Algebra Problems, Math Problems

Problem of the Day: 3/15/13

What is the probability that a term selected at random from the binomial expansion of (a+b)^{21} has a coefficient that is a multiple of 2? (edited)

Solution to yesterday’s problem:

The trick to sum of sequence problems is to find the pattern of the sums. In this problem, notice that the sum of the first and third terms, 71, is the same as the sum of the second and fourth terms. It’s also the sum of the third-last and last terms. So we can assume that the sum of two terms that have one term between them is 71.

The next step is to find the number of these pairs in the sequence. Well, we notice that the first two numbers starts counting down from 35, and each set of two numbers separated by two numbers continues this pattern down to 1 at the end. The numbers between the decreasing numbers start at 36 and increase to 70. It takes 35 numbers to count from 35 to 1, and 35 numbers to count from 36 to 70. Therefore there are 70 numbers and 35 pairs. Multiplying the number of pairs by the sum of each pair, 35 x 71 = 2485.

Another way this problem can be solved is to notice that the series essentially just contains the integers from 1 to 70. The sum of these integers is half the product of the biggest number and the next number after that.

frac{(70)(71)}{2}

frac{70}{2}times71

35times71

We wind up with the same answer, which is 2485.

Math Problems, Number Theory Problems

Problem of the Day: 3/14/13

What is the sum of the terms in the sequence 35+34+36+37+33+32+38+39+ … +2+1+69+70?

Solution to yesterday’s problem:

We can assume that triangle PQR has integer sides because more information was not specified. This would mean that PQR is a Pythagorean triple with a short side of 9: a 9-12-15. So the hypotenuse of PQR is 15 cm.

Now we’re going to do some more estimating! You can sort of see that the small triangle QUT takes about a third of the hypotenuse. This goes along with our theory about integer lengths because it would make the hypotenuse 5, which leads us to believe that QUT is a 3-4-5 triangle. Also, the larger triangle RST takes up two-thirds of the hypotenuse, or 10 cm. This suggests that RST is a 6-8-10 triangle. So let’s show the triangle again with the lengths labeled:

20130314-091142.jpg

The length we actually want to find is US, conveniently the only segment we don’t know. We can use the Pythagorean theorem for triangle STU.

4^2+6^2=x^2

16+36=x^2

52=x^2

\sqrt{52}=x

Simplifying the radical,

\sqrt{4\times13}=x

\sqrt{4}\times\sqrt{13}=x

2\sqrt{13}=x

Geometry Problems, Math Problems

Problem of the Day: 3/13/13

In the figure below showing three right triangles, segment PR measures 9 cm. What is the measure of segment US? Express your answer in simplest radical form.

20130313-072539.jpg

Solution to yesterday’s problem:

Let’s use the blanks method to solve this problem. We have 10 people to choose from and 10 seats.

__10__ __9__ __8__ __7__ __6__ __5__ __4__ __3__ __2__ __1__

However, we have to divide by 10 because each arrangement could start at any of the 10 seats around the table. Multiplying the blanks and dividing by 10, we find a total of 362,880 ways.

Math Problems, Probability Problems

Problem of the Day: 3/12/13

In how many different ways can 10 people be seated at a round table? Note that arranging the people the same way at different positions around the table does not count.

Solution to yesterday’s problem:

Since the common factor of the two numbers is 4, both numbers are divisible by 4. We can therefore say that the first number is 4 times a number, or 4m. The second number is 4 times another number, or 4n. The problem tells us that

4m+4n=52

Factoring out a 4 on the left side,

4(m+n)=52

We can now divide by 4:

m+n=13

So the range of possible values for m and n is such that their sum is 13: 1 and 12, 2 and 11. However, logic tells us that since we want the smallest possible difference between the numbers, we should choose numbers close to 13/2; namely, 6 and 7. Using 6 and 7, our original numbers would be 24 and 28. These do have a sum of 52 and a GCF of 4, and their difference is 4.

Math Problems, Number Theory Problems

Problem of the Day: 3/11/13

Two positive integers have a sum of 52. If their greatest common factor is 4, what is the least possible positive difference between the numbers?

Solution to yesterday’s problem:

If Carlo needs two cans of paint to coat three walls, then his rate of “paint usage” is \frac{2}{3} cans per wall. If he needs to paint eight 4-walled rooms, he has to paint 32 walls total. He also needs to paint two coats per wall, which is equivalent to painting 32 x 2 = 64 walls total. Using the rate mentioned above, we can write

\frac{2}{3}\times64

to represent the number of cans of paint he needs. This evaluates to \frac{128}{3}, or 42\frac{2}{3}. Remember that Carlo couldn’t very well buy two-thirds of a can of paint, so we round up to 43 cans.

Math Problems, Proportions Problems

Problem of the Day: 3/10/13

Carlo needs two cans of paint to coat every three walls in his house. How many cans does he need to paint two coats on each of his eight four-walled rooms?

Solution to yesterday’s problem:

Let’s find how many different ways we can subdivide one beat. First of all, we can have one quarter note. Then we can subdivide that into two eighth notes. That’s two different ways so far. Next, we can subdivide either of those two eighth notes into sixteenth notes, which adds two more ways. Finally we can subdivide both eighth notes into sixteenth notes, adding one more way. So there are five possible ways to subdivide a beat in this fashion. Using the blanks method to find how many possible ways there are for four beats:

__5__ __5__ __5__ __5__

Multiplying these numbers together gives a total of 625 ways.

Math Problems, Probability Problems

Problem of the Day: 3/9/13

In music, a single beat consists of one quarter note, which can be subdivided into two eighth notes. Each eighth note can be further subdivided into two sixteenth notes. How many four-beat rhythms can be constructed from quarter notes, eighth notes, and sixteenth notes?

Solution to yesterday’s problem:

All you need to do in this problem is write the factorial out as its entire product, like this:

14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1

Then split each factor into its prime factors:

7\times2\times13\times3\times2\times2\times11\times...

We don’t need to go any further because we already have the greatest prime factor we’re going to find here: 13.