (1) = 28.

 

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November's Problem
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Jennifer buys a smartphone with a case for $200. The smartphone costs $180 more than the case. What is the cost of the case?

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METHOD 1: Use a property of sums and differences.
For any two numbers, their sum minus the difference between them equals twice the smaller while their sum plus the difference between them equals twice the larger. (Consider 3 and 50: Their sum is 53, the difference between them is 47, subtracting 53 - 47 = 6 which is twice the smaller and adding 53 + 47 = 100 which is twice the larger.)
The sum of the phone and case is $200 while their prices differ by $180. 200 - 180 = 20 which is twice the price of the case. The case costs $10. (Note, 200 + 180 = 380 which is twice 190. The phone costs $190.)


METHOD 2: Make a table.
Make a table of pairs of numbers which are 180 apart until you find the pair that have a sum of 200. The smartphone costs $190 and the case costs $10.

METHOD 2: Use algebra.
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For many additional problems we highly recommend the following books:

Math Olympiad Contest Problems for Elementary and Middle Schools by Dr. G. Lenchner
and
Math Olympiad Contest Problems Volume 2 edited by Richard Kalman
and
MOEMS® Contest Problems Volume 3
edited by Richard Kalman & Nicholas J. Restivo.
are sources of many such problems.

Creative Problem Solving in School Mathematics 2nd Edition by Dr. George Lenchner
can help you to teach solving these types of problems