Lesson Problems with consecutive integer numbers; odd even consecutive integer numbers

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Problems with consecutive integer numbers; odd/even consecutive integer numbers


Problem 1

The sum of three consecutive integers is  240.  Find the integers.

Solution 1

This is a very simple problem.
Let  x  be the second  (the middle)  of these three consecutive integers.
Then the first one is  x-1,  while the third one is  x%2B1.
Since the sum of these three consecutive integers is equal to  240,  you can write the equation
%28x-1%29+%2B+x+%2B+%28x%2B1%29+=+240.

Simplify this equation.  You will get
3x+=+240   (after combining like terms at the left side),
x+=+80      (after dividing both sides by 3).

Thus,  the three consecutive integers are  79,  80  and  81.
You can easily  check  this solution:
79+%2B+80+%2B+81+=+240.

Solution 2

This time let  x  be the first  (the smallest)  of these three consecutive integers.
Then the second one is  x%2B1,  while the third one is  x%2B2.
Since the sum of these three consecutive integers is equal to  240,  you can write the equation
x+%2B+%28x%2B1%29+%2B+%28x%2B2%29+=+240.

Simplify this equation step by step.  You will get
3x+%2B+3+=+240   (after combining like terms at the left side),
3x+=+237       (after moving the constant term  3  to the right side with the opposite sign and combining like terms at the right side),
x+=+79          (after dividing both sides by  3).

Thus,  the solution is the same:  the three consecutive integers are  79,  80  and  81.

Answer.  The three consecutive integers are  79,  80  and  81.


Problem 2

One third of the sum of five consecutive integers is  15.  Find the integers.

Solution

Again,  this is a simple problem.
Since one third of the sum of five consecutive integers is  15,  the entire sum of five consecutive integers is  45.
Let  x  be the third  (the middle)  of these five consecutive integers.
Then the first integer is equal to  x-2,
the second integer is equal to       x-1,
the fourth integer is equal to       x%2B1,  and
the fifth integer is equal to         x%2B2.
Since the sum of these five consecutive integers is equal to  45,  you can write the equation
%28x-2%29+%2B+%28x-1%29+%2B+x+%2B+%28x%2B1%29+%2B+%28x%2B2%29+=+45.

Simplify this equation.  You will get
5x+=+45   (after combining like terms at the left side),
x+=+9      (after dividing both sides by  5).

Thus,  the five consecutive integers are  7,  8,  9,  10  and  11.

You can easily  check  that
7+%2B++9+%2B+10+%2B+11+=+45,
hence one third of the sum is  9.  The solution is correct.

Answer.  The five consecutive integers are  7,  8,  9,  10  and  11.


Problem 3

The sum of three consecutive even integers is  18.  Find the integers.

Solution

Let  x  be the second  (the middle)  of these three consecutive even integers.
Then the first one is  x-2,  while the third one is  x%2B2.
Since the sum of these three consecutive even integers  x-2,  x  and x%2B2  is equal to  18,  you can write the equation
%28x-2%29+%2B+x+%2B+%28x%2B2%29+=+18.

Simplify this equation.  You will get
3x+=+18   (after combining like terms at the left side),
x+=+6      (after dividing both sides by  3).

Thus,  the three consecutive even integers are  6-2=4,  6,  and  6+2=8.
Easy  check  shows that the solution is correct:
4+%2B+6+%2B+8+=+18.

Answer.  The three consecutive even integers are  4,  6 and  8.


Problem 4

The sum of three consecutive odd integers is  27.  Find the integers.

Solution

Let  x  be the second  (the middle)  of these three consecutive odd integers.
Then the first one is  x-2,  while the third one is  x%2B2.
Since the sum of these three consecutive odd integers is equal to  27,  you can write the equation
%28x-2%29+%2B+x+%2B+%28x%2B2%29+=+27.

Simplify this equation.  You will get
3x+=+27   (after combining like terms at the left side),
x+=+9      (after dividing both sides by 3).

Thus,  the three consecutive odd integers are  9-2=7,  9,  and  9+2=11.
Easy  check  shows that the solution is correct:
7+%2B+9+%2B+11+=+27.

Answer.  The three consecutive odd integers are  7,  9  and  11.


Problem 5

The sum of four consecutive odd integers is  40.  Find the integers.

Solution

This time there is no the unique middle term of four consecutive odd integers.
Therefore let us introduce the unknown  x  as the first  (minimal)  of these four consecutive odd integers.
Then the second number is  x%2B2,  the third one is  x%2B4  and the fourth one is  x%2B6.
Since the sum of these four consecutive odd integers is equal to  40,  you can write the equation
x+%2B+%28x%2B2%29+%2B+%28x%2B4%29+%2B+%28x%2B6%29=+40.

Simplify this equation.  You will get
4x+%2B+12+=+40   (after combining like terms at the left side),
4x+=+28         (after moving the constant term  12  to the right side with the opposite sign and combining like terms at the right side),
x+=+7            (after dividing both sides by  4).

Thus,  the three consecutive odd integers are  7,  7+2=9,  9+2=11,  and  11+2=13.
Easy  check  shows that the solution is correct:
7+%2B+9+%2B+11+%2B+13+=+40.

Answer.  The four consecutive odd integers are  7,  9,  11  and  13.

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