Math / Numbers & Operations

Question33. Fill in the blank. The sum of three consecutive counting numbers always has a divisor (other than 1 ) of $\qquad$ . Prove.

Studdy Solution

STEP 1

1. We are dealing with three consecutive counting numbers.
2. We need to find a divisor (other than 1) that always divides the sum of these three numbers.
3. Counting numbers are positive integers starting from 1.

STEP 2

1. Define the three consecutive numbers.
2. Express the sum of these numbers.
3. Simplify the expression for the sum.
4. Identify the divisor and prove it.

STEP 3

Let the three consecutive counting numbers be $n$ , $n+1$ , and $n+2$ .

STEP 4

Express the sum of these three numbers:
$n + (n+1) + (n+2)$

STEP 5

Simplify the expression for the sum:
$n + (n+1) + (n+2) = n + n + 1 + n + 2 = 3n + 3$

STEP 6

Factor the expression:
$3n + 3 = 3(n + 1)$
Since the expression is factored as $3(n + 1)$ , it is clear that the sum $3n + 3$ is always divisible by 3.
The sum of three consecutive counting numbers always has a divisor (other than 1) of:
$\boxed{3}$

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Question33. Fill in the blank. The sum of three consecutive counting numbers always has a divisor (other than 1 ) of $\qquad$ . Prove.

Studdy Solution

STEP 1

1. We are dealing with three consecutive counting numbers.
2. We need to find a divisor (other than 1) that always divides the sum of these three numbers.
3. Counting numbers are positive integers starting from 1.

STEP 2

1. Define the three consecutive numbers.
2. Express the sum of these numbers.
3. Simplify the expression for the sum.
4. Identify the divisor and prove it.

STEP 3

Let the three consecutive counting numbers be $n$ , $n+1$ , and $n+2$ .

STEP 4

Express the sum of these three numbers:
$n + (n+1) + (n+2)$

STEP 5

Simplify the expression for the sum:
$n + (n+1) + (n+2) = n + n + 1 + n + 2 = 3n + 3$

STEP 6

Factor the expression:
$3n + 3 = 3(n + 1)$
Since the expression is factored as $3(n + 1)$ , it is clear that the sum $3n + 3$ is always divisible by 3.
The sum of three consecutive counting numbers always has a divisor (other than 1) of:
$\boxed{3}$

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Question33. Fill in the blank. The sum of three consecutive counting numbers always has a divisor (other than 1 ) of \qquad . Prove.

Studdy Solution

STEP 1

1. We are dealing with three consecutive counting numbers.2. We need to find a divisor (other than 1) that always divides the sum of these three numbers.3. Counting numbers are positive integers starting from 1.

STEP 2

1. Define the three consecutive numbers.2. Express the sum of these numbers.3. Simplify the expression for the sum.4. Identify the divisor and prove it.

STEP 3

Let the three consecutive counting numbers be n n n, n+1 n+1 n+1, and n+2 n+2 n+2.

STEP 4

Express the sum of these three numbers:n+(n+1)+(n+2) n + (n+1) + (n+2) n+(n+1)+(n+2)

STEP 5

Simplify the expression for the sum:n+(n+1)+(n+2)=n+n+1+n+2=3n+3 n + (n+1) + (n+2) = n + n + 1 + n + 2 = 3n + 3 n+(n+1)+(n+2)=n+n+1+n+2=3n+3

STEP 6

Factor the expression:3n+3=3(n+1) 3n + 3 = 3(n + 1) 3n+3=3(n+1)Since the expression is factored as 3(n+1) 3(n + 1) 3(n+1), it is clear that the sum 3n+3 3n + 3 3n+3 is always divisible by 3. The sum of three consecutive counting numbers always has a divisor (other than 1) of:3\boxed{3}3​

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Question33. Fill in the blank. The sum of three consecutive counting numbers always has a divisor (other than 1 ) of $\qquad$ . Prove.

1. We are dealing with three consecutive counting numbers.
2. We need to find a divisor (other than 1) that always divides the sum of these three numbers.
3. Counting numbers are positive integers starting from 1.

1. Define the three consecutive numbers.
2. Express the sum of these numbers.
3. Simplify the expression for the sum.
4. Identify the divisor and prove it.

Let the three consecutive counting numbers be $n$ , $n+1$ , and $n+2$ .

Express the sum of these three numbers:
$n + (n+1) + (n+2)$

Simplify the expression for the sum:
$n + (n+1) + (n+2) = n + n + 1 + n + 2 = 3n + 3$

Factor the expression:
$3n + 3 = 3(n + 1)$
Since the expression is factored as $3(n + 1)$ , it is clear that the sum $3n + 3$ is always divisible by 3.
The sum of three consecutive counting numbers always has a divisor (other than 1) of:
$\boxed{3}$