Here’s a mathematical puzzle for you. I found the following in an article about how American kids are getting so much fatter. Can you see what’s wrong with it?
“According to the Centers for Disease Control and Prevention, approximately 17 percent or 12.5 million of children and adolescents aged 2 to 19 years are obese, as opposed to merely overweight.
Obesity in children is defined by the CDC as having a body mass index (BMI) at or above the 95th percentile for children of the same age and sex. ‘Overweight’ is defined as a BMI at or above the 85th percentile.”
Comments welcome.
Here’s the most important clue to cutting the candy bar (last week’s problem). Think about the area of a triangle.
It’s 1/2 (base x height), right? So what happens if you start at one corner of the candy bar and cut across to some point on the other side? You’ve cut off a triangle, and the height of the triangle is the width of the candy bar, which is two inches. Now make another cut starting from the new corner you made on the other side. Your triangles will all have the same height, but they’ll have different bases. Cut back and forth, and you’ll end up with a kind of zigzag pattern.
How can you end up with two piles of triangles that add up to the same area?
I got this problem from an e-journal called Girls’ Angle Bulletin. Check page 16. I was checking out what kind of girls are doing. It’s pretty intimidating, but this problem is one you can solve, even though I’ve changed it around.
So here’s the deal. You’re on a quest with a partner, and you come to a magical candy bar that is 10 inches long and only two inches wide. (It’s even and perfect, much more so than the irregular picture above.) The next stage of your journey is going to take you through a valley inhabited by evil and vengeful elves who hate strangers and use poison arrows. But the candy gives protection against the poison arrows to anyone you eats at least half a bar.
Okay, obviously you each need to eat exactly half the candy bar. But how could you ever cut it so precisely? Luckily, next to the candy bar is a magic knife exactly 3.9 inches long. It’s perfectly thin and straight, a perfect cutter.
Unfortunately, it’s too short to cut the candy bar with one single cut, or else you could cut from corner to corner.
So what do you do? Of course, you could try just eyeballing it and trying to cut, and then putting the pieces on top of each other and trying to trim and divide any extra… but remember, if you make even a tiny mistake, one of you is going to die in the Valley of the Elves.
Give it a try. When you get your answer, check the Girls’ Angle Bulletin to see if you’re right. If you try for a few days and aren’t getting anywhere, check Girls’ Angle anyway to get started (but don’t read the whole solution).
Here’s a problem that’s easier than it looks.
Two bicyclists, Sam and Joe, start riding toward each other. Each one is riding 10 miles per hour. A fly, flying at 36 miles per hour, starts at Sam, flies straight to Joe, turns around and immediately flies back to Sam, turns around again, etcetera, flying shorter and shorter distances from one to the other until Sam and Joe meet half an hour later.
How far does the fly fly?
There’s a whole set of problems about pouring water from one glass or jug to another. To make it more fun, let’s try lemonade. Warning: these problems are kind of hard. You have to figure out a way to draw them and keep track.
The rules are like this. You start with lemonade all in one or glass, and just by pouring from one glass to another, you divide the lemonade into certain volumes. Every time you pour from one glass to another, you have to pour as much as you can without spilling. So say you pour from a full 8-ounce glass into an empty 3-ounce glass. When you finish pouring, you have 5 ounces in the first glass and 3 in the second.
1. Okay, first problem. You have a full 7-ounce glass, an empty 4-ounce glass, and an empty 3-ounce glass. Pour from one to another until you have 3 ounces, 2 ounces, and 2 ounces. (I worked and worked on this one, and I finally got it in 7 pours.)
2. Second problem. In this one you can throw any leftover lemonade down the sink, and you can also refill any glass from the lemonade pitcher in the fridge. (But you don’t have any other glasses: you can’t pour, say, one ounce at a time into a third glass we don’t know about.) Okay, you have a 4-ounce glass and a 5-ounce glass. How can you measure out exactly 2 ounces? (For this one I filled a glass twice and emptied one down the sink once. I poured from one glass to another three times, I think.)
