MATH COUNTS® Problem of the Week Archive It All Adds to Ten – August 2, 2010
Problems
Thank you to Tom Butts for contributing this week's set of problems!!
Today’s date is 8/2/10 – a date where the sum of the month and the day equals the number formed by the last two digits of the year. How many such dates are there in 2010?
How many ordered sets of three positive integers have a sum of 10? Note: The set 8, 1, 1 is different from the set 1, 8, 1.
How many ordered sets of positive integers with two or more elements have a sum of 10? Solutions
1/9, 2/8, 3/7, 4/6, 5/5, 6/4, 7/3, 8/2, 9/1
So, 9 is our answer.
This one can be solved by making an organized list, as follows:
# of possible arrangements
x1 x2 x3
8 1 1 3
7 2 1 6
6 3 1 6
6 2 2 3
5 4 1 6
5 3 2 6
4 4 2 3
4 3 3 3
Total poss. ordered sets = 36
Another way to solve this would be by thinking of a row of ten 1s
1_1_1_1_1_1_1_1_1_1
Each possible arrangement corresponds to inserting a “+” sign in two of the nine “_” between the 1s. For example 1_1 + 1_1_1_1_1 + 1_1_1 corresponds to the arrangement 2 + 5 + 3 = 10. Thus there are 9 choose 2 ways to choose the two spaces, which is 36.
Problem 1 shows there are 9 choose 1 (9C1) sets of two positive integers whose sum is 10. Problem 2 shows there are 9C2 sets of three positive integers whose sum is 10. Generalizing the secondary solution method for problem 2, there are 9 choose 3 sets of 4
positive integers whose sum is 10. Continuing, the total number of sets of positive numbers whose sum is 10 is 9C1 + 9C2 +… + 9C9 = 29 – 1 = 511。