Math Problems and Solutions on Integers

Problems related to integer numbers in mathematics are presented along with their solutions.

Problem 1: Find two consecutive integers whose sum is equal 129.

Solution to Problem 1:
Let x and x + 1 (consecutive integers differ by 1) be the two numbers. Use the fact that their sum is equal to 129 to write the equation
x + (x + 1) = 129
Solve for x to obtain
x = 64
The two numbers are
x = 64 and x + 1 = 65
We can see that the sum of the two numbers is 129.

Problem 2: Find three consecutive integers whose sum is equal to 366.

Solution to Problem 2:
Let the three numbers be x, x + 1 and x + 2. their sum is equal to 366, hence
x + (x + 1) + (x + 2) = 366
Solve for x and find the three numbers
x = 121 , x + 1 = 122 and x + 2 = 123

Problem 3: The sum of three consecutive even integers is equal to 84. Find the numbers.

Solution to Problem 3:
The difference between two even integers is equal to 2. let x, x + 2 and x + 4 be the three numbers. Their sum is equal to 84, hence
x + (x + 2) + (x + 4) = 84
Solve for x and find the three numbers
x = 26 , x + 2 = 28 and x + 4 = 30
The three numbers are even. Check that their sum is equal to 84.

Problem 4: The sum on an odd integer and twice its consecutive is equal to equal to 3757. Find the number.

Solution to Problem 4:
The difference between two odd integers is equal to 2. let x be an odd integer and x + 2 be its consecutive. The sum of x and twice its consecutive is equal to 3757 gives an equation of the form
x + 2(x + 2) = 3757
Solve for x
x = 1251
Check that the sum of 1251 and 2(1251 + 2) is equal to 3757.

Problem 5: The sum of the first and third of three consecutive odd integers is 131 less than three times the second integer. Find the three integers.

Solution to Problem 5:
Let x, x + 2 and x + 4 be three integers. The sum of the first x and third x + 4 is given by
x + (x + 4)
131 less than three times the second 3(x + 2) is given by
3(x + 2) - 131
"The sum of the first and third is 131 less than three times the second" gives
x + (x + 4) = 3(x + 2) - 131
Solve for x and find all three numbers
x = 129 , x + 2 = 131 , x + 4 = 133

As an exercise, check that the sum of the first and third is 131 less than three times

Problem 6: The product of two consecutive odd integers is equal to 675. Find the two integers.


Let x, x + 2 be the two integers. Their product is equak to 144
x (x + 2) = 675
Expand to obtain a quadratic equation.
x
2 + 2 x - 675 = 0
Solve for x to obtain two solutions
x = 25 or x = -27
if x = 25 then x + 2 = 27
if x = -27 then x + 2 = -25
We have two solutions. The two numbers are either
25 and 27

or
-27 and -25
Check that in both cases the product is equal to 675.

Problem 7: Find four consecutive even integers so that the sum of the first two added to twice the sum of the last two is equal to 742.

Solution to Problem 7:
Let x, x + 2, x + 4 and x + 6 be the four integers. The sum of the first two
x + (x + 2)
twice the sum of the last two is written as
2 ((x + 4) + (x + 6)) = 4 x + 20
sum of the first two added to twice the sum of the last two is equal to 742 is written as
x + (x + 2) + 4 x + 20 = 742
Solve for x and find all four numbers
x = 120 , x + 2 = 122 , x + 4 = 124 , x + 6 = 126
As an exrcise, check that the sum of the first two added to twice the sum of the last two is equal to 742

Problem 8: When the smallest of three consecutive odd integers is added to four times the largest, it produces a result 729 more than four times the middle integer. Find the numbers and check your answer.

Solution to Problem 8:
Let x, x + 2 and x + 4 be the three integers. "The smallest added to four times the largest is written as follows"
x + 4 (x + 4)
"729 more than four times the middle integer" is written as follows
729 + 4 (x + 2)
"When the smallest is added to four times the largest, it produces a result 729 more than four times the middle" is written as follows
x + 4 (x + 4) = 729 + 4 (x + 2)
Solve for x and find all three numbers
x + 4 x + 16 = 729 + 4 x + 8
x = 721
x + 2 = 723
x + 4 = 725
Check: the smallest is added to four times the largest
721 + 4 * 725 = 3621
four times the middle
4 723 = 2892
3621 is more than 2892 by
3621 - 2892 = 729

The answer to the problem is correct.

More math problems with detailed solutions in this site.