Question 1

Step 1 · Analyze Powers of 2

Let $S = 2^k$ (a power of $2$) for an integer $k \ge 0$.

$2S = 2^{k+1}$

The only factors of $2^{k+1}$ are powers of $2$, so its only odd factor is $1$.

Since $n$ and $Y = 2a + n - 1$ must have different parities, one of them must equal $1$:

• Case 1 ($n = 1$): This represents a single term ($a$), not a sum of two or more consecutive numbers.
• Case 2 ($Y = 1$): For natural numbers, $a \ge 1$ and $n \ge 2$: $Y = 2a + n - 1 \ge 2(1) + 2 - 1 = 3 \ne 1$

Thus, powers of $2$ ($1, 2, 4, 8, 16, \dots$) cannot be written as a sum of two or more consecutive natural numbers.

Step 1 · Determine Condition on Odd Factors

The number of ways to write a natural number $S$ as a sum of two or more consecutive natural numbers equals: $\tau_o(S) - 1$ where $\tau_o(S)$ is the total number of odd factors of $S$ (including $1$).

For more than one way, we require: $\tau_o(S) - 1 > 1 \implies \tau_o(S) > 2$

This means the odd part $M$ of $S$ (where $S = 2^k \cdot M$) must have more than $2$ factors, so $M$ must be an odd composite number.

Examples:

• For $S = 9$ (odd factors: $1, 3, 9 \implies 3 - 1 = 2\text{ ways}$): $9 = 4 + 5$ $9 = 2 + 3 + 4$
• For $S = 15$ (odd factors: $1, 3, 5, 15 \implies 4 - 1 = 3\text{ ways}$): $15 = 7 + 8$ $15 = 4 + 5 + 6$ $15 = 1 + 2 + 3 + 4 + 5$

Step 1 · Express Odd and Even Numbers

For Odd Numbers: Let $N$ be an odd natural number written as the sum of two consecutive integers: $N = a + (a + 1) = 2a + 1$ $a = \dfrac{N - 1}{2}$ For $a \ge 1$, we need $N - 1 \ge 2 \implies N \ge 3$. Thus, every odd number greater than $1$ can be expressed as a sum of two consecutive natural numbers (e.g., $3 = 1+2$, $5 = 2+3$, $7 = 3+4$).

For Even Numbers: Even numbers that are powers of $2$ ($2, 4, 8, 16, \dots$) have no odd factors other than $1$, so they cannot be written as a sum of consecutive natural numbers. Even numbers with an odd factor $> 1$ can be expressed (e.g., $6 = 1 + 2 + 3$, $10 = 1 + 2 + 3 + 4$, $12 = 3 + 4 + 5$).

Step 1 · Set $S = 0$ in the General Formula

Using $2S = n(2a + n - 1)$ with $S = 0$: $n(2a + n - 1) = 0$

Since $n > 0$, we have: $2a + n - 1 = 0 \implies a = \dfrac{1 - n}{2}$

For $a$ to be an integer, $n$ must be an odd integer:

• For $n = 3$: $a = \dfrac{1 - 3}{2} = -1$ $(-1) + 0 + 1 = 0$
• For $n = 5$: $a = \dfrac{1 - 5}{2} = -2$ $(-2) + (-1) + 0 + 1 + 2 = 0$

Step 1 · Generalize for Any Integer

Allowing negative terms and zero:

• For any positive integer $S$: Choose $n = 2S$ terms starting at $a = 1 - S$: $Y = \dfrac{2S}{2S} = 1 \implies a = \dfrac{1 - 2S + 1}{2} = 1 - S$ For example, $S = 4$: $n = 8$, $a = -3 \implies (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 = 4$.

• For $S = 0$: Valid for any odd number of terms centered at $0$.

• For any negative integer $S = -X$ ($X > 0$): Choose $n = 2X$ terms starting at $a = -X$: For example, $S = -4$: $n = 8$, $a = -4 \implies (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 = -4$.