QuestionFind three consecutive integers that add up to 345. Let the smallest be $n$ , so the integers are $n$ , $n+1$ , $n+2$ .

Studdy Solution

STEP 1

Assumptions1. The three integers are consecutive. . The sum of the three integers is345.
3. If $n$ is the smallest integer, then the other two integers are $n+1$ and $n+$ .

STEP 2

We can write the sum of the three consecutive integers as an equation.
$n + (n+1) + (n+2) =345$

STEP 3

implify the left side of the equation by combining like terms.
$3n +3 =345$

STEP 4

Subtract3 from both sides of the equation to isolate the term with $n$ .
$3n =345 -3$

STEP 5

Calculate the right side of the equation.
$3n =342$

STEP 6

Divide both sides of the equation by3 to solve for $n$ .
$n =342 /3$

STEP 7

Calculate the value of $n$ .
$n =342 /3 =114$

STEP 8

Now that we have the value of $n$ , we can find the other two integers by adding1 and2 to $n$ .
$n+1 =114 +1$ $n+2 =114 +2$

STEP 9

Calculate the values of $n+$ and $n+2$ .
$n+ =114 + =115$ $n+2 =114 +2 =116$ The three consecutive integers are114,115, and116.

Was this helpful?

QuestionFind three consecutive integers that add up to 345. Let the smallest be $n$ , so the integers are $n$ , $n+1$ , $n+2$ .

Studdy Solution

STEP 1

Assumptions1. The three integers are consecutive. . The sum of the three integers is345.
3. If $n$ is the smallest integer, then the other two integers are $n+1$ and $n+$ .

STEP 2

We can write the sum of the three consecutive integers as an equation.
$n + (n+1) + (n+2) =345$

STEP 3

implify the left side of the equation by combining like terms.
$3n +3 =345$

STEP 4

Subtract3 from both sides of the equation to isolate the term with $n$ .
$3n =345 -3$

STEP 5

Calculate the right side of the equation.
$3n =342$

STEP 6

Divide both sides of the equation by3 to solve for $n$ .
$n =342 /3$

STEP 7

Calculate the value of $n$ .
$n =342 /3 =114$

STEP 8

Now that we have the value of $n$ , we can find the other two integers by adding1 and2 to $n$ .
$n+1 =114 +1$ $n+2 =114 +2$

STEP 9

Calculate the values of $n+$ and $n+2$ .
$n+ =114 + =115$ $n+2 =114 +2 =116$ The three consecutive integers are114,115, and116.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind three consecutive integers that add up to 345. Let the smallest be nnn, so the integers are nnn, n+1n+1n+1, n+2n+2n+2.

Studdy Solution

STEP 1

Assumptions1. The three integers are consecutive. . The sum of the three integers is345.3. If nnn is the smallest integer, then the other two integers are n+1n+1n+1 and n+n+n+.

STEP 2

We can write the sum of the three consecutive integers as an equation.n+(n+1)+(n+2)=345n + (n+1) + (n+2) =345n+(n+1)+(n+2)=345

STEP 3

implify the left side of the equation by combining like terms.3n+3=3453n +3 =3453n+3=345

STEP 4

Subtract3 from both sides of the equation to isolate the term with nnn.3n=345−33n =345 -33n=345−3

STEP 5

Calculate the right side of the equation.3n=3423n =3423n=342

STEP 6

Divide both sides of the equation by3 to solve for nnn.n=342/3n =342 /3n=342/3

STEP 7

Calculate the value of nnn.n=342/3=114n =342 /3 =114n=342/3=114

STEP 8

Now that we have the value of nnn, we can find the other two integers by adding1 and2 to nnn.n+1=114+1n+1 =114 +1n+1=114+1n+2=114+2n+2 =114 +2n+2=114+2

STEP 9

Calculate the values of n+n+n+ and n+2n+2n+2.n+=114+=115n+ =114 + =115n+=114+=115n+2=114+2=116n+2 =114 +2 =116n+2=114+2=116The three consecutive integers are114,115, and116.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind three consecutive integers that add up to 345. Let the smallest be $n$ , so the integers are $n$ , $n+1$ , $n+2$ .

Assumptions1. The three integers are consecutive. . The sum of the three integers is345.
3. If $n$ is the smallest integer, then the other two integers are $n+1$ and $n+$ .

We can write the sum of the three consecutive integers as an equation.
$n + (n+1) + (n+2) =345$

implify the left side of the equation by combining like terms.
$3n +3 =345$

Subtract3 from both sides of the equation to isolate the term with $n$ .
$3n =345 -3$

Calculate the right side of the equation.
$3n =342$

Divide both sides of the equation by3 to solve for $n$ .
$n =342 /3$

Calculate the value of $n$ .
$n =342 /3 =114$

Now that we have the value of $n$ , we can find the other two integers by adding1 and2 to $n$ .
$n+1 =114 +1$ $n+2 =114 +2$

Calculate the values of $n+$ and $n+2$ .
$n+ =114 + =115$ $n+2 =114 +2 =116$ The three consecutive integers are114,115, and116.