Math Olympiads for Elementary and Middle Schools

Problem of the Month

Ten Fibonacci numbers are split into three sets:
A = {1, 2}, B = {3, 5, 8}, and C = {13, 21, 34, 55, 89}.
One number is chosen at random from each set.
What is the probability that their sum is odd?

 

Answer= 15/30

 

Each choice from set A can be paired with any of 3 choices in set B and each of the resulting 6 choices can then be paired with any of 5 choices in set C. Thus, 2 × 3 × 5 = 30 different results are possible. An odd sum is the result of adding 3 odd numbers or of adding 2 evens and one odd.

Which Combination?

   A        B        C   

Resulting Probability

Odd A, Odd B, Odd C (O, O, O)

    1/2  x  2/3  x  4/5

      8/30

Odd A, Even B, Even C (O, E, E)

    1/2  x  1/3  x  1/5

      1/30

Even A, Odd B, Even C (E, O, E)

    1/2  x  2/3  x  1/5

      2/30

Solutions to this problem will appear along with next month’s problem.

February's Problem

February's Problem

The rectangle shown consists of three large congruent squares, four small congruent squares, and a shaded rectangle. AB = 7 m and CD = 5 m. What is the area of the shaded rectangle, in square meters?

Answer is: 14

METHOD


1: Find the dimensions of the shaded rectangle.

The side lengths of the small and large squares are 2m and 5 m, respectively, and the dimensions of the entire figure are 7m and 15 m. The shaded region measures 15 – 4×2 = 15 – 8 = 7 m long and 2 m wide, and its area is 2 × 7 = 14 sq m

METHOD

2: Use all the known areas.

The side lengths of the small and large squares are 2 m and 5 m, respectively, and the dimensions of the entire figure are 7m by 15 m. The total area of the three large squares is 3 × 52 = 75 sq m and the total area of the four small squares is 4 ×22 = 16 sqm. The total area of the unshaded squares is 75 + 16 = 91 sq m, making the area of the shaded region 105 – 91 = 14sq m.

January's Problem

Sara and Josh each threw 5 darts at the target shown. Sara scored 19 points by landing 2 darts in A and 3 in B. Josh scored 17 points by landing 1 dart in A and 4 in B. How many points are given for a dart landing in A?

Answer is: 5

METHOD


1: Find the difference between A and B

Notice that of the 5 darts each  both Sara and Josh threw one dart in A and 3 darts in B. The remaining dart was in area A for Sara and area B for Josh. Thus the two point difference between their scores is a result of those remaining darts. That means a dart in A scores 2 more points than a dart in B (i.e. B = A -2) The second sentence leads to the equation 5A – 6 = 19. (Or the third sentence leads to 5A – 8 = 17). Solving either equation results in A = 5.

METHOD

 

2: Use algebra.

Let a and b represent the points for one dart in regions A and B, respectively. Then: (1) Sara’s point-score is 2a + 3b = 19 and (2) Josh’s point-score is 1a + 4b = 17. Multiply equation (1) by 4 and equation (2) by 3. The resulting equations are: (3) Sara: 8a + 12b = 76 and (4) Josh: 3a + 12b = 51 Subtract (4) from (3) to get: (5) 5a = 25 (6) Then a = 5. A dart landing in A scores 5 points. Be sure to check.

December's Problem

A concession stand sells movie popcorn in only two sizes. Their prices are $4 and $7 per serving. What is the greatest popcorn-sales total, in dollars, that is NOT POSSIBLE?

Answer is: 17

COMMON TO BOTH METHODS: Find the first 4-consecutive possible numbers.

The smaller number involved is 4, so once four consecutive possible numbers is found, all following whole numbers are possible at least by the one way of adding 4’s. The following methods are used to determine the lowest set of 4 consecutive whole numbers that IS possible.

METHOD 1: Use combinations of the two numbers.

1
2
3
4 (=4)
5
6
7 (=7)
8 (=4+4)
9
10
11 (=4+7)
12 (=4+4+4)
13
14 (=7+7)
15 (=4+4+7)
16 (=4+4+4+4)
17
18 (=4+7+7)
19 (=4+4+4+7)
20 (=4+4+4+4+4)
21 (=7+7+7)

The first set of 4-consecutive whole numbers is 18, 19, 20, 21. Thus the greatest number that is NOT possible is 17.


METHOD 2: Use remainders.


All whole numbers leave a remainder of 0, 1, 2, or 3 upon division by the price of the smaller size, $4, so all
whole numbers can be partitioned into four independent sets by remainder:
Remainder = 0 (No $7 servings + any number of $4 servings)
0, 4, 8, 12, 16, 20, 24, 28, 32, …
Remainder = 1 (3 $7 servings + any number of $4 servings)
21, 25, 29, 33, …
Remainder = 2 (2 $7 servings + any number of $4 servings)
14, 18, 22, 26, …
Remainder = 3 (1 $7 serving + any number of $4 servings)
7, 11, 15, 19, 23, 27, …

Once again the first set of 4-consecutive whole numbers is 18, 19, 20, 21; and the greatest number that is NOT possible is 17.