is there a way to distribute xor over multication?

exe t*t^58 = t^? *t^?

Started By
cookietran
, Jul 28 2020 10:11 PM

5 replies to this topic

Posted 28 July 2020 - 10:11 PM

is there a way to distribute xor over multication?

exe t*t^58 = t^? *t^?

Posted 29 July 2020 - 08:22 AM

What are you asking, I don't understand your formula?

Posted 29 July 2020 - 09:13 AM

is there a way to distribute xor over multication?

exe t*t^58 = t^? *t^?

t^{59} is the solution as t*t = t^{2 }, the exponent just shortens the list of multiplication t * t * t = t^{3 }and so on. t^{58} is just 58 sequences of t*t which is the same as saying ∏_{0}^{58}_{ }t * t, thus multiplying t * t^{58} = t^{59}

**Edited by VictorMedvil, 29 July 2020 - 09:28 AM.**

Posted 29 July 2020 - 09:32 AM

Oh that's what he meant. The exe threw me off, we usually write or either as above of exp.

Posted 29 July 2020 - 09:36 AM

Oh that's what he meant. The exe threw me off, we usually write or either as above of exp.

Well I assume that's what he means, i dunno what the EXE means. t^{t }would be something different if he meant that or t^t^58 = ∏_{0}^{t }t^{58 }* t^{58 } or t^{58} * t^{58} , t number of times, which when t = 2 would be 2^{58 }* 2^{58}

**Edited by VictorMedvil, 29 July 2020 - 09:45 AM.**

Posted 29 July 2020 - 04:07 PM

Well I assume that's what he means, i dunno what the EXE means. t

^{t }would be something different if he meant that or t^t^58 = ∏_{0}^{t }t^{58 }* t^{58 }or t^{58}* t^{58}, t number of times, which when t = 2 would be 2^{58 }* 2^{58}

Indeed. We can relax I suppose, but the poster should be more clear.