Maths Brains! Can a number in the powers-of-2 sequence ever be divisible by 3?

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powers-of-2 sequence being 1 - 2 - 4 - 8 - 16 - 32 - 64 - 128 - etc etc

And, if not, why not?

NO!! Because all numbers in the sequence can be written as:

2
2x2
2x2x2
2x2x2x2
2x2x2x2x2

etc. If they were able to be divided by three you'd have to have a factor of 3 in there and there aren't any....

Why do you want to know?

Disagreement between me & ds - or not so much a disagreement - me not being able to understand why ds was so definite about its not being possible.

He tried to explain with a sort of air-sketch gesture of a pyramid standing on its tip & all the sides going out evenly by 2 ... but I still didn't understand why you wouldn't eventually hit on a number which happens to be divisible both by 2 and 3. He also said "it's because none of 1, 2 or 4 are divisible by 3" - but I didn't really understand that either.

To be honest, I still don't get it. Can you do an even simpler explanation? Please?

I tried to explain it in english and couldn't.

Yes - light is dawning, thank you everyone.

I of course was thinking that at some point you could clump together those x2x2x2 that Worzella set out into something that also had 3 as a common denominator - but of course you can only clump them into 2x2 = 4 or 2x2x2 = 8 (etc.) so the same non-divisible-by-3 units keep repeating themselves. (That probably only makes sense to me, but now I understand.)

Lovely MN.

Gald to have helped a little- apologies for looking like I was shouting NO! caps lock went on....

you aint seen nothing yet littlelapin!!