I am having trouble with a riddle?!


Question: I am having trouble with a riddle!?
25314 = !?

You must figure out what 25314 equals!. Here is a hint:

24351 = 51324
43125 = 34215

Both of these equations use the same pattern!. You must find the pattern and apply it to the original number!.

If you figure it out, please tell me because I am stumped!Www@Enter-QA@Com


Answers:
Ok!. For this riddle you need to look at the positioning of the numbers!.

For example, take 43125!. "1" is in the 3rd digit place, "2" is in the 4th digit place, "3" is in the 2nd digit place, "4" is in the 1st digit place, and "5" is in the 5th digit place!. So then you get 34215!. Then you use the same process with 24351 and you get 51324!.

So, for 25314, 1 is 4th, 2 is 1st, 3 is 3rd, 4 is 5th, and 5 is 1st!. So then the answer is 41325!.Www@Enter-QA@Com

24351 = 51324
43125 = 34215

If you look at both of these you'll see that one number always stays in the same position!. The 3 on the first line and the 5 on the next line!. For the first line the 24 and 51 simply swap places!.

In the second line the 5 is on the end!. So the numbers can't symmetrically swap places!. Thus they swap digits!.!.!.(43 becomes 34 and 12 becomes 21!.)

I got stuck here for a while until I noticed that the 3 in 25314 is in the same position as the first line!. And if this number follows the pattern as the others, then!.!.!.

25314 = 14325

Phew, this one was a tuffy!.Www@Enter-QA@Com

OK this probably isn't the right answer, but I've been looking at all the different ways you could switch the numbers around, and this is all I've come up with!.

If one of the digits is in that digits' place!. (i!.e!. in the first hint equation, the 3 is in the 3rd place) that digit stays put!. Ignoring that digit, then the first odd switches with the first even digit, and the second odd switches with the second even digit!.

The answer would then be 14325!.

It does work but may not be what the asker was looking for!.



By the way, for those of you who don't seem to understand this, try looking at it this way:

F(24351) = 51324 and F(43125) = 34215!.
Solve for F(25314) where F(x) can be any kind of function!.Www@Enter-QA@Com

there is no equasions shown!. 25314=!? means nothing!. 24315 does not equal 51324, nor does 43125 equal 34125!. Without a basis for solving, IE looking for an answer required, this is a waste of time!. I'm sure it's a cute riddle but, pointless!. maybe you presented it incorectly!. I'm real good at math,but this means nothing!. Sorry!.I must be stupid!.Www@Enter-QA@Com

First number, add 3 to 2 and get 5, subtract 3 from 4 get 1, leave the 3 alone, then subtract 3 from 5 and get 2, then add 3 to the 1 and get 4!.

+3, -3, 0, -3, +3Www@Enter-QA@Com

25314=34152
ok so this is true bc 2+4+3+5+1=5+1+3+2+4 and 4+3+1+2+5=3+4+2+1+5!. If works in any order bc u are using the same #s!.Www@Enter-QA@Com

25314=41325


my brain hurts now but ur welcome!

i would explain but its complexWww@Enter-QA@Com

25314=14325

53431=35341Www@Enter-QA@Com

2+4+3+5+1=5+1+3+2+4
4+3+1+2+5=3+4+2+1+5

Each side of both equations is 15!.Www@Enter-QA@Com

25314=25314

it does not equal anything else but itselfWww@Enter-QA@Com

24351 = 51324
43125 = 34215

25314=52327!.!.!. i think!. it goes +3, -3!. 3, -3, +3Www@Enter-QA@Com

41352 i thinkWww@Enter-QA@Com

25314 = 12345Www@Enter-QA@Com

41352Www@Enter-QA@Com

12345Www@Enter-QA@Com

41352Www@Enter-QA@Com

41325!.!.!.!.!.or!.!.!.!.!.52134Www@Enter-QA@Com

Hmm!.!.!. I'm stumped!. When you find out please email me!.


PS!. Is this your math homework!? lolWww@Enter-QA@Com

this is not a riddle

it's just hard MathWww@Enter-QA@Com

Wow!Www@Enter-QA@Com

try looking it up on google!Www@Enter-QA@Com

WOW!.!. Man, I am stumped too!.!. Looks like a job for GOOGLE!! Lol!.!.Www@Enter-QA@Com



The answer content post by the user, if contains the copyright content please contact us, we will immediately remove it.
Copyright © 2007 enter-qa.com -   Contact us

Entertainment Categories