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 \[ x = 64 \quad \text{and} \quad x + 1 = 65 \] 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 \(x, \; x + 1 \; \text{and} \; x + 2\). Their sum is equal to 366, hence \[ x + (x + 1) + (x + 2) = 366 \] Solve for \(x\) and find the three numbers. \[ x = 121, \quad x + 1 = 122, \quad x + 2 = 123 \]

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 \(x, \; x + 2,\; \text{and}\; x + 4\) be the three numbers. Their sum is equal to 84, hence \[ x + (x + 2) + (x + 4) = 84 \] Now solve for \(x\): \[ 3x + 6 = 84 \quad \Rightarrow \quad 3x = 78 \quad \Rightarrow \quad x = 26 \] So the three numbers are \[ x = 26, \quad x + 2 = 28, \quad x + 4 = 30 \] The three numbers are even. Check: \[ 26 + 28 + 30 = 84 \]

Problem 4:

The sum of an odd integer and twice its consecutive is 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 \(x, \, x + 2, \, x + 4\) 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: \[ x = 129, \quad x + 2 = 131, \quad x + 4 = 133 \]

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

Problem 6:

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

Solution to Problem 6:

Let \(x, \; x + 2\) be the two integers. Their product is equal to 675. \[ x(x + 2) = 675 \] Expand to obtain a quadratic equation: \[ x^2 + 2x - 675 = 0 \] Solve for \(x\) to obtain two solutions: \[ x = 25 \quad \text{or} \quad x = -27 \] If \(x = 25\), then \(x + 2 = 27\).
If \(x = -27\), then \(x + 2 = -25\).
We have two solutions. The two numbers are either: \[ 25 \quad \text{and} \quad 27 \] or \[ -27 \quad \text{and} \quad -25 \] 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 the four consecutive integers be \(x, \; x+2, \; x+4, \; x+6\). The sum of the first two is \[ x + (x+2). \] Twice the sum of the last two is \[ 2\big((x+4) + (x+6)\big) = 4x + 20. \] Hence, the condition that the sum of the first two added to twice the sum of the last two equals 742 is written as \[ x + (x+2) + 4x + 20 = 742. \] Solving for \(x\), we find: \[ x = 120, \quad x+2 = 122, \quad x+4 = 124, \quad x+6 = 126. \] As an exercise, check that the sum of the first two added to twice the sum of the last two is indeed 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 \(x, \; x + 2, \; x + 4\) be the three integers. "The smallest added to four times the largest" is written as: \[ x + 4(x+4) \] "729 more than four times the middle integer" is written as: \[ 729 + 4(x+2) \] So the equation becomes: \[ x + 4(x+4) = 729 + 4(x+2) \] Solve for \(x\): \[ x + 4x + 16 = 729 + 4x + 8 \] \[ x = 721 \] Hence, the three numbers are: \[ x = 721, \quad x+2 = 723, \quad x+4 = 725 \] Check: The smallest added to four times the largest: \[ 721 + 4 \times 725 = 3621 \] Four times the middle: \[ 4 \times 723 = 2892 \] The difference is: \[ 3621 - 2892 = 729 \] Therefore, the answer to the problem is correct.

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