# Math is Hard ! What is the Rule?

These terms are all decoded in a certain way or they follow a certain rule. Your job is to figure out how to encode the last term.

1 = 22
3 = 38
6 = 34
7 = 41
14 = 68
11 = ?


Edit
There was a mistake in the earlier version of the question, which has been rectified.

Hint 1:

Math isn't the important here ;) It is however required

• @CodeNewbie Thanks. I can always count on you when it come to my bad english :) Aug 12, 2015 at 12:38
• Happy to help :) Aug 12, 2015 at 12:42
• Maybe 16? I'm really unsure about this so I will wait until you probably say it's wrong :P (Not worth an answer or explanation for now)
– user14478
Aug 13, 2015 at 14:19
• @LuxxMiner Its Wrong :) Nevertheless i would like to know how you came up with this Aug 13, 2015 at 14:21
• Ok... (This will be very hard to explain since I'm not a native-english-speaker). First I calculated the "difference" between the next two numbers that follow each other. (so 22 -> 38, 38 -> 34,etc.). We get a numerical sequence of (16,-4,7,27,a). Now I did the same with these numbers, we get: (-20,+11,+20,b). This looked like an ordered sequence to me, so b must be (-11). Thus a must be 16 (because 27-11) . I just realised that the answer, when following my theory, would be 84 (because 68+16), but the chances are very low that this is right. :P
– user14478
Aug 13, 2015 at 14:34

I think I found the answer.

Maybe it's a programmer exercise: you have to perform following conversions:
decimal => string => sum of alphabet position of each letter => conversion in hexadecimal format

In this way the list is:

1 = "one" = 15+14+5 = 34 (decimal) = 0x22 (hexadecimal)
3 = "three" = 20+8+18+5+5 = 56 (decimal) = 0x38 (hexadecimal)
6 = "six" = 19+9+24 = 52 (decimal) = 0x34 (hexadecimal)
7 = "seven" = 19+5+22+5+14 = 65 (decimal) = 0x41 (hexadecimal)
14 = "fourteen" = 6+15+21+18+20+5+5+14 = 104 (decimal) = 0x68 (hexadecimal)

And obviously,

11 = "eleven" = 5+12+5+22+5+14 = 63 (decimal) = 0x3F (hexadecimal)

• I don't know how to hide solution. I hate spoilers.... Aug 13, 2015 at 15:08
• @Stefano No worries, got that for you :) Good answer! Aug 13, 2015 at 15:13
• thx @luxmi12, good hint for future solutions ;) Aug 13, 2015 at 15:15

I am very unsure about my solution, but I found something interesting. First I observed that

the positional value (that is a=1,b=2, etc., then all added up) of the word of the number on the left hand side has the same remainder after integer division by 6. I will write $a(n)$ for the alpha-numerical value of the word of the number $n$. Then we have $a(1)=34=22+2 \cdot 6, a(3)=56=38+3\cdot 6, a(6)=52=34+3\cdot 6, a(7)=65=41+4\cdot 6, a(14)=104=68+6 \cdot 6$. For $n=11$ this would give us $a(11)=63=33+5 \cdot 6$.

Maybe this is all bull, but so far I think that the answer is

"11 = 33" - but maybe its wrong and my idea can be used otherwise.

• Nice! But thats not what i had in mind. But you are on the right track. You just pushed it a little too far :) Aug 12, 2015 at 18:41