Question 1

The sum of consecutive numbers can be explored using a general formula.

Step 1 — Understanding the Formula

Let the first number in a sequence of consecutive integers be $a$. Let the number of terms in the sequence be $n$. The sequence of numbers is $a, a+1, a+2, \dots, a+(n-1)$. The sum $S$ of these $n$ consecutive integers is given by the formula for an arithmetic progression: $S = \frac{n}{2}(\text{first term} + \text{last term})$ $S = \frac{n}{2}(a + a + n - 1)$ $S = \frac{n}{2}(2a + n - 1)$ To make it easier to work with, we can multiply both sides by 2: $2S = n(2a + n - 1)$ Let's call the second factor $Y$. So, $Y = 2a + n - 1$. Now we have $2S = nY$. Consider the difference between $Y$ and $n$: $Y - n = (2a + n - 1) - n$ $Y - n = 2a - 1$ Since $a$ is an integer, $2a$ is always an even integer. So, $2a - 1$ is always an odd integer. This means that $Y$ and $n$ must have different parities (one must be odd and the other must be even).

Step 2 — Can every natural number be written as a sum of consecutive numbers?

Anshu is exploring sums of consecutive numbers. The examples given use natural numbers (positive integers) and involve sums of two or more terms ($n \ge 2$). Let's assume this context. We know from Step 1 that $2S = nY$, where $n$ and $Y$ have different parities. Consider natural numbers that are powers of 2, such as $1, 2, 4, 8, 16, \dots$. Let $S = 2^k$ for some integer $k \ge 0$. Then $2S = 2 \cdot 2^k = 2^{k+1}$. The factors of $2^{k+1}$ are $1, 2, 4, \dots, 2^{k+1}$. All these factors are even, except for $1$. For $n$ and $Y$ to have different parities, one of them must be 1.

• Case 1: $n=1$. This means the "sum" is just the number $a$ itself. For example, $7=7$. This is not a sum of two or more terms.
• Case 2: $Y=1$. This means $2a+n-1=1$. Since we are looking for sums of natural numbers, $a \ge 1$. So $2a \ge 2$. Since we are looking for sums of two or more terms, $n \ge 2$. So $n-1 \ge 1$. Therefore, $2a+n-1 \ge 2+1 = 3$. This contradicts $Y=1$. So $Y$ cannot be 1. Since neither $n=1$ nor $Y=1$ leads to a valid sum of two or more consecutive natural numbers for powers of 2, we conclude that numbers that are powers of 2 cannot be written as a sum of two or more consecutive natural numbers. For example, 4 cannot be written as $a+(a+1)$ or $a+(a+1)+(a+2)$, etc., using natural numbers.

So, the answer to Anshu's first question is no.

Step 3 — Which numbers can be written in more than one way?

A natural number $S$ can be written as a sum of two or more consecutive natural numbers if and only if $S$ is not a power of 2. For $S$ to be written in more than one way, there must be multiple distinct pairs of $(n, Y)$ factors for $2S$ that satisfy the conditions:

1. $n \ge 2$ (sum of two or more terms)
2. $a \ge 1$ (terms are natural numbers), which implies $Y > n$ (since $Y-n = 2a-1 \ge 1$)
3. $n$ and $Y$ have different parities.

The number of ways to write $S$ as a sum of two or more consecutive natural numbers is equal to the number of odd factors of $S$ (including 1), minus 1. Let's call this $\tau_o(S)-1$. For $S$ to be written in more than one way, we need $\tau_o(S)-1 > 1$, which means $\tau_o(S) > 2$. This means $S$ must have at least three odd factors (including 1). If we write $S = 2^k \cdot M$, where $M$ is the odd part of $S$, then $\tau_o(S)$ is the number of factors of $M$. So, $M$ must have more than two factors. This means $M$ must be a composite number (not 1 and not a prime number). For example, $S=9$. Its odd part is $M=9$. The factors of 9 are $1, 3, 9$. Since $9$ is a composite number, $S=9$ can be written in $\tau_o(9)-1 = 3-1 = \textbf{2 ways}$: $9 = 4+5$ $9 = 2+3+4$ Another example is $S=15$. Its odd part is $M=15$. The factors of 15 are $1, 3, 5, 15$. Since $15$ is a composite number, $S=15$ can be written in $\tau_o(15)-1 = 4-1 = \textbf{3 ways}$: $15 = 7+8$ $15 = 4+5+6$ $15 = 1+2+3+4+5$ So, numbers whose odd part is a composite number can be written as a sum of consecutive natural numbers in more than one way.

Step 4 — Odd numbers as sum of two consecutive numbers; Even numbers as sum of consecutive numbers?

