Solve Exponents Question: 333333/33 Remainder Value

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Discussion Overview

The discussion revolves around finding the remainder when dividing 333333 by 33. Participants explore various mathematical approaches, including logarithms and prime factorization, as well as concepts related to modular arithmetic.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about how to approach the problem, suggesting the use of logarithms to simplify the number.
  • Another participant recommends writing 333 in prime number factors as a potential method to solve the problem.
  • A repeated post reiterates the initial confusion about using logarithms, mentioning that taking the logarithm of 333 raised to the power of 333 and dividing by 3 could lead to complications due to rounding errors in calculators.
  • A participant introduces the concept of modulo arithmetic as a simpler alternative for tackling the problem.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the best approach to solve the problem, with multiple methods being suggested and explored.

Contextual Notes

There is mention of potential limitations related to calculator precision when using logarithmic methods, as well as the need for clarity on the application of modular arithmetic.

TheExibo
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If you divide 333333 by 33, what is the value of the remainder?

I'm not sure where to starte since this number can't be put into a calculator. Is there something with logs? I was thinking of bringing the number down with logs to 333log333, but I'm confused as to where that will lead me.
 
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Write 333 in prime number factors and see if that helps.
 
TheExibo said:
If you divide 333333 by 33, what is the value of the remainder?

I'm not sure where to starte since this number can't be put into a calculator. Is there something with logs? I was thinking of bringing the number down with logs to 333log333, but I'm confused as to where that will lead me.
Taking the logarithm of [itex]333^{333}[/itex], divided by 3, would give 333log(333)- log(3). The difficulty is that your calculator can only give a limited number of decimal places for the logarithm and multiplying by 333, then taking the exponential of the result, will make the "round off error" worse.
 
Do you know modulo arithmetic? You should look into that. (It's not very difficult, it's mainly just a handy notational system to make problems like these easier)
 

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