# 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?

**Solution**

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**

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**2.** For positive integers *x* and *y*, (*x* + *y*)^{2}=576 and *x ^{2}* +

*y*=290. What is the positive difference between

^{2}*x*and

*y*?

**Solution**

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

Another way to solve it:

(*x* + *y*)^{2}=x^{2 }+ 2xy + y^{2}=576=*x ^{2}* +

*y*+ 286

^{2 }Subtract

*x*and

^{2}*y*from both sides, you get 2xy =286 or xy=143. 11 x 13=143. 13-11=2.

^{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 1^{st} this year?

**Solution**

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

**Answer: Friday**

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

**Solution**

Let a = 2^{x} – 4 and b = 4^{x} – 2. Then we have a^{3 }+ b^{3 }= (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?

**Solution**

**6.** A 7-digit telephone number d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} is called memorable if the prefix d_{1} d_{2} d_{3 }is exactly the same as either d_{4} d_{5} d_{6} or d_{5} d_{6} d_{7 }(possibly both). Assume each digit d_{i} can be any of the ten decimal digits 0,1,2…9. How many distinct memorable telephone numbers are out there?

**Solution**

There are 1,000 possibilities for d_{1} d_{2} d_{3. }The telephone number can become memorable through equality with d_{4} d_{5} d_{6} or d_{5} d_{6} d_{7 }or both. If the equality is achieved through d_{4} d_{5} d_{6}, then there are still 10 possibilities for d_{7}, and there are a total of 10,000 of such memorable numbers. If the equality is achieved through d_{5} d_{6} d_{7}, then there are still 10 possibilities for d_{4}, 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 180^{0} · (*n*-2)]

**Solution**

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 = 2^{3 }·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(2*n*)= *n* ·a(*n*)

What is the value of a(2^{100}) ?

**Solution**

a(2^{100}) ^{ }= 2^{99 }· a(2^{99}) = 2^{99 }· 2^{98 }·a(2^{98}) = ….2^{99 }· 2^{98 }·…..2· 1 = 2 ^{1+2+…98+99 }= 2^{4950}

^{ }

**Answer: 2 ^{4950}**

**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?

**Solution**

Let D = total distance between cities.

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

Prior to the 2^{nd} 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 2^{nd} meeting is 3 times larger than the time that passed until the 1^{st} meeting.

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

**Answer: 12 miles. **