Monday, February 27, 2012

Answer for the 2581=?? riddle

This problem can be solved by pre-school children in 5-10 minutes, by programmers – in 1 hour, by people with higher education… well, check it yourself!

8809 = 6
7111 = 0
2172 = 0
6666 = 4
1111 = 0
3213 = 0
7662 = 2
9312 = 1
0000 = 4
2222 = 0
3333 = 0
5555 = 0
8193 = 3
8096 = 5
7777 = 0
9999 = 4
7756 = 1
6855 = 3
9881 = 5
5531 = 0

2581 = ?

recently this riddle have been circulating around the internet and i am about to tell you how i crack the code MY WAY. It took me about less than 30 minutes and I am only a first degree commerce student (yeah, i am bragging!)

The first method which is not my method was the easiest and took the 'pre-school' as a hint. Pre-school student have limited mathematical knowledge so diving deep into all sorts of mathematical figure like arithmatic, differentiation, qudratic and whatever else is a trap and if you fell for it it shows your knowledge is not flexible and rigid to a certain form of thinking.

Anyways, the method was simply to count the 'circles' in the numbers. like '6' has one circle in it so its value is 1. 6666 is therefore has the value of 4. numbers like '8' has two circles so its value is 2. the illustration below from baka-baka should be helpful.



Another method is similar to mine but way too complicated for me to tell you how it goes. it uses mathematical functioning and algebra which i doubt pre-school kids knew about it. if you want to learn that then here is the link.

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Now for my method. The advantage of my method is that even if the quad series are replaced with a specific correspondence numbers (eg. ABBC=2).

it is actually quite simple but you need a paper and pen. The first thing i did was list all the quads (series of four numbers) with identical numbers into this:

1111=0
2222=0
3333=0
4444=0
5555=0
7777=0
6666=4
9999=4
0000=4

the only thing missing was 8888. from here i used the logi gate method (something in physics but even without that knowledge you can still solve it with this method). i remained all 0=0 and all 4=1. eg.

7777=0
6666=1

i checked it with some of the quads without the number 8. here are some example.

2172: 2222=0, 1111=0, 7777=0, 2222=0
therefore its final value its '0'

then i tried 7662: 7777=0, 6666=4, 6666=4, 2222=0
since there is two 4 it is 8 and since 4=1, therefore 8=2
the final value is '2'

i know comfirmed my hypothesis is true. now for the next step is to find the value of '8888' since number 8 is present in the 2581=??

i took the quad 8809=6.
we know that 0000=4 and 9999=4
since 4=1
0000=1 and 9999=1
1+1 = 2
6-2= 4

there are two 8s. that means we split the remaining value (4) to each 8 equally.
therefore 8888=2. (REMEMBER: This is in the [1,0] form and not the [4,0] form)

hope you are still with me.

now for the conclusion:

2581:
2222=0=0
5555=0=0
8888=8=2
1111=0=0

0+0+2+0 = 2

the answer is 2!!!

3 comments:

  1. the correct answer is to count the number of circles in the numbers... 8 has 2 and a 6 for example has 1 circle a 7 has none...

    ReplyDelete
  2. pokalipse.Mnogo je bitan i glossary dock.ga pokusava prisvojiti+

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  3. your logic is flawed with baseless assumptions , it is clear to me you are not particularly good with abstract thinking . for example : 2172: 2222=0, 1111=0, 7777=0, 2222=0
    therefore its final value its '0' has no logical base , it is just an assumption which you cannot demonstrate , very closely linked to kindergarden thinking.

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