# Solving Math Problems About Consecutive Integers

Consecutive integer problems are math puzzles that can be solved with simple algebra. They are usually expressed along the lines of "Four consecutive whole numbers have a sum of 70, what are the four numbers?" or "Three consecutive odd numbers have a product of 693, what are the three numbers?" At first glance they may seem challenging or even impossible to figure out because you have to solve for several numbers rather than just one. You might attempt the first problem by setting up the equation A + B + C + D = 70, or the second problem by setting up the equation ABC = 693. But where do you go from there?

Fortunately, in consecutive integer problems every number in the solution set can be expressed in terms of the smallest number. This reduces every problem to a single-variable algebra problem. For example, if you are given a statement about five consecutive integers and you let the smallest be N, then the five numbers are

N, N+1, N+2, N+3, and N+4

This is because consecutive integers differ by 1, so to get the next you simply add 1 to the previous. Consecutive even integers differ by 2, thus, if you are given a problem about six consecutive even numbers you can call them

N, N+2, N+4, N+6, N+8, and N+10

Consecutive odd numbers also differ by 2, so the list above also works for puzzles about odd numbers. This tutorial shows how to solve a variety of consecutive integer puzzles with six examples of varying difficulty.

## Problem 1: Easy

The sum of five consecutive odd integers is 245. What are the five numbers?

**Solution:** The basic principle in solving consecutive integer problems is to let the smallest number be N. If they are all odd, then each consecutive pair differs by 2. Thus, the five odd integers are N, N+2, N+4, N+6, and N+8. The sum of the five odd numbers is

N + (N+2) + (N+4) + (N+6) + (N+8) = 5N + 20

Now we solve 5N + 20 = 245. This gives us N = 45, which means the smallest of the five odd number is 45. The full set is 45, 47, 49, 51, and 53. You can double check that 45 + 47 + 49 + 51 = 53 = 245.

## Problem 2: Medium

The sum of some consecutive positive integers is 42. If there are at least three positive integers, how many could there be?

**Solution:** Let's say that the sum of K positive integers is 42. We want to find all the values of K for which this could be true. There could be more than one solution. This problem can be solved systematically by testing values of K starting with K = 3 and continuing on with K = 4, K = 5, etc. If for some value of K you cannot find K consecutive integers that add up to 42, then that value of K can be discarded. As in the previous problem, we will call the smallest of the consecutive integers N.

- With K = 3, the three consecutive integers are N, N+1, and N+2. Their sum is 3N + 3. The equation 3N + 3 = 42 works out to N = 13, so this case yields a solution.
- With K = 4, the four consecutive integers are N, N+1, N+2, and N+3. Their sum is 4N + 6. The equation 4N + 6 = 42 works out to N = 9, so this case also yields a solution.
- With K = 5, the five consecutive integers are N, N+1, N+2, N+3, and N+4. Their sum is 5N + 10. The equation 5N + 10 = 42 does not have an integer solution, so K = 5 is discarded.
- With K = 6, the six consecutive integers are N, N+1, N+2, N+3, N+4, and N+5. Their sum is 6N + 15. The equation 6N + 15 = 42 does not have an integer solution, so K = 6 is discarded as well.
- With K = 7, the seven consecutive integers are N, N+1, N+2, N+3, N+4, N+5, and N+6. Their sum is 7N + 21. The equation 7N + 21 = 42 works out to N = 3, so this case yields a solution.
- With K = 8, the eight consecutive integers are N, N+1, N+2, N+3, N+4, N+5, N+6, and N+7. Their sum is 8N + 28. The equation 8N + 28 = 42 does not have an integer solution, so K = 8 can be discarded.

Going higher than K = 8 you will not find any more solutions. Therefore, if a set of some consecutive integers has a sum of 42, then there could be either three, four, or seven integers in that set:

- 13 + 14 + 15 = 42
- 9 + 10 + 11 + 12 = 42
- 3 + 4 + 5 + 6 + 7 + 8 + 9 = 42

## Problem 3: Medium

The sum of eight consecutive positive integers is 10 times the smallest of them. What are the eight numbers?

**Solution:** Let the eight consecutive integers be N, N+1, N+2, N+3, N+4, N+5, N+6, and N+7. Their sum is 8N + 28. The equation 8N + 28 = 10N works out to

8N + 28 = 10N

28 = 2N

14 = N.

Therefore, the eight consecutive numbers are 14, 15, 16, 17, 18, 19, 20, and 21. You can check that their sum is 140, which is 10 times the smallest number 14.

