# Jan 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.

Happy New Year!

Week of Jan 13 – Jan 19.

Jan 13: The sum of three numbers is 20. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

Jan 14: Patty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?

Jan 15: Some boys and girls are having a car wash to raise money for a class trip to China. Initially 40% of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then 30% of the group are girls. How many girls were initially in the group?

Jan 16: The area of a rectangle remains unchanged when it is made 2 1/2 inches longer and 2/3 inches narrower, or when it is made 2 1/2 inches shorter and 4/3 inches wider. Its area, in square inches, is:

Jan 17: The number of geese in a flock increases so that the difference between the populations in year n+2 and year n is directly proportional to the population in year n+1. If the populations in the years 1994, 1995, and 1997 were 39,60, and 123,respectively, then the population in 1996 was:

Jan 18: Elmo makes N sandwiches for a fundraiser. For each sandwich he uses B globs of peanut butter at 4 cents per glob and J globs of jam at 5 cents per glob. The cost of the peanut butter and jam to make all the sandwiches is \$2.53. Assume that B,J, and N are all positive integers with N>1. What is the cost of the jam Elmo uses to make the sandwiches?

Jan 19: Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

Week of Jan 6 – Jan 12. The seven problems this week are related to arithmetic and geometric sequences. If $a_1, a_2, a_3, \dots$ is an arithmetic sequence, then the neighboring terms have a common difference $a_i-a_{i-1}$; if it is a geometric sequence, then the neighboring terms have a common ratio $a_i/a_{i-1}$.

Jan 6: In a given arithmetic sequence, the first term is 2, the last term is 29, and the sum of all the terms is 155. The common difference is:

Answer: 3. The average of the terms is 31/2, so there are 10 terms in total.

Jan 7: If the sum of the first 10 terms and the sum of the first 100 terms of given arithmetic progression are 100 and 10, respectively, then what is the sum of the first 110 terms?

Answer: (a + (a+9d))*5 = 100, (a + (a+99d))*50 = 10. Therefore, d=-0.22 and 2a=21.98 We want (a+(a+109d))*55 = -110

Jan 8: The measures of the interior angles of a convex polygon of n sides are in arithmetic progression. If the common difference is 5∘ and the largest angle is 160∘, then n equals:

Answer: n=9.    (n-2)*180 = (160 + 160-5(n-1))*n/2.

Jan 9: In the sequence 2001, 2002, 2003, …, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001+2002−2003=2000. What is the 2004th term in this sequence?

Jan 10: A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

Answer: Let the three numbers in the arithmetic progression be 9, 9+d, 9+2d. The three numbers in the geometric sequence are 9, 11+d, 29+2d. So (11+d)^2 = 9(29+2d). Therefore, d=-14 or d=10. The smallest possible value is 29+2(-14)=1.

Jan 11: The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?

Answer: 2005, 133, 55, 250, 133, 55, 250… So the 2005th term is 250.

Jan 12: Suppose that $u_n$ is a sequence of real numbers satisfying $u_{n+2}=2u_{n+1}+u_n$, and that $u_3=9$ and $u_6=128$. What is $u_5$?

Answer: $u_6 = 2u_5+u4$, $u_5=2u_4+u_3$. This translates to $128=2u_5+u_4$, $u_5=2u_4+9$$u_5=53$.