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chrono
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 Posted: Wed Apr 09, 2008 9:36 pm Post subject: Need help on a program [Any Language] |
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Hi, it's kind of rare to see a noob programmer in this section, but I need some assistance.
I have a project in math which goes as follows:
| Quote: |
The product of the digits in the number 126 is 12: 1*2*6 = 12
a. How many different three digit numbers have a product equal to 12? Explain how you know
b. How many three digit numbers have a product equal to 18? Explain.
c. Compare your answers to parts a and b and explain why they work out that way.
d. Extra Credit: How many four digit numbers have a product equal to 12? Explain.
e. Extra Credit 2: Explain alternate methods to solve this problem
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Basically, I can solve the problem easily through factorization, but I really need to find a method to answer "Extra Credit 2".
So I thought, why not make a program that can solve which numbers work for me?
Of course, I'm a programming noob, so if anybody can give me some assistance, that'd be great!
Also, If this can be done in VB, that would be even easier for me, but if anybody has any other programming methods to solve this problem, let me know!!
Thanks again~
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Negima
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| Posted: Wed Apr 09, 2008 9:42 pm Post subject: |
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| If this is due anytime soon then I wouldn't count on learning a language for it can take a while.
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Overload
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| Posted: Wed Apr 09, 2008 9:54 pm Post subject: |
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| Negima wrote: | | If this is due anytime soon then I wouldn't count on learning a language for it can take a while. |
ahh, but learning how to use math operators is not hard at all. At least, not in most languages.
Then you just have to figure out the correct formula or method for yourself.
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hcavolsdsadgadsg
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| Posted: Wed Apr 09, 2008 10:06 pm Post subject: |
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I don't think there's really a fast way to do factoring.
The only way I can think of is looping through and trying to sort things out.
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oib111
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| Posted: Wed Apr 09, 2008 10:15 pm Post subject: |
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yea, what slovach said. I would most likely make an triple array or whatever its called (to hold the 3 numbers that multiply to twelve), then like loop each number. Then, you would have to look through those numbers, and see which ones come out to twelve. Which would additionally add to the time it takes. Hehe, I wonder if you could make threads to sort out numbers at the same time? lol. Anyway for example, for something like 1:
| Code: |
1*1*1
1*1*2
1*1*3
1*1*4
1*1*5
1*1*6
1*1*7
1*1*8
1*1*9
1*1*10
1*1*11
1*1*12
1*2*1
1*2*2
1*2*3
1*2*4
1*2*5
1*2*6
1*2*7
1*2*8
1*2*9
1*2*10
1*2*11
1*2*12
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And so forth...So you can see, even for one number it would take you a LONG time. But if you really want, I could help you to write something
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HolyBlah
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| Posted: Thu Apr 10, 2008 12:05 am Post subject: |
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(In Delphi)
| Code: | for i:=100 to 999 do
begin
units:=i mod 10;
teens:= (i div 10) mod 10;
hundreds := i div 100;
if units*teens*hundreds = product then //The product you want to check
//Adding the number to the list of good numbers.
end; |
@oib111: You don't have to try 1*1*10, 1*1*11 and 1*1*12 because its three digit numbers and 1110 is 4 digit number.
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Estx
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| Posted: Thu Apr 10, 2008 7:37 am Post subject: |
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HolyBlah's code in VB6:
| Code: | Private Sub Form_Load()
Dim units As Integer
Dim tens As Integer
Dim hundreds As Integer
Dim product As Integer
product = 12
For i = 100 To 999
units = i Mod 10
tens = (i / 10) Mod 10
hundreds = i / 100
If units * tens * hundreds = product Then
Call List1.AddItem(units & " x " & tens & " x " & hundreds & " = " & product)
End If
Next i
End Sub |
The code above assumes you have a ListBox named List1.
Returns the following:
| Code: | 6 x 2 x 1 = 12
4 x 3 x 1 = 12
3 x 4 x 1 = 12
1 x 6 x 2 = 12
6 x 1 x 2 = 12
3 x 2 x 2 = 12
2 x 3 x 2 = 12
4 x 1 x 3 = 12
2 x 2 x 3 = 12
1 x 4 x 3 = 12
3 x 1 x 4 = 12
1 x 3 x 4 = 12
2 x 1 x 6 = 12
1 x 2 x 6 = 12 |
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oib111
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| Posted: Thu Apr 10, 2008 8:56 am Post subject: |
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| HolyBlah wrote: | (In Delphi)
| Code: | for i:=100 to 999 do
begin
units:=i mod 10;
teens:= (i div 10) mod 10;
hundreds := i div 100;
if units*teens*hundreds = product then //The product you want to check
//Adding the number to the list of good numbers.
end; |
@oib111: You don't have to try 1*1*10, 1*1*11 and 1*1*12 because its three digit numbers and 1110 is 4 digit number. |
Oh get whats hes saying. Now, at first I thought he meant three numbers, but he menat three digits.
