# Eureka 2012 | Level A – Problems and Solutions

**Level A – Grades 7-8**

- 1. The sum of all digits in the numbers 34, 35, and 36 is 24 because (3+4) + (3+5) + (3+6) = 24. Find the sum of all digits in the first twenty-five natural numbers: 1, 2, 3, 4, 5, …., 23, 24, 25.

**Correct Answer: 127**

Solution:

Single digits: 1+2+3…+9 = 45

Teens: 10 + 45 = 55

Twenties: 6×2 + 1+2+3+4+5 = 12+15=27

Total: 45+55+27=127

- 2. When Jason crossed the finish line of a 60 meter race, he was ahead of Robert by 10 meters and ahead of Mike by 20 meters. Suppose Robert and Mike continue to race to the finish line without changing their rates of speed. By how many meters will Robert beat Mike?

**Correct Answer: By 12 meters**

Solution:

In the amount of time that it took Robert to cover 50 meters, Mike only covered 40. As such, his running speed is 4/5 that of Robert’s. While Robert covers the next 10 meters, Mike will only cover 8. He will then be at the 48 meter line, which is 12 meters away from the finish line.

- 3. A9543B represents a six-digit number in which A and B are digits different from each other. The number is divisible by 11 and also by 8. What digit does A represent?

**Correct Answer: 7**

Solution:

(Process of substitution was an acceptable way to solve this problem.)

The value of A9543B can be written in the form A95×1000+43B. A95×1000 is divisible by 8 because 1000 is a multiple of 8. 43B also must be divisible by 8. As such, B = 2. To solve for A, we’ll need to apply the divisibility by 11. For a number to be divisible by 11, the difference between the sum of its odd-place digits and the sum of its even-place digits must be 0 or a multiple of 11. A95432: the sum of odd-digits is 2+4+9=15; the sum of even-digit is A+5+3=A+8. The difference 15 = A – 8. A=7.

- 4. What is the smallest integer number with the sum of digits equal to 30?

**Correct Answer: 3999.**

Solution:

When looking for the lowest integer, you want to keep the highest numbers at right most positions. Try to get as many 9’s as possibly can from right to left**.**

- 5. After walking 2/5 of the bridge, Michael noticed his dog running towards him at the speed of 30 mph. If he walks back, he will meet his dog exactly at the beginning of the bridge. If he continues to walk forward, he will meet his dog exactly at the end of the bridge. What is Michael’s walking speed?

**Correct Answer: 6mph**

Solution:

The dog is clearly outside of the bridge. From the first condition we know that by the time the dog runs to the beginning of the bridge, Michael can cover 2/5 of the bridge. We can also deduce, that by the time the dog makes it to the beginning of the bridge, Michael can cover 2/5 of the bridge in the forward direction and be at the 4/5 mark. Therefore, Michael has only 1/5 of the bridge to cover, while the dog will be able to cover the full size of the bridge in the same time. Hence, the dog is 5 times faster than Michael. Michael’s walking speed is 6 mph.

- 6. Beauty and the Beast are playing a game. They are taking turns at breaking a chocolate bar, 6×9 squares, along the straight lines. Whoever can’t break the chocolate any further loses the game. Beauty started first. Who is going to win the game? How many turns will they take before the game is over?

**Correct Answer: Beauty will win after 53 turns**

**Solution:**

The game starts with exactly 1 piece of Chocolate. They must break it along straight lines. Key Observation: upon each move, a piece of chocolate is broken into 2, so the total number of pieces of chocolate increases by 1 at each turn. At the end of the game there will be 54 single squares of chocolate. Therefore, it will take 53 turns before the game is over (regardless of how they play it). Since Beauty took the 1^{st} turn, she will also take the 53^{rd} one. Beauty will win.

- 7. What is larger 1/1001 + 1/1002 + 1/1003 + 1/1004 + ….+1/2000 or ½? Explain your answer.

**Correct Answer: The sum of fractions is larger than 1/2**

Solution:

Every single number added is larger than 1/2000. Since there are 1000 of such numbers, they should add up to more than 1000*1/2000 = ½.

- 8. You have some number of tennis balls less than 100. If you put them in cases, 2 in each case, there will be one extra left. If you put them in cases, 3 in each case, there will be two extra left. If you put them in cases, 4 in each case, there will be three extra left. If you put them in cases, 5 in each case, there will be four extra left. Finally, if you put them in cases, 6 in each case, there will be five extra left. How many tennis balls do you have?

**Correct Answer: 59**

Solution:

The amount of balls remaining is always 1 more than the amount of cases used. As such, if we were to add 1 ball, and make it X+1, the total number of balls would now be divisible by 2,3,4,5, and 6. There is only one such number between 0 and 100, and it is 60. Therefore, we had 59 balls originally.

- 9. Abe is twice as old as Sarah was when he was as old as Sarah is. Sum of their ages is 49. How old are they now?

**Correct Answer: Abe is 28 and Sarah is 21**

**Solution:**

A = Abe’s age now. S = Sarah’s age now. When Abe was S years old, it was (A-S) years ago. At that point in time, Sarah was S – (A – S) = (2S –A) years old. The first condition of the puzzle becomes A = 2 x (2S – A). It simplifies to 3A = 4S. The second condition implies A+S = 49. This system of equations is easily solved: A = 28, S = 21.

- 10. Athos, Porthos, Aramis, and d’Artagnan were playing Tug of War. Athos and Porthos easily beat Aramis and d’Artagnan. Porthos and d’Artagnan just barely beat Aramis and Athos. Finally, Porthos and Aramis were tied for a victory with Athos and d’Artagnan. Please rank all 4 musketeers by strength from strongest to weakest.

**Correct Answer: (From strongest to weakest) Porthos, Athos, d’Artagnan, Aramis.**

Solution:

A = Athos, P = Porthos, R = Aramis, and D = d’Artagnan.

What do we know?

A + P >> R +D, P + D >≈ R + A, and P + R = A + D

These are the only 3 ways in which 4 musketeers can be split into 2 teams of 2. Porthos is clearly the strongest, because he never lost. Aramis (R) is the weakest, because he never won. Since the strength of the Porthos’s team became lower after we swapped A and D, we can conclude that A >D. Therefore, P>A>D>R.