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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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A solution to this problem will appear along with next month’s problem.
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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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October's Problem
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As shown, three 3-digit numbers have a 4-digit sum. Different letters represent different digits. What is the greatest prime factor of the sum, ACCA?

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METHOD 1: Reason using addition number facts.
For the ones column to sum to a number ending in A, B must be either 0 or 5.
In the hundreds column, neither A nor B can be 0, so B = 5.
In the thousands column A is either 1 or 2. Since A + 5 + 5 < 20, in the thousands column A = 1.
Thus, 155 + 515 + 551 = 1221. Use the test of divisibility for 11: 1221 = 11 × 111 = 11 × 3 × 37. Or use the test of divisibility for 3: 1221 = 3 × 407 = 3 × 11 × 37. Either way, the greatest prime factor is 37.

METHOD 2: Use algebra.
The sum in each place is A + 2B. The entire sum is 100(A + 2B) + 10(A + 2B) + (A + 2B) which simplifies to 111(A + 2B). 111 = 3x37, both of which are prime. In the original sum ACCA, the thousands digit A is the result of carrying and cannot be larger than 2 (upon further investigation, we can determine that it must be 1 and that the sum A + 2B must = 11). Even if (A + 2B) is the largest 2-digit prime sum possible beginning with 2 (i.e. 29), this is still a prime smaller than 37. So the largest prime factor is still 37.

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September's Problem
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An inexpensive toy normally sells for 20 cents each. At a special reduced price, a store sold all of its toys, one at a time, for a total of $3.91. How many of these toys were sold if the reduced price is a whole number of cents?

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METHOD: Find the new price.
Divide 391 cents by the new price which is an odd whole number less than 20. The result must also be a whole number. Dividing by 19 leaves a remainder. 391 divided by 17 is 23. Thus 23 toys sold for 17 cents each.

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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