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cigs dont make people high, which weed does, which causes them to do bad stuff. like killing |
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HalfPrime
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| Posted: Thu Apr 10, 2008 2:10 pm Post subject: |
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| Code: | 6 x 2 x 1 = 12
4 x 3 x 1 = 12
3 x 4 x 1 = 12
1 x 6 x 2 = 12
6 x 1 x 2 = 12
3 x 2 x 2 = 12
2 x 3 x 2 = 12
4 x 1 x 3 = 12
2 x 2 x 3 = 12
1 x 4 x 3 = 12
3 x 1 x 4 = 12
1 x 3 x 4 = 12
2 x 1 x 6 = 12
1 x 2 x 6 = 12 |
It's missing 261 o.O
Last edited by HalfPrime on Thu Apr 10, 2008 2:11 pm; edited 1 time in total |
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chrono
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| Posted: Thu Apr 10, 2008 2:10 pm Post subject: |
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| thanks a million~ i owe you guys an A+ ~
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killersamurai
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| Posted: Thu Apr 10, 2008 5:41 pm Post subject: |
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Here's how I thought of it (if I were to make a program), if anyone is interested
If you go over 6, your result is over 12. Which we don't want.
| Code: |
6 / 2 = 3
3 = end of algorithm
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You'll see why.
Since there is three, we would want to increase the second number and decrease the third. The third will start at max and the second will start with two. When you find that the number that equals 12, you keep the index of the third and use that when the second is increased. If the number didn't equal 12, remember the third number you started with.
| Code: |
1 * 2 * 6 = 12 12 == 12
1 * 3 * 6 = 18 18 != 12
1 * 3 * 5 = 15 15 != 12
1 * 3 * 4 = 12 12 == 12
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If I were to expand it so that the third number goes all the way down to 1 and always starts at 6, you will notice when the second number is 4, you will just repeat. It makes for less of a loop.
So, for this one, there is really only two solutions. Unless you count the ones where you use the same number more than once. The algorithm will still apply to that as well.
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oib111
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| Posted: Thu Apr 10, 2008 7:45 pm Post subject: |
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| HolyBlah wrote: | (In Delphi)
| Code: | for i:=100 to 999 do
begin
units:=i mod 10;
teens:= (i div 10) mod 10;
hundreds := i div 100;
if units*teens*hundreds = product then //The product you want to check
//Adding the number to the list of good numbers.
end; |
@oib111: You don't have to try 1*1*10, 1*1*11 and 1*1*12 because its three digit numbers and 1110 is 4 digit number. |
Its fine to use this code. But if you're going to use code that isn't made by you (don't get confused with stealing, the code thats given right now is basically an outline) it's always good to know how it works instead of just copying and pasting. And seeing as your new to programming, it helps you to understand. I looked at HolyBlah's code and thought, man that's smart. Basically, he's using simple division and remainder tactics to get the digits of the number. Take the 3-digit number 126.
126 divided by 10 equals 12 remainder 6. Mod returns the remainder. So now you have the variable units (aliasing the ones digit place) to equal 6.
Now you divide 126 by 10 and that equals as we've stated 12 with a remainder of 6. The div function throws away the remainder, so now you have twelve. Then you divide again using the mod function, which we know returns the remainder. So 12 divided by 10 equals 1 with a remainder of 2. So now in teens (I think it was a typo, and HolyBlah meant tens) we have the value 2.
Now we just divided 126 by 100 and throw away the remainder. 126 divided by 100 equals 1. So now in the variable hundreds you have the value 1. Now look at the number you got.
Hundreds=1
Teens=2
Units=6
126. Then if you multiply each one of those variables: 1 times 2 equals 2. 2 times 6 equals 12. So then you can store that in whatever. As Estx used, he did it in a list box.
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| 8D wrote: |
cigs dont make people high, which weed does, which causes them to do bad stuff. like killing |
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