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huldu

Simple math question

14 posts in this topic

#1 ·  Posted (edited)

Its been a while since i were in school, and i werent that good at math even in school to start with :o. I ran into a problem calculating the following...

For example: 1x1 + 2x1 + 3x1 + 4x1 + 5x1 = 15

The thing is i just dont remember how to do this math. Im pretty sure it involved something like this (x * y) and something else hehe.

$x = 10
$y = 2

$sum = <some formula here>

Hope someone knows how it worked!

Edited by huldu

"I'm paper, rock is fine, nerf scissors!!!"

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$x = 10
$y = 2

$sum = $x*$y
MsgBox(0,"Total",$sum)

This seems too easy so I have probably misunderstood what you are asking



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It doesnt give the result however.

This is how the math looks before the "forumla": 1x1 + 1x2 + 1x3 + 1x4 + 1x5 = 15

The formula made the above math much more simple as you only needed to insert x and y to get the sum.


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It doesnt give the result however.

This is how the math looks before the "forumla": 1x1 + 1x2 + 1x3 + 1x4 + 1x5 = 15

The formula made the above math much more simple as you only needed to insert x and y to get the sum.

Try

$sum = (1*1)+(1*2)+(1*3)+(1*4)+(1*5)
MsgBox(0,"Total",$sum)


Get Beta versions Here Get latest SciTE editor Here AutoIt 1-2-3 by Valuater - A great starting point.

Time you enjoyed wasting is not wasted time ......T.S. Elliot
Suspense is worse than disappointment................Robert Burns
God help the man who won't help himself, because no-one else will...........My Grandmother

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#5 ·  Posted (edited)

Ye hehe i understand that :o

Lets say the user adds a number x and wants it added with y (like below). How would the formula look?

$sum = (1*1)+(1*2)+(1*3)+(1*4)+(1*5)

Since x is always the same, thats the easy part. However y is changing as you add up.

$x = 1

$y = 5

$sum = ($x * $y)^2 / 2 .. ugh (just cant get it to work)

When this "caluclation" is complete the total should be 15, thats all i know :geek:

Edited by huldu

"I'm paper, rock is fine, nerf scissors!!!"

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This any good to you?

$x = 1
$y = 5
$Sum = 0

For $i = 1 To $y
    $Sum = $Sum + ($x * $i)
Next

MsgBox(0,'Sum',$Sum)

----[ SandyD ]---

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Ye this seems right, thanks


"I'm paper, rock is fine, nerf scissors!!!"

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$x = 1
$y = 5

$sum = ($x * $y)^2 / 2
MsgBox(0,"", $sum)

this is correct

1 x 5 = 5

5 Squared = 25

25 divided by 2 = 12.5 ( not 15 as you were looking for)

and the message box reflects the correct answer of 12.5

8)


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#9 ·  Posted (edited)

$x = 1
$y = 5

$sum = ($x * $y)^2 / 2
MsgBox(0,"", $sum)

this is correct

1 x 5 = 5

5 Squared = 25

25 divided by 2 = 12.5 ( not 15 as you were looking for)

and the message box reflects the correct answer of 12.5

8)

it's triangular numbers guys... Nth triangular number = N + (N-1) + (N-2) + (N-3) ...

when i was playing with recursion i made a little script for figuring triangular numbers recursively...

HotKeySet("{CAPSLOCK}","endit")
Func endit()
    Exit
EndFunc
while 1
$blah = Number(InputBox("Enter a number","Enter a number N to find the Nth triangular number",6))
MsgBox(0,"triangular numbers",triangle($blah))
WEnd
func triangle($blah)
    if $blah = 1 Then
        Return(1)
    Else
        Return($blah + triangle($blah-1))
    EndIf
EndFunc
Edited by cameronsdad

1100111 00001011101111 00011101101111 00010111100100 00001111110100 00110111110010 00101101111001 0011100i didn't make up this form of encryption, but i like it.credit to the lvl 6 challenge on arcanum.co.nz

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Although you may have to use Mathematical Induction to prove this is the answer it it is not necessary to use recursion to calculate 1 + 2 + ... + N. The answer is N * (N + 1 ) / 2. The solution to the problem

x*1 + x * 2 + ... + x * y

y a positive integer, is x * y * (y + 1) / 2.

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#11 ·  Posted (edited)

Although you may have to use Mathematical Induction to prove this is the answer it it is not necessary to use recursion to calculate 1 + 2 + ... + N. The answer is N * (N + 1 ) / 2. The solution to the problem

x*1 + x * 2 + ... + x * y

y a positive integer, is x * y * (y + 1) / 2.

HI,

yeah that is although what I can remember from school

$x = 1
$y = 5
MsgBox(0, "", $x * $y * ($y + 1) / 2)

So long,

Mega

Edited by th.meger

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Although you may have to use Mathematical Induction to prove this is the answer it it is not necessary to use recursion to calculate 1 + 2 + ... + N. The answer is N * (N + 1 ) / 2. The solution to the problem

x*1 + x * 2 + ... + x * y

y a positive integer, is x * y * (y + 1) / 2.

i didn't mean to imply that i thought recursion was the only way to go; i was on a little bit of a recursion kick for a little while last month or the month before, and that was just one implementation of it.


1100111 00001011101111 00011101101111 00010111100100 00001111110100 00110111110010 00101101111001 0011100i didn't make up this form of encryption, but i like it.credit to the lvl 6 challenge on arcanum.co.nz

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"Simple math question"

not so simple!!!

lol

8)


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"Simple math question"

not so simple!!!

You're right, but it takes me back (retired math lecturer). This problem is one of the first proofs for which we used Mathematical Induction (which is the backbone of recursion) but I also remember doing this at secondary school before I came across Mathematical Induction and the method is neat, simple when you see it, but not something you would necessarily think of immediately.

Let

S = 1 + 2 + .... + N

turn the order of the terms round, then

S = N + (N-1) + ... + 1

add these two sums up term-by-term

2S = (N + 1) + (N -1 + 2) + .... + (1 + N) = N * (N + 1)

all the brackets are equal to N + 1 and there are N of them, so the right hand side adds up to N * (N + 1). Divide both sides by 2

S = N * (N + 1) / 2

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