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Math Problems and Solutions on Integers

Problems related to integer numbers 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.


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Updated: 2 April 2013

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