# Sum of power of two exact 2010 different ways

Hello,

I look for a solution for the following problen.

Determine all numbers, which can be written in exact 2010 different ways as a sum of powers of two with non-negative exponent, while all exponents are only allowed to appear maximal three times in one sum.
(All sums where only the order of the summands is changed are count as one sum.)

My problem is that I do not have any idea, how to find a solution to this task.
Therefore I started to determine in how many different sums I can write for example 2^4:
1. 2^4
2. 2*2^3
3. 2*2^2+2^3
4. 2*2+2^2+2^3
5. 2*2^0+2+2^2+2^3
6. 2*2^0+3*2+2^3
7. 2*2^0+3*2+2*2^3

Nevertheless, I did not find any coherenz zu the number and the number of different sums.

Do you have any tips?

Christian

No idea?

Mark44
Mentor
Hello,

I look for a solution for the following problen.

Determine all numbers, which can be written in exact 2010 different ways as a sum of powers of two with non-negative exponent, while all exponents are only allowed to appear maximal three times in one sum.
(All sums where only the order of the summands is changed are count as one sum.)
I don't understand what you're asking. Is it that you want all numbers that can be written in exactly 2010 different ways?

Or is it that you want to find the number of ways that 2010 can be written as a sum of powers of 2?
My problem is that I do not have any idea, how to find a solution to this task.
Therefore I started to determine in how many different sums I can write for example 2^4:
1. 2^4
2. 2*2^3
3. 2*2^2+2^3
4. 2*2+2^2+2^3
5. 2*2^0+2+2^2+2^3
6. 2*2^0+3*2+2^3
7. 2*2^0+3*2+2*2^3

Nevertheless, I did not find any coherenz zu the number and the number of different sums.

Do you have any tips?

Christian

Hello,

I look for a solution for the following problen.

Determine all numbers, which can be written in exact 2010 different ways as a sum of powers of two with non-negative exponent, while all exponents are only allowed to appear maximal three times in one sum.
(All sums where only the order of the summands is changed are count as one sum.)

My problem is that I do not have any idea, how to find a solution to this task.
Therefore I started to determine in how many different sums I can write for example 2^4:
1. 2^4
2. 2*2^3
3. 2*2^2+2^3
4. 2*2+2^2+2^3
5. 2*2^0+2+2^2+2^3
6. 2*2^0+3*2+2^3
7. 2*2^0+3*2+2*2^3

Nevertheless, I did not find any coherenz zu the number and the number of different sums.

Do you have any tips?

Christian

Hi there!
I believe you quote the question from Columbus state university .

This is an easy question.
Assume there is four blocks since the two non zero integer are similar ,
we label one of the a and b. one with a selection of 3 integers another with 4.
For a>2.

Therefore the blocks of 4 integer will look like this:
a00b
a0b0
ab00

I'll leave the rest to you.
ans is thirty-_____ .

Hi there!
I believe you quote the question from Columbus state university .

This is an easy question.
Assume there is four blocks since the two non zero integer are similar ,
we label one of the a and b. one with a selection of 3 integers another with 4.
For a>2.

Therefore the blocks of 4 integer will look like this:
a00b
a0b0
ab00

I'll leave the rest to you.
ans is thirty-_____ .

Huh? You're sure this is the answer to the right question?

Hello,

I look for a solution for the following problen.

Determine all numbers, which can be written in exact 2010 different ways as a sum of powers of two with non-negative exponent, while all exponents are only allowed to appear maximal three times in one sum.
(All sums where only the order of the summands is changed are count as one sum.)

My problem is that I do not have any idea, how to find a solution to this task.
Therefore I started to determine in how many different sums I can write for example 2^4:
1. 2^4
2. 2*2^3
3. 2*2^2+2^3
4. 2*2+2^2+2^3
5. 2*2^0+2+2^2+2^3
6. 2*2^0+3*2+2^3
7. 2*2^0+3*2+2*2^3

Nevertheless, I did not find any coherenz zu the number and the number of different sums.

there are actually 9 ways to write 2^4

writing them in 'binary'

10000
2000
1200
1120
1112
1032
320 = 3*2^2 + 2 * 2^2
312 = 3*2^2 + 1 * 2^1 + 2*2^0
232 = 2*2^2 + 3 * 2^1 + 2*2^0

you missed the last 3 and your number 7 isn't valid.

It appears that if f(n) is the number of ways to write n, then f(n) = 1 + floor(n/2)

(floor(x) is the greatest integer <= x)

I haven't been able to prove this so far. It's easy to prove that f(2n+1) = f(2n) so you
only have to prove f(n) = 1+n/2 for even numbers.