Every number from $1$ to $n$ gets multiplied by $2^n$

Question

Every natural number from $1$ to $n$ ($n >0)$ was multiplied by a power of two (the exponent can be different). After that all the numbers were summed. At what $n$ will the sum also be a power of $2$? Something tells me that it can be at any $n$. I think we have to start with summarizing numbers from $1$ to $n$ by using the formula $$\frac{n(n+1)}{2}.$$

Answer 1

Are you talking about the following series : $$1.2^1,2.2^2,...n.2^n$$

It is an Arithmetico Geometric series. Summation of its terms can be found out as follows:

Let $$S = 1.2^1+2.2^2+3.2^3...+n.2^n $$ Now multiplying both sides by the common ratio $2$, we get

$$2.S = 1.2^2+2.2^3+3.2^4...+(n-1).2^n+n.2^{n+1}$$

Subtracting $2^{nd}$ equation from $1^{st}$ equation, we get

$$-S = 1.2^1+1.2^2+1.2^3+...+1.2^n-n.2^{n+1}$$

Now leaving the last term of our series aside, we observe a G.P. with common ratio $2$ and first term $2$, upon summing it we get,

$$-S = 2(2^n-1)-n.2^{n+1}$$ $$-S = 2^{n+1}-n.2^{n+1}-2$$ $$S=(n-1)2^{n+1}+2$$

$$S = 2(1+(n-1)2^n)$$ Now, we can see that the term $(n-1)2^n$ is even, so adding $1$ to it which is the case will make it odd. Now, for being power of $2$ all the factors of the numbers must be $2$ which is even. Hence, $S$ can't be power of $2$ except for $n = 1$ in which case $S = 2$.

So, the only solution is $n = 1$.

Answer 2

It is always possible. Let $T_n=\frac 12n(n+1)$ be the sum of the numbers from $1$ to $n$. For $n \gt 1$ we can always find $k$ such that $2^{k-1} \lt T_n \lt 2^k$. We just have to find a set of numbers between $1$ and $n$ that sum to $2^k-T_n$ and multiply them all by $2$, multiplying all the rest by $1$. We can do this by taking $n, n-1, n-2 \ldots$ until the sum gets too large, deleting the last one, and picking the smaller number we need to come out even. We can then achieve $2^m$ for $m \gt k$ by multiplying all the multipliers by $2^{m-k}$

Answer 3

Assume $n$ is an "$m$-bit number", i.e., $2^{m-1}\le n<2^{m}$. Let $S$ be the sum of all numbers from $1$ to $n$ that are neither powers of $2$ nor $3$ times powers of $2$. Then $$ S\le \frac{n(n+1)}{2}-2^{m}+1\le(2^m-1)(2^{m-1}-1)<2^{2m-1},$$ So $S$ is at most a $2m-1$-bit number. By picking $2^m$ as coefficient for each non-power-of-2 summand, these therefore contribute $2^mS$. But we still have $m$ summands of the form $2^k$ and at least $m-1$ sumands of the form $3\cdot 2^k$. We will use these wisely:

Suppose our sum so far ends in $r$ binary zeroes, i.e., it is $2^r s$ where $s$ is odd (and $r\ge m$).

If $s\equiv 1\pmod 4$ and we still have a number $3\cdot2^k$ available, we can pick a suitable power of two as coefficient to arrive at $$ 2^rs+3\cdot 2^{r-k}\cdot 3\cdot 2^k=2^r(s+3)=2^{r+2}\cdot\frac{s+3}4,$$ a number with at least two more zeroes at the end and not more bits in total (unless we had $s=1$ and thus now have $2^{r+2}$, a power of two, as sum).

On the other hand, if $s\equiv 3\pmod 4$ and we still have a number $2^k$ available, w can achieve $$2^rs+2^{r-k}\cdot 2^k=2^{r+2}\cdot\frac{s+1}4$$ as next sum, again a number with at least two more zeroes at the end an not more bits in total (unless we now have a power of two).

Thus we can at least $m-1$ times (namely, as long as we can guarantee to still have a $2^k$ or $3\cdot 2^k$ available, whatever is needed) increase the number of trailing $0$ digits by two without increasing the bit length of the sum (unless we achieve a power of $2$ as sum). As $S$ has at most $2m-1$ bits, this means that we can zero $2(m-1)$ of these, i.e., we can arrive at a power of $2$, possibly without even using all our summands. The remaining sumamnds, if any, can be added one by one without destroying the power-of-two-ness of the sum. So finally we arrive at a sum that is a power of $2$.