| Author | Who is the cleverest of 'em all |
|---|
The increases imply a squared function ,
squared +1 works foir teh first 2 numbers, but then cannot find a link to make the third |
29.
5 X 26 - 3 = 127
3 x 10 - 1 = 29 |
it's atricky one... i don't get it =(
but the last numbers (3 and 127) are primenumbers, so i'd go for 19 or 29 - though i don't know the way to get there... |
for Zarebrant:
This math problems have unique solutions. Whenever you get 2 solutions with a way of thinking the problem, it means the way you thought the problem is not correct. :) |
well... sure.
i didn't give the answer, just some thoughts about it... |
| just some thoughts about it... and i was trying to read your mind and to continue your thoughts... am i a good telepath? :) |
| hehe... =) |
29
cube the first number then add on twice the difference between the square of the first and the second.
eg
5^3= 125
26-(5^2)=1
(5^3) + 2*(26-(5^2)) = 127
(1^3) + 2*(2-(1^2)) = 3
(3^3) + 2*(10-(3^2)) = 29 |
| the "second intelegent one" |
I also got 29, but by a different method.
1, 2, 3... from 1 to 2, add 1, from 2 to three add 1. so the formula I used was...
2 (second number of series) + 1*1 (first number of series multiplied by difference of second number minus first number) - (1-1) (first number of series -1)
So you get...
2+1(1)-(1-1) = 2+1-0=3
Third line
26 + 5(21) -(5-1) = 26 + 105 - 4 = 127
Second line
10 + 3(7) - (3-1) = 10 + 21 - 2 = 29 |
divit is a genius in Q maker ;)
2+2-1*0 :p |
another way(much easier):
second numbers r squared first numbers+1: 1^2+1=2; 5^2+1=26; 3^2+1=10
third numbers r third degeree of first +2: 1^3+2=3; 5^3+2=127; 3^3+2=29
The easiest and i think the rightest way, but don't forget that mathematical problems have enourmous quantity of ways of solving. |
wow, after studying it, I can simplify rozatron's answer
First number = number
Second number = first number squared plus 1
Third number = first number cubed plus 2
Or even... (and I realized this as I typed the above)
First number = number
to get to each subsequent number, take first number raised to the power of (the position of the number in the series [ie. the second number is position 2, third number is position 3, etc]) minus however many positions this number is after the first number (second number is 1 position after first, 4th number would be 3rd position from 1st)
such that...
first number = x, x=1 in first line, 3 in second, and 5 in third line
any number there after would be...
x^n - (n-1)
where n= the number in the series (1 for first, 6 for 6th number in series, etc) |
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(")_(")into you profile to help him gain world domination with Bunny.
\__/)
(='.'=) This is Bunny. Copy and paste bunny
(")_(") into your profile to help him gain world domination.
(')_(')
(=^.^=)This is brother bunny. copy and paste brother bunny
(")_(")into you profile to help him gain world domination with Bunny.
(')_(')
(^.^)This is brother bunny. copy and paste brother bunny
(")_(")into you profile to help him gain world domination with Bunny.
how many bunnies are there in this post |
| 2 |
| i think we should reward divit for such a good and interesting math task. |
16
The word bunny is stated 16 times. |
| 17+ if you include bunnies from how many bunnies are there in this post |
for Arsuha:
Agree, just a Q, does 2^2 equals to 8? I want to test this "^" since I've never used it before |
'^' means involution or exponentiation
2^2=4; 3^2=9;4^2=16 etc.
2^3=8; 3^3=27;4^3=64 etc. |
