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=64andx+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+1andx+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,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,andx+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⇒3x=78⇒x=26 So the three numbers are x=26,x+2=28,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,x+2=131,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: x2+2x−675=0 Solve for x to obtain two solutions: x=25orx=−27 If x=25, then x+2=27.
If x=−27, then x+2=−25.
We have two solutions. The two numbers are either: 25and27 or −27and−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((x+4)+(x+6))=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,x+2=122,x+4=124,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,x+2=723,x+4=725 Check: The smallest added to four times the largest: 721+4×725=3621 Four times the middle: 4×723=2892 The difference is: 3621−2892=729 Therefore, the answer to the problem is correct.

More math problems with detailed solutions in this site.