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 10n 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?
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.
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 said:However you could just calculate it:
4^(5^6) + 6^(5^4) = 153...3920275853921484800000
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.
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.
A factorial is a mathematical function denoted by an exclamation mark (!). It is used to calculate the number of ways in which a set of objects can be arranged. For example, 5! (pronounced "five factorial") is equal to 5 x 4 x 3 x 2 x 1, which is 120.
To find the number of zeros at the end of a factorial, you need to count the number of times the factorial is divisible by 10. This is equivalent to counting the number of times the factorial is divisible by 5 and 2. The number of zeros at the end of a factorial is determined by the smaller of the two counts. For example, if a factorial is divisible by 5 three times and 2 four times, then it will have three zeros at the end.
Finding zeros at the end of a factorial is useful in a variety of mathematical and scientific fields. For example, it can help in calculating the number of trailing zeros in a large number, which is important in data storage and compression. It is also used in analyzing the efficiency of algorithms and in solving problems in combinatorics and number theory.
Yes, there is a faster way to find the zeros at the end of a factorial without having to calculate the entire factorial. This method involves counting the number of multiples of 5, 25, 125, and so on, that are present in the factorial. The sum of these counts will give the total number of zeros at the end of the factorial. This method is more efficient for larger factorials.
Yes, many scientific calculators have a function that can calculate the number of zeros at the end of a factorial. However, it is important to note that the factorial function on a calculator may have a limit on the size of numbers it can handle. For larger factorials, it may be necessary to use a computer or a specialized factorial calculator.