March 2019 (Mathcounts)

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.
For previously posted problems, click on the “Archive” tab.
For previous prizes, click on the “News” tab.


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 2^5\cdot 5^8.

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) \sqrt 13    B) \sqrt 74   C) \sqrt 2     D) \sqrt 15   E) \sqrt 8.

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 6^3). 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 216^8

March 11: Find x if x = \sqrt {2 \sqrt{2 \sqrt { 2\sqrt{\dots} }} }. (Hint: square both sides and write an equation in terms of x.)

Answer: x^2 = 2x. Therefore, x=2.

March 12: Find the value of {1+{1\over{1+{1\over{1+{1\over {1+\dots}}} }}}}

Answer: x = 1+1/x. Therefore, x = (-1+\sqrt{5})/2.

March 13: B and 9 are two solutions to the equation x^2-4x=A.  What is A+B?

Answer: B+9 = 4 and 9B = -A. So B=-5 and A=45. A+b=40.

March 14: If x+{1\over x} = 3, what is x^2 + {1\over x^2}? (Hint: you don’t need to solve for x first.)

Answer: Square both sizes of the equation. x^2 + 2 + {1\over x^2} = 9. Therefore, x^2+{1\over x^2} = 7.

March 15: If x+{1\over x} = 3, what is x^4 + {1\over x^4}? (Hint: use your answer from the previous problem.)

Answer: Square both sides again. x^4 +2 + {1\over x^4} = 49. Therefore, x^4+{1\over x^4} = 47

March 16: If x+{1\over x} = 3, what is x^3 + {1\over x^3}?

Answer: (x^2+{1\over x^2}) ( x+{1\over x}) = x^3 + x + {1\over x}+ {1\over x^3} = 21. Therefore, x^3 + {1\over x^3} = 18.  

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 \sqrt{7^2+24^2} = 25, 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\sqrt{3}. The area is 18\sqrt{3}.

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. a+b=49, a^2+b^2=41^2. We have ab=(49^2-41^2)/2 = 360. a= 40, b=9, and the area is 360.

March 6: Find the area of \triangle ABC if AB=BC=12 and \angle ABC=120^o.

Answer: The height AH = 6, BC = 12\sqrt 3. So the area is 36\sqrt 3.

March 7: A regular hexagon has perimeter p and area A. Compute p^2/A.

Answer: The side length is {p\over 6}. The area A= 6\cdot {1\over 2} \cdot {p\over 6}\cdot {\sqrt{3}p\over 6} = {\sqrt{3} p^2\over 12}. Therefore, {p^2\over A} = {12\over \sqrt{3} } = 4\sqrt{3}.

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 44^o. 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.