If you google “water-pouring problem,” you can find more of these. There are even some animations if you have Java. See what you can figure out, and then come back and tell me!
I just sat down to solve the problem I gave you last week.
R E A D
+ R E A L
__________
L E A R N
It was really, really hard. I think maybe I didn’t give you enough information. I’m going to make it a lot easier by telling you that E=7.
Remember, you have to find what digit each letter stands for. Each letter stands for a different digit. Try it yourself before you read down to the answer.
* * * * * *
OK, now here goes. If E is 7, what can A be? 7 + 7 is 14, so A can be 4, and you have to carry a one into the R+R column. So that means R+R+1= 7… only not really, because you have that L in there. So R+R+1= 17. L=1. R has to be 8. Are you writing this down?
Now we have:
8 7 4 D
+ 8 7 4 1
__________
1 7 4 8 N
You can choose what you like for D and N. I chose D=5 and N=6.
* * * * * *
Good. Now how about the other one I gave you?
S I N G
+ S O N G
__________
C O I N S
Well, it turns out you can prove this one is impossible. How? Take a look.
If N+N=N, then when you first think about it, it seems that N has to be zero. But does it? What if N is 9? 9+9=18, but if G is 5 or more, you’ll also be adding a one carried from the G+G column. 9+9+1=19, so that could work. But N can’t possibly be any number other than 0 or 9. Try it for yourself.
So now we know N is either 0 or 9. How about O? I+O=I. That can only work if O is actually zero. Or… it could work if O=9 and there was a one carried from the N+N column.
Okay, let’s check the possibilities. If N=0, then there’s no 1 carried over to the “hundreds” column. So I+O=I, and O has to be zero, which it can’t be, because N is zero.
If N=9, then a one is carried over to the the hundreds column, and I+O+1=I. That is, I+10 is 1I. For that to work, O has to be 9, but it can’t be, because N is already 9.
So we’ve just proved that SING+SONG=COINS can’t be solved!
Sorry it’s so confusing. It’s sort of annoying that O and I look so much like zero and one. Maybe I should have used lower-case letters. But then the L would look like a one.
* * * * * *
Now that we’ve gone through those two, are you ready for another? I’ll give you a hint to make it not quite so bad. Here goes:
T O L L
+ T I M E
_________
M I L E S
Here’s the clue, which will help a lot. L=4. Try it!
Have you ever read Sideways Arithmetic from Wayside School by Louis Sachar? You should. The math the kids do in Mrs. Jewls’s class is all puzzles. For example, they have a different kind of word problem. Here’s the first one:
ELF
+ELF
_____
FOOL
Instead of figuring out the answer, you have to figure out what numbers the different letters stand for. In each problem, a letter can only stand for one digit, and no two letters stand for the same digit. So how do you solve it?
Well, in ELF + ELF = FOOL, you see you get a 4-digit number as the answer. After you think about it for a while, you realize that no matter what E is, the first digit F can only be one. Once you know what F is, you can figure out what L is, then what O is… and then what E is. I won’t give you the answer because it’s really Louis Sachar’s answer.
So this week I thought I’d make up a sideways word problem of my own for you. It turned out to be really hard to do! You have to use enough letters more than once to make it possible to figure out, but I kept coming up with impossible problems that used the same digit twice or something.
Anyway, I finally came up with one that works. Here it is:
READ + REAL = LEARN.
Go ahead, give it a try! Send in your answers and explanations. Then, if you like, try this one. If it can’t work, tell me why it can’t:
SING + SONG = COINS
Have fun!
Someone wrote:
I found a hard math problem on my math test. It took a while but i figured it out. Here it is.
“The Empire State Building has 104 floors. The elevator operator on the bottom floor took a group of people up 73 floors, down 26 floors, up 52 floors, down 47 floors, up 15 floors, and down 9 floors to an observatory. What level was the observatory?”
I was just fooling around and I found out some cool things about numbers. Have you heard of triangle numbers? The first triangle number is one. You can imagine a really small equilateral triangle with only one point on each side, like this: *
The second triangle number is 3, like this: *
* * See how it has two dots on each side?
