Hard Problem

[MathStudent1999]

You select an integer N from 1 to 100 inclusive, and calculate the sum from k=1 to k=N of K^4, noticing the units digit of that sum. What is the average value of these units digits? Express your answer correct to 1 digit after the decimal point.

I don't get what from k=1 to k=N of K^4 means

 

[srmichael]

You select an integer N from 1 to 100 inclusive, and calculate the sum from k=1 to k=N of K^4, noticing the units digit of that sum. What is the average value of these units digits? Express your answer correct to 1 digit after the decimal point.

I don't get what from k=1 to k=N of K^4 means

It means \(\displaystyle 1^4+2^4+3^4+4^4+...+N^4\)

 

[JeffM]

You select an integer N from 1 to 100 inclusive, and calculate the sum from k=1 to k=N of K^4, noticing the units digit of that sum. What is the average value of these units digits? Express your answer correct to 1 digit after the decimal point.

I don't get what from k=1 to k=N of K^4 means

Let's say you choose 4 as N

\(\displaystyle 1^4 = 1.\)

\(\displaystyle 1^4 + 2^4 = 1 + 16 = 17.\)

\(\displaystyle 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98.\)

\(\displaystyle 1^4 + 2^4 + 3^4 + 4^4 = 1 + 16 + 81 + 256 = 354.\)

So, using sigma notation: \(\displaystyle \displaystyle \sum_{k = 1}^Nk^4.\)

 

[soroban]

Hello, MathStudent1999!

You select an integer N from 1 to 100 inclusive,
and calculate the sum from k=1 to k=N of k^4, noting the units digit of that sum.
What is the average value of these units digits?
Express your answer correct to 1 digit after the decimal point.

I don't get what from k=1 to k=N of k^4 means.

They should have said: .\(\displaystyle \text{the sum of }k^4\text{, from }k=1\text{ to }k=N.\)

 

[DrPhil]

Not sure what this is all about but if you go from 1 to 100,
sum of the last digit of each is surprisingly 450.
(1+7+8+4+9+ ............ +2+3+9+0+0 = 450)

And 10(sum 0 to 9) = 450

And this is a matter for Dr Phil (Bob is too old!)

This calculation could get tedious [Life gits tejous, don't it!] if you don't take just the units digit of everything. Also I used Excel - is that cheating?

I was also interested to note it comes out with the same result as assuming every digit from 0 to 9 is equally probable:
Avg{0,1,2,3,4,5,6,7,8,9} = 4.5

What constitutes "too old"?

 

[Deleted member 4993]

Let's say you choose 4 as N

\(\displaystyle 1^4 = 1.\)

\(\displaystyle 1^4 + 2^4 = 1 + 16 = 17.\)

\(\displaystyle 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98.\)

\(\displaystyle 1^4 + 2^4 + 3^4 + 4^4 = 1 + 16 + 81 + 256 = 354.\)

So, using sigma notation: \(\displaystyle \displaystyle \sum_{k = 1}^Nk^4.\)

\(\displaystyle \displaystyle \sum_{k = 1}^nk^4 \ = \frac{6n^5 + 15n^4 + 10n^3 - n}{30}.\)

Reference: http://mathforum.org/library/drmath/view/56383.html

 

[MathStudent1999]

Thank you all for their help. I got 4.5 is this correct?

 

[DrPhil]

Thank you all for their help. I got 4.5 is this correct?

Yes, precisely.