Easy
3 $ 2 = 15
7 $ 4 = 311
62 $ 13 = 4975
51 $ 43 = ???
Harder:
15 $ 11 = 104
18 $ 7 = 275
62 $ 13 = 3675
24 $ 1 = ???
Difficult:
31 $ 41 = 592
27 $ 18 = 281
16 $ 180 = 339
33 $ 21 = ???
A $ B denotes a function of A and B, unchanged within a section.
As this post has been for several months without solution I suppose nobody is working on it. So I explain the solution.
The first section is very easy (3 $ 2 = 15) because (3-2 =1; 3+2 =5).
The second section is not difficult (15 $ 11 = 104) because (15+11=26 and 15-11=4; 26*4=104)
The third section is simple but a posteriori. When you notice that 31 $ 14 = 592 is derived from π = 3,141592.
31 $ 41 = 592 is from π = 3,141592…
27 $ 18 = 281 is from e = 2,718281… (Euler’s number)
16 $ 180 = 339 is from φ = 1,6180339… (Golden’s number).
So, the solution for 33 $ 21 = ? would be a number whose first digits are: 3,321… The best candidate to this I suppose is number x that 2^x=10 (logarithm of 10 in base 2).
x= 3,321928… and then 33 $ 21 = 928
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When I was thinking on it I thought the question posted was 21 & 33 (instead of 33 &21).
I found a surprising and quite different solution for that.
Because 2,133… is 32/15 that could came from 3 $ 2 = 15, just the first sequence of the first section.
In that case the title of the post would have been: From E to D and From D to E.
Edited on November 19, 2015, 9:22 am
Edited on November 19, 2015, 9:26 am
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Posted by armando
on 2015-11-19 09:21:03 |
Solution explained
