* I will post* seven new problems every weekend, one for each day of the week. The idea is that you work on your own during the week and I will post answers the following Saturday. For those of you who practiced with me last year, the process is similar.

* For instructions* on submitting weekly answers, click on the “Logistics” tab.

*posted problems, click on the “Archive” tab.*

**For previously****, click on the “News” tab.**

*For previous prizes*

Week of March 17 – March 23.

Click GoogleForm to submit answers by Sat, March 23.

March 17: Find the sum of the digits in the product .

March 18: If 1 is added to the numerator of a fraction its value becomes 1/3. If 1 is subtracted from the denominator of the same fraction, its value becomes 1/4. That is the original fraction?

March 19: How many 3-digit numbers consist of even digits? (Zero cannot be the leading digit of a 3-digit number.)

March 20: Find 3 consecutive numbers such that if they are divided by 2, 3 and 4 respectively, the sum of the quotients will be the next higher number.

March 21: There are 5 men in a car. Find the sum of the ages of the 5 men if their ages in all possible pairs are given as 96, 79, 84, 82, 95, 100, 98, 83, 81 and 86.

March 22: A fruit market received 540 apples packed in boxes, where each box contains the same number of apples. The number of boxes was 6 less than twice the number of apples in each box. How many boxes were used to ship the 540 apples?

March 23: Points X and Y both have integer coordinates. Which one of the following cannot be the distance from X to Y?

A) B) C) D) E) .

Week of March 10 – March 16.

Click GoogleForm to submit answers by Sat, March 16.

March 10: There are 16 positive integer factors of 216 (or ). What is the product of all 16 factors?

Answer: We can pair up the 16 factors: 1 and 216, 2 and 108, 3 and 72… such that the product of each pair is 216. There are 8 such pairs. So the product is .

March 11: Find if . (Hint: square both sides and write an equation in terms of .)

Answer: . Therefore, x=2.

March 12: Find the value of

Answer: . Therefore, .

March 13: and 9 are two solutions to the equation . What is ?

Answer: and . So and . .

March 14: If , what is ? (Hint: you don’t need to solve for first.)

Answer: Square both sizes of the equation. . Therefore, .

March 15: If , what is ? (Hint: use your answer from the previous problem.)

Answer: Square both sides again. . Therefore, .

March 16: If , what is ?

Answer: . Therefore, .

Week of March 3 – March 9.

Click GoogleForm to submit answers by Sat, March 9.

March 3: In quadrilateral ABCD, AB=BC=CD=DA, AC=14 and BD=48. Find the perimeter of ABCD.

Answer: This quadrilateral ABCD is a rhombus, and so the diagonals are perpendicular to each other. Each side is , and the perimeter is 100.

March 4: Find the area of a rhombus with side length 6 and an interior angle with 120 degrees.

Answer: The diagonals are 6 and 6. The area is 18.

March 5: A diagonal of a rectangle has length 41, and the perimeter is 98. Find the area of the rectangle.

Answer: Let a and b two sides of the rectangle. , . We have . , and the area is 360.

March 6: Find the area of if AB=BC=12 and .

Answer: The height AH = 6, BC = 12. So the area is 36.

March 7: A regular hexagon has perimeter and area . Compute .

Answer: The side length is . The area . Therefore, .

March 8: The number of diagonals in a certain regular polygon is equal to five times the number of sides. How many sides does this polygon have?

Answer: 5n = n(n-3)/2. The number of sides is n=13.

March 9: The interior angle measures of a pentagon form an arithmetic progression. The difference between the largest and smallest angle measures is . Find the measure of the smallest angle, in degrees.

Answer: Let the 5 angles be a-22, a-11, a, a+11, a+22. The total angle 5a = 540. Therefore, a=108. The smallest angle is therefore 108-22=86.