Euler Problems
Miano Njoka
mianonjoka at gmail.com
Mon Jul 14 08:08:23 BST 2008
On Jul 12, 2008, at 8:15 PM, kinuthiA muchanE wrote:
> This email is from a ubuntu-ke subscriber to other subscribers
> inluding you:)
> Habari wote,
>
> Now, this is getting lively, and I think Zacharia is the teacher all
> of
> us have been longing for :-) Step by step, you elucidate all the
> underling principles. I thought nobody ever reads this, I am mistaken.
> Zacharia, why dont you subscribe to project Euler
> (http://projecteuler.net ) and at least I will not be leading in the
> list Kenyans who have solved the most problems, only 52, the last
> time I
> looked.
> And to Andrew (ha... you are at 17 ), you ask a question and then
> disappear!! Hmmmm, and where is Miano??
I've been busy the past two weeks, but I'll catch up with you guys
soon. :)
Miano.
>
>
> Cheers!!
> Kinuthia...
>
>
>
> On Sat, 2008-07-12 at 07:16 -0700, Zacharia M. Rugongo wrote:
>> This email is from a ubuntu-ke subscriber to other subscribers
>> inluding you:)
>> My solution to Euler #12 runs in less than 1 minute.
>>
>>
>> Let us list the factors of the first eight triangle numbers:
>> 1: 1
>> 3: 1,3
>> 6: 1,2,3,6
>> 10: 1,2,5,10
>> 15: 1,3,5,15
>> 21: 1,3,7,21
>> 28: 1,2,4,7,14,28
>> 36: 1,2,3,4,6,8,9,12,18,36
>> Note the numbers that are underlined in each sequence. This is the
>> half way point.
>> a) If you divide the triangle number by the first number that is
>> underlined the result will be greater than the divisor.
>> b)
>> On the other hand if you divide the triangle number by the second
>> number that is underlined the result will be less than the divisor.
>> c)
>> The triangle number 36 has only one number at the centre of the
>> sequence, because that centre number is the square root of the
>> triangle
>> number 36.
>>
>> In
>> cases a) and b) if you wanted to know the total number of divisors,
>> you
>> look for the position of the number underlined as described in b),
>> then
>> subtract 1 from it, then multiply it by 2. For example, for triangle
>> number 28, value 7 is in position 4. So (4-1) * 2 = 6
>>
>> In
>> cases c) if you wanted to know the total number of divisors, you look
>> for the position of the center number underlined, then multiply it by
>> 2. For example, for triangle number 36, value 6 is in position 5.
>> So 5 * 2 = 10
>>
>> Here is the program.
>>
>>
>> Dim triagNo As Integer ' Triangle number
>> Dim counter As Integer ' for calculating the next
>> triangle number
>> Dim varValue As Integer ' Incremented in search for a
>> Divisor
>> Dim NoDivisors As Integer ' number of Divisors found
>> Dim Divisor As Integer ' Value of found Divisor
>>
>> Do
>> counter += 1
>> triagNo = triagNo + counter
>> varValue = 0
>> Divisor = 0
>> NoDivisors = 0
>>
>> Do
>> varValue += 1
>> If triagNo Mod varValue = 0 Then ' is
>> varValue a Divisor
>> Divisor = varValue
>> NoDivisors += 1
>> End If
>> Loop Until (triagNo / Divisor) <= Divisor ' Find
>> when you reach the second centre value
>>
>> If (triagNo / Divisor) = Divisor Then
>> NoDivisors = (NoDivisors - 1) * 2 + 1 '
>> cases a) and b)
>> Else
>> NoDivisors = (NoDivisors - 1) * 2 '
>> case c)
>> End If
>>
>> Loop Until (NoDivisors > 500)
>>
>> TextBox1.Text = triagNo
>>
>>
>> Z. Mwangi
>>
>>
>>
>> ----- Original Message ----
>> From: Andrew Mathenge <mathenge at gmail.com>
>> To: ubuntu Kenya <ubuntu-ke at lists.ubuntu.com>
>> Sent: Friday, July 11, 2008 3:24:23 AM
>> Subject: Re: Euler Problems
>>
>> This email is from a ubuntu-ke subscriber to other subscribers
>> inluding you:)
>> I've been working on #12 and the performance is simply not
>> acceptable.
>> Has anyone been able to get this running as fast as some of the posts
>> on the site claim?
>>
>> Andrew.
>>
>> On Tue, Jul 8, 2008 at 12:40 PM, kinuthiA muchanE
>> <muchanek at gmail.com> wrote:
>>> Andrew,
>>> For #10, if you do not use a sieve to find the primes, it will run
>>> forever!
>>> There is one called the Sieve of Erastothenes. Check it out on
>>> Wikipedia.
>>>
>>> Good luck :-)
>>> Kinuthia...
>>> On Tue, 2008-07-08 at 10:47 -0400, Andrew Mathenge wrote:
>>>> I'm still trying them. I started at #1 after the one you proposed
>>>> (#26) and I'm at #10 now. This one should be very simple but I
>>>> don't
>>>> know why it's taking so long to run.
>>>>
>>>> Andrew.
>>>>
>>>> On Tue, Jul 8, 2008 at 6:52 AM, kinuthiA muchanE <muchanek at gmail.com
>>>> > wrote:
>>>>> Habari,
>>>>> Anybody still trying these problems, you are all so, so silent...
>>>>>
>>>>> Kinuthia...
>>>>>
>>>>>
>>>
>>>
>>
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