Cannot freeken get this one!

Sven · 2005-11-08 04:38:53

THe puzzle is this:

When is half of 13 equal 8? (Hint: different number system)

I have tried this one with different bases (Base 3 Base 5 etc) and in Hex and Octal...I am either missig something or I am way off track.

Anyone get this?

mathsyperson · 2005-11-08 04:44:31

Hmm. In base x, 13 would be x + 3 (we have base 10, so 13 is 10 + 3), which means that your thing would be (x + 3)/2 = 8.
Shifting the 2 gives x + 3 = 16, so the base is 13.

Sven · 2005-11-08 05:07:37

I understand the basics there in that equation. HOwever, maybe I am just overthinking:

But is not 13 in base 13 = 1?

That being the case, how can the answer be Half of 13 in base 13 is 8?

mathsyperson · 2005-11-08 05:12:37

No. 10 in base 10 isn't 1. It's 10.

In base 10, numbers are made up of 1's, 10's, 100's, 1000's and so on, because they are 10¹, 10², 10³ and so on.
Base 13 is similar, but it's made up of 13°'s, 13¹'s, 13²'s and so on.

That means that 13 in base 13 is 1*13+3 = 16, half of which is 8.

Sven · 2005-11-08 05:22:12

Ahh, I WAS overthinking it. Missed the equation meaning.

Problems arise frequently when you are right-brained and tend to see math in visual form.

Yeah....13 in base 13 means one 13 and 3 ones.

I forgot to see it in a columnar format as opposed to a number of a whole.

Yeah, got it. Thanks alot.

COnverting that number to 10 base would give me

13/10=1 R 3
1/10=0 R 1

13.

Gee, sometimes I really need to get that left side to kick in sooner.

MathsIsFun · 2005-11-08 08:31:57

LOL! And what a great illustration of bases.

Math Is Fun Forum

#1 2005-11-08 04:38:53

Cannot freeken get this one!

#2 2005-11-08 04:44:31

Re: Cannot freeken get this one!

#3 2005-11-08 05:07:37

Re: Cannot freeken get this one!

#4 2005-11-08 05:12:37

Re: Cannot freeken get this one!

#5 2005-11-08 05:22:12

Re: Cannot freeken get this one!

#6 2005-11-08 08:31:57

Re: Cannot freeken get this one!

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