Eureka 2013 | Problems and Solutions | Level B

1. Jack and John are trying to fill a pool with buckets of water. Working alone, Jack can fill the entire pool in 4 hours. It takes John 5 hours to do the same. Jill wants to empty the pool. She can empty an entire pool by herself in 6 hours. If the pool is already half full when Jack and John begin filling it and Jill begins emptying it, how many minutes (nearest whole number) will it take for the pool to get filled?



If all 3 of them are working in the same time, the pool is being filled at the speed of 1/4 + 1/5 – 1/6 = 17/60 per hour. How quickly can they fill ½ (or 30/60) of a pool? In 30/17 hr = 106 minutes.

Answer: 106 minutes


2. For positive integers x and y, (x + y)2=576 and x2 + y2=290. What is the positive difference between x and y?


(x-y)2 = (x+y)2 – 4xy = (x+y)2 – 2[ (x+y)2  –  x2 – y2] = 2(x2+y2) – (x+y)2 = 2*290 – 576 = 4. Therefore, positive (x-y) = 2.

Another way to solve it:

(x + y)2=x+ 2xy + y2=576=x2 + y+ 286
Subtract x2 and y2 from both sides, you get 2xy =286 or xy=143. 11 x 13=143. 13-11=2.

Answer: 2


3. Last year, the 300th day of the year was a Monday. This year, the 200th day is also a Monday. What day of the week was January 1st this year?



If last year was regular, then 265 days passed between the two Mondays. If last year was a leap year, then 266 days passed between the two Mondays. Since it had to be a whole number of weeks and 7 divides 266, we know that last year was a leap year.

January 1st of this year was 67th day after the 300th day of last year. 67 divided by 7 leaves a remainder of 4. So January 1st must have been Friday, the 4th day after Monday.

Answer: Friday


4. Find all real numbers x such that (2x – 4)3 + (4x – 2)3 = (4x +2x  – 6)3



Let a = 2x – 4 and b = 4x – 2. Then we have a3 + b3 = (a+b)3. This leads to

3ab(a+b) = 0. So either a=0, b=0 or a+b = 0.

When a=0, x= 2

When b=0, x= ½

When a+b=0, x=1. Other solutions don’t exist for this case, because a+b is a strictly increasing function of x.

Answer: ½, 1, and 2


5. In rectangle ABCD we have AD = 1, P is on the side AB, and lines DB and DP trisect angle ADC. What is the perimeter of triangle BDP?








6. A 7-digit telephone number d1 d2 d3 d4 d5 d6 d7 is called memorable if the prefix d1 d2 d3 is exactly the same as either d4 d5 d6  or d5 d6 d7 (possibly both). Assume each digit di can be any of the ten decimal digits 0,1,2…9. How many distinct memorable telephone numbers are out there?



There are 1,000 possibilities for d1 d2 d3. The telephone number can become memorable through equality with d4 d5 d6  or d5 d6 d7 or both. If the equality is achieved through d4 d5 d6, then there are still 10 possibilities for d7, and there are a total of 10,000 of such memorable numbers. If the equality is achieved through d5 d6 d7, then there are still 10 possibilities for d4, and there are a total of 10,000 of such memorable numbers. The overlap comes from 10 telephone numbers where all 7 digits are all the same. Total number of possibilities is 10,000 + 10,000 – 10 = 19,990

Answer: 19,990



7. For how many values of n will an n-sided regular polygon have interior angles with integer degree measure? [Sum of angles of n-gon is 1800 · (n-2)]



Each angle of such n-sided polygon would be 180(n-2) / n = 180 – 360 /n. Clearly, the angle measure would only be integer if n is a divisor of 360 (and n≥3, because it is a triangle or a larger number of sides).

The prime decomposition is 360  = 23 ·5·9. Therefore, total number of positive divisors is 4·2·2 = 16. If we subtract those that are less than 3, we get 14 distinct divisors.


Answer: 14.



8. Let a(1), a(2), …., be a sequence with the following properties


(i)                a(1)= 1

(ii)              a(2n)= n ·a(n)

What is the value of a(2100) ?



a(2100)  = 299 · a(299) = 299 · 298 ·a(298) = ….299 · 298 ·…..2· 1 = 2 1+2+…98+99 = 24950


Answer: 24950






10. Abraham left city A to do a delivery in city S. Sarah left city S to do a delivery in city A at the same time as Abraham. They met each other 5 miles away from city A. Once Abraham and Sarah reached their respective destinations, they immediately turned around and started walking back. They met again 3 miles away from city S. What is the distance between cities A and S?



Let D = total distance between cities.

Prior to the 1st meeting, they jointly covered an entire distance D, while Abraham only covered 5 miles.

Prior to the 2nd meeting, they jointly covered an entire distance 3D (each covered a full distance D and they also shared 1 such such distance). Therefore, the time that passed from the beginning until the 2nd meeting is 3 times larger than the time that passed until the 1st meeting.


This means that Abraham covered a distance of 3*5 = 15 miles prior to the 2nd meeting. Since he was already 3 miles away from city S during 2nd meeting, the distance between cities D = 15 – 3 = 12. Algebraic solution is also available.


Answer: 12 miles.