Problem of the Week

__11/12/18__

**Classmates' Birthdays**

Your mathematics teacher displays a monthly calendar, which highlights all the birthdays of your classmates. Your mathematics class has 25 students. What is the probability that 3 or more students were born in the same month?

__11/5/18__

**Cards in a Hat**

Cards numbered 1 to 50 are put in a hat. What is the probability that the first two cards chosen at random have prime numbers on them?

**Solution**

3/35, or about 0.086. There are 15 prime numbers in the range of 1 to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Remember that 1 is not a prime number. The probability of selecting a prime on the first pick is 15 out of 50; for the second pick, the probability is 14/49. The probabilities are combined by multiplying them, so 15/50 × 14/49 = 210/2450, or 3/35, or about 0.086.

__10/29/18__

**How Many?**

How many two-digit numbers exist such that when the product of its digits is added to the sum of its digits, the result is equal to the original two-digit number?

**Solution**

Nine two-digit numbers: 19, 29, 39, 49, 59, 69, 79, 89, and 99. Any two-digit number can be represented by 10*a + b*. The product of the digits is *ab*, and the sum is *a + b*. When does *ab* + (*a + b*) = 10*a + b*? Solving, when *ab* = 9*a*, or when *b* = 9.

__10/22/18__

**Old Books**

A mathematics teacher collects old books. One day, a student asked him how many old math books he has. He replied, “If I divide the books into two unequal whole numbers, then 64 times the difference between the two numbers equals the difference between the squares of the two numbers.” How many old math books does the teacher have?

**Solution**

64. Let *a*and *b*represent the two unequal whole numbers.
Then 64(*a* – *b*) = *a*^{2} – *b*^{2}.
Factoring the right side yields 64(*a* – *b*) = (*a* – *b*)(*a* + *b*).
Since *a*and *b*are not equal, (*a* – *b*) does not equal zero.
Therefore, you can divide both sides by (*a* – *b*),
which yields 64 = *a* + *b*.
To solve the problem,
you do not have to know the values of *a*and *b*,
just the sum of *a*and *b*,
which is 64, the total number of books.

__10/15/18__

**Mouse Race**

Two mice are racing around the edges of a square whose sides are 2 feet in length. They start at the same vertex (corner) and both go in a clockwise direction. One mouse travels at a constant rate of 1 foot per second, and the second mouse travels at a constant rate of 2 feet per second. After 22 seconds, how far apart will the mice be from each other?

**Solution**

2 feet apart. Label a square with vertices *A*,*B*, *C*, and *D*in clockwise order. Suppose the mice start at vertex *A*. After 22 seconds, the first mouse will be on vertex *D*, while the second mouse will be on vertex *C*. Therefore, they will be two feet apart.

__10/8/18__

**New Operation**

Imagine that a new mathematical operation is being used. Its symbol is #. See the following equations:

1 # 1 = 2

3 # 5 = 34

6 # 9 = 117

10 # 14 = 296

Find the value of 15 # 19, and explain your reasoning.

**Solution**

586

a # b = a^{2}+ b^{2}. For example, 3 # 5 = 3^{2}+ 5^{2}= 9 + 25 = 34. Therefore, 15^{2}+ 19^{2}= 225 + 361 = 586.

__10/1/18__

**Digging a Hole**

If it takes two workers 4 hours and 35 minutes to dig a hole 3 meters long, 3 meters wide, and 3 meters deep, how long would it take five workers to dig a hole 6 meters long, 6 meters wide, and 6 meters deep if they worked at the same rate?

**Solution**

About 880 minutes or 14:40. The first hole has a volume of 3 × 3 × 3 = 27 cubic units. The new hole has a volume of 6 × 6 × 6 = 216 cubic units; 216/27 shows that 8 smaller holes would fit into the larger hole. In other words, the new hole is twice as wide, twice as long, and twice as deep as the original, so it would take 2 × 2 × 2 = 8 times as long. The smaller hole took two workers 4 hours and 35 minutes or 275 minutes, so it would take two workers 2,200 minutes to dig the larger hole. With 5 workers, it would take 2/5 * 2200 minutes or 880 minutes to dig the larger hole.

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