How many 3 digit positive integers have exactly 1 even digit?
(A) 350 (B) 450 (C) 375 (D) 75 (E) 125
Well, we know that we can only use the numbers between 100 and 999. We also know that 0 cannot be used as the first number because that would result in a 2 digit number. The easiest was to find the answer is to write out the possibilities.
The first group is going to be the integers that start with an even number.
First number: 2, 4, 6, 8
Second number: 1, 3, 5, 7, 9
Third number: 1, 3, 5, 7, 9
This means that the number of possibilities equal 4 x 5 x 5 which equals 100.
The second group is going to be integers in which the second number is even.
First number: 1, 3, 5, 7, 9
Second number: 0, 2, 4, 6, 8
Third number: 1, 3, 5, 7, 9
This means that the number of possibilities equal 5 x 5 x 5 which equals 125.
The second group is going to be integers in which the third number is even.
First number: 1, 3, 5, 7, 9
Second number: 1, 3, 5, 7, 9
Third number: 0, 2, 4, 6, 8
This means that the number of possibilities equal 5 x 5 x 5 which equals 125.
Now we just add the numbers to find the answer
100 + 125 + 125 = 350
The answer to this equation is (A) 350.
At first I had no idea how to approach this question. The first thing I thought of was to list all the numbers but that would take a very long time to do and I thought that there must be an easier, more organized way to do it (I remembered doing something like this in elementary school when they gave us a few numbers and we were supposed to find out how many combinations there were). So I thought to categorize them by which number was even and then writing all the possibilities and multiplying them. This quickly gave me the answer in an organized fashion that's easy to understand.
In the process of problem solving I learned that the way I first think about things may not always be the best way to approach a question. I should always keep my mind open to other solutions to each question.
Friday, November 12, 2010
Friday, October 22, 2010
Who Wants to be a Mathematician Field Trip
Some things I learned on this field trip are:
1. a squared plus b squared will always equal c squared
2. There are an infinite amount of fractions between 0 and 1
3. In 2000, people made these math equations that are worth 1 000 000 if you solve one
4. P=NP? is actually a math equation
5. Pythagoras may not have actually created the Pythagorean Theorem
6. Pythagorean Theorem is always true on a plain
7. There are an infinite amount of Pythagorean Triples
8. Gauss astonished his teacher by adding all of the numbers from 1 to 100 very quickly
9. "Chos" or "Choss" or something that sounds like that is a math term
10. i and ! are also math terms
In the talk, I learned some things that are listed above. For the most part, I didn't understand exactly what he was saying like when he started the whole thing about the circle and m and all that stuff. I don't really think that it helped me to learn that much because what I did understand I already knew and what I didn't understand wasn't explained well enough.
The Workshop could help me in future math learning because I learned that it is easier to work with different people because you have different ideas on how to answer the question. I also learned that hard questions could be solved through much thinking and guess and check.
1. a squared plus b squared will always equal c squared
2. There are an infinite amount of fractions between 0 and 1
3. In 2000, people made these math equations that are worth 1 000 000 if you solve one
4. P=NP? is actually a math equation
5. Pythagoras may not have actually created the Pythagorean Theorem
6. Pythagorean Theorem is always true on a plain
7. There are an infinite amount of Pythagorean Triples
8. Gauss astonished his teacher by adding all of the numbers from 1 to 100 very quickly
9. "Chos" or "Choss" or something that sounds like that is a math term
10. i and ! are also math terms
In the talk, I learned some things that are listed above. For the most part, I didn't understand exactly what he was saying like when he started the whole thing about the circle and m and all that stuff. I don't really think that it helped me to learn that much because what I did understand I already knew and what I didn't understand wasn't explained well enough.
The Workshop could help me in future math learning because I learned that it is easier to work with different people because you have different ideas on how to answer the question. I also learned that hard questions could be solved through much thinking and guess and check.
Friday, October 15, 2010
Problem Set #3
In the diagram, the sum of the numbers in each quarter circle is the same. the value of x + y + z is
a. 75
b. 64
c. 54
d. 171
e. 300
So this question is asking what the three unknown numbers are.
To solve this question, you first need to know how much is in each quadrant. In quadrant a (12 o' clock to 3 o'clock), you have all the numbers. 13 + 17 + 45 is equal to 75. If all the quadrants are equal in value, we can find out what x, y, and z are.
In the second quadrant (3 o'clock to 6 o'clock), the numbers present are 19, 50, and x. In this quadrant,19 + 50 + x = 75. Now, using simple algebra, we can take the two numbers move them to the other side, and subtract them from 75, therefore resulting in x.
19 + 50 + x = 75
75 - 50 - 19 = x
6 = x
In order to find y and z, we repeat this process with the other 2 remaining quadrants.
63 + 3 + y = 75
75 - 3 - 63 = y
9 = y
28 + 8 + z = 75
75 - 8 - 28 = z
39 = z
Now that we know the values of x, y, and z, we can add them together to get the answer to the equation which is 6 + 9 + 39 = 54. So now we know the answer is C.
I liked this equation because at first it looks difficult but if you try hard enough, it is actually really simple and straightforward. It was one of those questions you barely get but once you finish it its like, "YES!" I thought this question was kinda fun.
In the process of problem solving, I learned that it may seem hard at first but when you know what to do, it gets easier. I learned that math can be fun even though it doesn't sound like it.
a. 75
b. 64
c. 54
d. 171
e. 300
So this question is asking what the three unknown numbers are.
To solve this question, you first need to know how much is in each quadrant. In quadrant a (12 o' clock to 3 o'clock), you have all the numbers. 13 + 17 + 45 is equal to 75. If all the quadrants are equal in value, we can find out what x, y, and z are.
In the second quadrant (3 o'clock to 6 o'clock), the numbers present are 19, 50, and x. In this quadrant,19 + 50 + x = 75. Now, using simple algebra, we can take the two numbers move them to the other side, and subtract them from 75, therefore resulting in x.
19 + 50 + x = 75
75 - 50 - 19 = x
6 = x
In order to find y and z, we repeat this process with the other 2 remaining quadrants.
63 + 3 + y = 75
75 - 3 - 63 = y
9 = y
28 + 8 + z = 75
75 - 8 - 28 = z
39 = z
Now that we know the values of x, y, and z, we can add them together to get the answer to the equation which is 6 + 9 + 39 = 54. So now we know the answer is C.
I liked this equation because at first it looks difficult but if you try hard enough, it is actually really simple and straightforward. It was one of those questions you barely get but once you finish it its like, "YES!" I thought this question was kinda fun.
In the process of problem solving, I learned that it may seem hard at first but when you know what to do, it gets easier. I learned that math can be fun even though it doesn't sound like it.
Subscribe to:
Posts (Atom)