• "Ohh, I know all odd numbers can be written as a sum of two consecutive numbers." Let an odd number be $N$. We want to write $N = a + (a+1)$. $N = 2a + 1$ Solving for $a$: $2a = N - 1$ $a = \frac{N-1}{2}$ For $a$ to be a natural number (i.e., $a \ge 1$), we need $\frac{N-1}{2} \ge 1$. This means $N-1 \ge 2$, so $N \ge 3$. Therefore, all odd numbers greater than 1 can be written as a sum of two consecutive natural numbers. For example, $3 = 1+2$, $5 = 2+3$, $7 = 3+4$. The number $1$ cannot be written this way using natural numbers.

• "Can we write all even numbers as a sum of consecutive numbers?" From Step 2, we found that powers of 2 (e.g., $2, 4, 8, 16, \dots$) cannot be written as a sum of two or more consecutive natural numbers. Since these are even numbers, the answer is no. Not all even numbers can be written as a sum of two or more consecutive natural numbers. However, any even number that is not a power of 2 can be written as a sum of consecutive natural numbers. For example, $6 = 1+2+3$, $10 = 1+2+3+4$, $12 = 3+4+5$.

Step 5 — Can 0 be written as a sum of consecutive numbers using negative numbers?

Anshu asks if 0 can be written as a sum of consecutive numbers if negative numbers are allowed. If we allow the terms to be any integers (positive, negative, or zero), then yes, 0 can be written as a sum of consecutive integers. Using our formula $2S = n(2a+n-1)$, if $S=0$: $0 = n(2a+n-1)$ Since $n$ (the number of terms) must be positive, $n \ne 0$. So, we must have $2a+n-1=0$. $2a = 1-n$ $a = \frac{1-n}{2}$ For $a$ to be an integer, $1-n$ must be an even number. This means $n$ must be an odd number.

• If $n=1$, then $a=(1-1)/2 = 0$. The sum is just $\textbf{0}$.
• If $n=3$, then $a=(1-3)/2 = -1$. The sum is $(-1) + 0 + 1 = \textbf{0}$.
• If $n=5$, then $a=(1-5)/2 = -2$. The sum is $(-2) + (-1) + 0 + 1 + 2 = \textbf{0}$. So, yes, 0 can be written as a sum of consecutive integers in infinitely many ways, using any odd number of terms $n \ge 1$.

Step 6 — Other explorations: Can every integer be written as a sum of consecutive integers (allowing negative terms)?

If we allow the terms to be any integers (positive, negative, or zero), then every integer can be written as a sum of consecutive integers. The condition from Step 1 that $n$ and $Y=2a+n-1$ must have different parities still holds.

• For any natural number $S$ (including powers of 2): Let $S$ be any natural number. We can choose $n=2S$ (which is an even number). Then $Y = 2S/n = 2S/(2S) = 1$. Since $n=2S$ is even and $Y=1$ is odd, they have different parities. Now we find $a$: $a = \frac{Y-n+1}{2} = \frac{1-2S+1}{2} = \frac{2-2S}{2} = 1-S$ So, any natural number $S$ can be written as a sum of $n=2S$ consecutive integers starting from $a=1-S$. For example, $S=4$. We choose $n=2(4)=8$. Then $a=1-4=-3$. The sum is $(-3)+(-2)+(-1)+0+1+2+3+4 = \textbf{4}$.
• For $S=0$: We showed in Step 5 that 0 can be written as a sum of consecutive integers.
• For any negative integer $S$: Let $S = -X$, where $X$ is a natural number. We can choose $n=2X$ (which is an even number). Then $Y = 2S/n = 2(-X)/(2X) = -1$. Since $n=2X$ is even and $Y=-1$ is odd, they have different parities. Now we find $a$: $a = \frac{Y-n+1}{2} = \frac{-1-2X+1}{2} = \frac{-2X}{2} = -X$ So, any negative integer $S=-X$ can be written as a sum of $n=2X$ consecutive integers starting from $a=-X$. For example, $S=-4$. We choose $n=2(4)=8$. Then $a=-4$. The sum is $(-4)+(-3)+(-2)+(-1)+0+1+2+3 = \textbf{-4}$. Therefore, every integer (positive, negative, or zero) can be written as a sum of consecutive integers if negative terms are allowed.

Answer

(i) No, not every natural number can be written as a sum of two or more consecutive natural numbers. Powers of 2 (like 1, 2, 4, 8, 16, ...) cannot be written in this way. (ii) Numbers whose odd part is a composite number can be written as a sum of consecutive natural numbers in more than one way. Examples include 9, 15, 18, 21, 25. (iii) All odd numbers greater than 1 can be written as a sum of two consecutive natural numbers. No, not all even numbers can be written as a sum of two or more consecutive natural numbers; specifically, powers of 2 cannot. (iv) Yes, 0 can be written as a sum of consecutive integers if negative numbers are allowed. For example, $\textbf{0 = (-1) + 0 + 1}$. (v) If negative numbers are allowed, then every integer (positive, negative, or zero) can be written as a sum of consecutive integers.