UND MATHEMATICS TRACK MEET Individual Test 3

University of North Dakota Grades 11/12
January 12, 2026
School Team Name
Calculators are allowed. Solutions Student Name

• 1.

  Find the value(s) of $a$ such that the equation $3x^2-5x+a=0$ has exactly one solution. Give an exact answer.

    (2 pts) 1. ${\displaystyle \frac{25}{12}}$

  Solution: We need the discriminant $b^2-4ac=0$ or $25-12a=0$, so we want $a=25/12$.

• 2.

  An investment account is setup with an initial investment and, after 25 years, contains one million dollars. If the amount in the account doubled every 7 years while increasing continuously, find the initial amount invested. Give your answer in dollars and cents.

    (3 pts) 2. $84,118.76

  Solution: If $s$ is the number of doubling times and $A$ is the initial amount in the account, there are $A{2}^s$ dollars in the account. After $s=25/7$ doubling times, there are a million dollars, so we solve $1,000,000=A{2}^{25/7}$ for $A$ to get $A=1,000,000\cdot 2^{-25/7}\approx 84118.76$

• 3.

  A bag contains 3 blue, 2 green, and 5 red marbles. What’s the probability that there are still green marbles in the bag after 4 marbles have been removed? Give the exact answer.

    (3 pts) 3. ${\displaystyle \frac{182}{210}}={\displaystyle \frac{13}{15}}$

  Solution: We need to find the probability that we chose zero or one green marble from our four. There are $\left(\genfrac{}{}{0pt}{}{10}{4}\right)=210$ total outcomes for the four marbles selected. There are $\left(\genfrac{}{}{0pt}{}{8}{4}\right)=70$ outcomes with zero green marbles selected and $2\left(\genfrac{}{}{0pt}{}{8}{3}\right)=112$ outcomes with one green marble selected. Therefore, the probability is $\frac{70+112}{210}=\frac{182}{210}=\frac{13}{15}$

• 4.

  A group of people is randomly selected to come to a soccer field. Find the minimum number of people required to ensure that it’s possible to form two teams of 11 people such that any two teammates have the same birthday (though are not necessarily the same age). Remember, February 29th is a valid birthday for those born on leap years!

    (3 pts) 4. 3672

  Solution: Since there are 366 possible birthdays, the worst case scenario where no team of 11 can be formed with the same birthday is 3660 people having 10 people per birthday. One more person ensures we have at least one team and we need to add 11 more after this since these 11 could all have the same birthday as the first team. So $3660+12=3672$ people are required to ensure this happens.

• 5.

  Find the units place of the sum $1+2026+{2026}^2+{2026}^3+\mathrm{\dots }+{2026}^{2026}$

    (3 pts) 5. $7$

  Solution: Since $2026\equiv 6(mod10)$, we have ${2026}^2\equiv 6^2\equiv 6(mod10)$, ${2026}^3\equiv 6^3\equiv 6^2\equiv 6(mod10)$, and so forth for every power of $2026$. Therefore, $1+2026+{2026}^2+\mathrm{\dots }+{2026}^{2026}\equiv 1+6+\mathrm{\dots }6(mod10)$. There are $2026$ copies of 6 for $1+2026\cdot 6\equiv 1+6\cdot 6\equiv 7(mod10)$.

• 6.

  A square is contained in an equilateral triangle with side length 2 such that the bottom edge of the square lies along the base of the triangle and the two top vertices of the square lie on the other two edges of the triangle as shown in the figure below. Find the side length of the square. You may round your answer to two decimal places.

    (3 pts) 6. $\frac{2\sqrt{3}}{2+\sqrt{3}}\approx 0.93$

  Solution: Let $s$ be the square side length. Since the top of the square is parallel to the base of the triangle, it cuts off a triangle on top of the square similar to the whole triangle and hence this upper triangle must also be equilateral. This upper equilateral triangle has side lengths $s$ and so has height $\sqrt{3}s/2$, meaning the height $\sqrt{3}$ of the large triangle equals $s+\frac{\sqrt{3}s}{2}$ allowing us to solve for $s$ as $s=\frac{2\sqrt{3}}{2+\sqrt{3}}$.

• 7.

  A spiral staircase climbs the outer wall of a 20 story circular tower with radius 10 meters. Given that the staircase completes one rotation every 3 stories and each story is 3 meters tall, find the (straight line) distance, in meters, between the bottom and top of the staircase. You may round your answer to two decimal places.

    (3 pts) 7. $\sqrt{3900}\approx 62.45$

  Solution: We model horizontal position on the staircase with $x^2+y^2=100$ with the bottom of the staircase at $(10,0)$. At the top, we are $2/3$ of the way through a rotation, meaning at angle $4\pi /3$ at position $(10\cos (4\pi /3),10\sin (4\pi /3))=(-5,-5\sqrt{3})$. Assuming the staircase starts at $z=0$, the initial and final positions for the staircase are $(10,0,0)$ and $(-5,-5\sqrt{3},60)$ which are separated by distance
  $\sqrt{{(10-(-5))}^2+{(0-(-5\sqrt{3}))}^2+{(60-0)}^2}=\sqrt{3900}\approx 62.45$