Your answer differs from that of Viet Dao, who rearranged the problem as (4^5)^6 + (6^5)^4. It seems to depend upon which interpetation is correct. Is it the same in every country? Also, I am not sure which is correct in the USA. I understand that multiplications and divisions are carried out from the left to the right. For instance 24/3*6 = 8*6 = 48. At first I thought it be the same for powers. But then I found that reiterated powers are evaluated from the right as in your post. See http://en.wikipedia.org/wiki/Order_of_operations . I am not sure whether this convention is international or not.Zurtex said:I'm afraid I don't know how to do this other than working it out mod 10^{n} and finding the largest value of n that is 0 and all previous n are 0. Which is fairly easy because of the form you've given.
However you could just calculate it:
4^(5^6) + 6^(5^4) =
There is no convention; there is no right in the USA compared to anywhere else. You're just supposed to put parenthesis around the binary operators that are non-associative.ramsey2879 said:Your answer differs from that of Viet Dao, who rearranged the problem as (4^5)^6 + (6^5)^4. It seems to depend upon which interpetation is correct. Is it the same in every country? Also, I am not sure which is correct in the USA. I understand that multiplications and divisions are carried out from the left to the right. For instance 24/3*6 = 8*6 = 48. At first I thought it be the same for powers. But then I found that reiterated powers are evaluated from the right as in your post. See http://en.wikipedia.org/wiki/Order_of_operations . I am not sure whether this convention is international or not.
Then there is a problem with LaTeX since I tried writing [tex]4^{5^6}[/tex] three different ways: 4^5^6, {4^5}^6, and 4^{5^6}; each within the LaTeX coding operators of course. The first gives the same result as 4^56 and the latter two forms give the same result for either choice. How would you add the parenthesis to the printed form in LaTeX?Zurtex said:There is no convention; there is no right in the USA compared to anywhere else. You're just supposed to put parenthesis around the binary operators that are non-associative.
Quote the original post and look at the LaTeX, you will see my interpretation of it is correct.
There isn't a problem with LaTeX, "4^5^6" is just really bad use of it. You could do either:ramsey2879 said:Then there is a problem with LaTeX since I tried writing [tex]4^{5^6}[/tex] three different ways: 4^5^6, {4^5}^6, and 4^{5^6}; each within the LaTeX coding operators of course. The first gives the same result as 4^56 and the latter two forms give the same result for either choice. How would you add the parenthesis to the printed form in LaTeX?
Of course there's a convention; exponentiation is right-associative. Implicitly, 2^3^4 should be interpreted as 2^(3^4). Of course, that still agrees with your interpretation.Zurtex said:There is no convention; there is no right in the USA compared to anywhere else. You're just supposed to put parenthesis around the binary operators that are non-associative.
Quote the original post and look at the LaTeX, you will see my interpretation of it is correct.
Really? That's a convention? That's what I intuitively thought, maybe I just noticed it so many times and it came to me without thinking.master_coda said:Of course there's a convention; exponentiation is right-associative. Implicitly, 2^3^4 should be interpreted as 2^(3^4). Of course, that still agrees with your interpretation.
Yes; it's even mentioned on the wikipedia page that ramsey2879 was referencing (and on the associativity page as well). It doesn't seem to be as well known as most other such conventions, probably because expressions like a^b^c don't occur all that often.Zurtex said:Really? That's a convention? That's what I intuitively thought, maybe I just noticed it so many times and it came to me without thinking.
Zurtex, what software did you use to get this answer, because it seems right that the answer would end in a two, as per Viet Dao's reply? Yet, your software has it ending in five zeros! To help validate your answer, calculate the two parts before you add them, inspect the last digits on each, and see if a rounding error happens before or after the final additon.Zurtex said:However you could just calculate it:
4^(5^6) + 6^(5^4) = 153...3920275853921484800000
Viet Dao is answering a different question- he interpreted the stacked exponentiation in a way different from the standard.SteveRives said:Zurtex, what software did you use to get this answer, because it seems right that the answer would end in a two, as per Viet Dao's reply? Yet, your software has it ending in five zeros! To help validate your answer, calculate the two parts before you add them, inspect the last digits on each, and see if a rounding error happens before or after the final additon.
VietDao29 was interpreting the numbers differently than Zurtex; so they got different results.SteveRives said:Zurtex, what software did you use to get this answer, because it seems right that the answer would end in a two, as per Viet Dao's reply? Yet, your software has it ending in five zeros! To help validate your answer, calculate the two parts before you add them, inspect the last digits on each, and see if a rounding error happens before or after the final additon.