Then the third triangle number is 6. *
* *
* * * Three dots on each side. See it?
The fourth triangle number is… Yes, you got it! Ten! Because each new row adds one more dot than the last one. So the triangle numbers are:
1
1+2=3
3+3=6
6+4=10
10+5=15 and so on.
Now let’s try square numbers. Yeah, you know these. 1, 4, 9, 16, 25, etc. Or look at a picture.
* * * * * * * * * *
* * * * * * * * *
* * * * * * *
* * * *
Okay, now look at how many new dots are added each time.
1
1+3 = 4
4+ ? = 9 ?=5
9 + ? = 16 ?=7
16 + ? = 25 ?=9
So you see, to get from one triangle number to the next, you add 1 more than you added last time. That is, You add 2, then 3, then 4…
To get from one square number to the next, you add 2 more than you added last time. To get from 1 to 4 you add 3, but then to get from 4 to 9 you add 5, then 7, then 9… Right?
So how do you get from one pentagon number to another? And what is a pentagon number anyway?
Well, the bad news is I really can’t draw one for you on the computer very well at all. So just take out your pencil and draw a dot. Now you have a pentagon, a five-side figure, with one dot per side. Next, expand it, using your dot as one corner, until you have two dots per side. It sort of looks like this:
*
* *
* *
How many dots? 5. That’s four more than the first triangle number. Now, based on the pattern for triangle numbers and square numbers, going from the second to the third pentagon number should take how much more than going from the first to the second? For triangle numbers, it went up by one more number each time. For square numbers it went up by 2 more numbers each time. So for pentagon numbers, it goes up by… yes, if my theory is correct and the pattern holds true, it should go up by 3 more numbers each time.
Going from 1 to 5 was going up by 4. To reach the next triangle number, we’ll have to go up from 5 by 3+4, which is 7. So the pentagon number should be 5+7, which is 12. Try it. Add to a five-dot pentagon by using one corner and two sides as your starting point, and add dots until there are 3 dots on each side. Count how many you get. Does it work? Can you take it further, to hexagon numbers or heptagon (7-sided shape) numbers? When you’ve figured out the 4th heptagon number, send me the answer!
Anonymous asked: Hi i have a lot of trouble with long divisions can you give me some tips?????:):):)
Okay, wow. I’m not sure how easy this will be to explain, but I thought about it a lot and I’m going to give it a try. Maybe if I can explain what you’re really doing in a way that makes sense, that will help you remember all those steps of dividing and multiplying and subtracting you do in long division.
The key is that in long division you have a large number to divide into parts, and you do it by chunks. For example, say you’re dividing 472 by 3. That is, you’re sorting 472 things into 3 equal piles. To make it easier first you divide the hundreds, then the tens, then the ones.
So you have:
4 hundreds
7 tens
4 ones.
First divide the hundreds into 3 piles. You put one hundred into each of your three piles, and you have one hundred left over. You can’t divide a single hundred into thirds very easily, so you break it apart into 10 tens and add it to the 7 tens you already had.
Now you have:
3 groups of 1 hundred
17 tens
4 ones.
Okay, now divide the 17 tens into your three groups. The closest you can get is to put 5 tens into each of your three groups, along with the 1 hundred that’s already there. Three 5’s are 15, so you have 2 tens left over. Just like when you were breaking a hundred into 10 tens, you break those 2 tens up into 20 ones and combine them with the 4 ones you had from the beginning.
Now you have:
3 groups, each with 1 hundred and 5 tens.
24 ones.
Now you divide the 24 ones into 3 equal groups. You get 8 ones in each group. So now you have:
Three equal groups, each containing 1 hundred, 5 tens, and 8 ones.
That is, 472 divided by 3 is 158!
Well, if you think about it carefully, what I’ve just explained to you is what you do in the steps of long division, except that you keep track of where you are by carefully lining up numerals instead of by writing out each step separately.