## Problem 4: Hard

The sum of the squares of six consecutive positive even integers is 8284. What are the six numbers?

**Solution:** If the smallest of the six is N, then the next five are N+2, N+4, N+6, N+8, and N+10. Remember, even integers go up by 2s. The sum of their squares is

N^2 + (N+2)^2 + (N+4)^2 +(N+6)^2 + (N+8)^2 +(N+10)^2

= 6N^2 + 60N + 220

Setting this sum equal to 8284 gives us a quadratic equation that we can solve using the quadratic formula.

6N^2 + 60N + 220 = 8284

6N^2 + 60N - 8064 = 0

6(N^2 + 10N - 1344) = 0

N^2 + 10N - 1344 = 0

N = [-10 ± sqrt(100 + 4*1344)]/2

N = [-10 ± 74]/2

N = 32 or -42

Since the problem specifies that the integers are positive, we can discard the solution N = -42. This means N = 32 and the six numbers are 32, 34, 36, 38, 40, and 42. You can double check that if you square these numbers and add the squares you get a sum of 8284.

## Problem 5: Harder

The product of five consecutive positive integers is is 6375600. What are the five numbers?

**Solution:** If the smallest number is N, then the other four are N+1, N+2, N+3, and N+4. Their product is the polynomial

N(N+1)(N+2)(N+3)(N+4) = N^5 + 10N^4 + 35N^3 + 50N^2 + 24N

The problem reduces to solving the quintic equation

N^5 + 10N^4 + 35N^3 + 50N^2 + 24N = 6375600

This is much more complicated than solving a quadratic as in the previous problem, or even a cubic. But, there is a way to narrow down the correct value of N to a small range, from which we can then use "guess-and-check" to pinpoint the solution. Notice that if N is a positive number then the quintic polynomial expression is bounded above and below by two simpler quintics:

N^5 < N(N+1)(N+2)(N+3)(N+4) < (N+4)^5

Since we know that N(N+1)(N+2)(N+3)(N+4) = 6375600, we can simplify this inequality to

N^5 < 6375600 < (N+4)^5

Taking the fifth root of each side gives us

N < 22.9564 < N + 4

This means that the correct value of N is less than 22.9564 and greater than 18.9564. And since N must be an integer, this tells us that N is either 19, 20, 21, or 22. Let's test each possibility starting with 19.

- If N = 19, then the product of the five consecutive integers is 19*20*21*22*23 = 4037880, which is too small.
- If N = 20, then the product of the five consecutive integers is 20*21*22*23*24 = 5100480, which is too small.
- If N = 21, then the product of the five consecutive integers is 21*22*23*24*25 = 6375600, which is just right.

Therefore, the solution is {21, 22, 23, 24, 25}. Note that we could have used guess-and-check from the beginning, but without a good a idea of where to start. This process takes some of the guesswork out of guess-and-check.

## Problem 6: Challenging

Pick any 10 consecutive positive integers, cube each number (i.e., raise each to the third power), and then add them all up. What is the last digit of the number you get? Will the last digit be the same no matter which 10 consecutive integers you pick?

**Solution:** No matter what set of 10 consecutive integers you start with, the sum of their cubes will always be a number that ends in 5. To prove this fact, let's start by calling the smallest number N and working out the expression for the sum of cubes. This is

N^3 + (N+1)^3 + (N+2)^3 + (N+3)^3 + (N+4)^3 + (N+5)^3 + (N+6)^3 + (N+7)^3 + (N+8)^3 + (N+9)^3 = 10N^3 + 135N^2 + 855N + 2025

Notice that the cubic polynomial has coefficients that are all divisible by 5. This means you can rewrite it as

10N^3 + 135N^2 + 855N + 2025 = 5 * (2N^3 + 27N^2 + 171N + 405)

All this proves is that the sum of the cubes of 10 consecutive integers is *divisible by 5*, but not that the last digit of the sum is 5. A number that is divisible by 5 always ends in either 0 or 5. It ends in 0 if the number is even and 5 if the number is odd. Equivalently, 5 times an even number always ends in 0, and 5 times an odd number always ends in 5.

What we need to prove next is that the cubic expression 5 * (2N^3 + 27N^2 + 171N + 405) always represents a number that ends in 5. This is equivalent to proving that 2N^3 + 27N^2 + 171N + 405 is an odd number for any integer value of N. There are two cases to consider: N is an even number and N is an odd number.

- If N is even, then the expression 2N^3 + 27N^2 + 171N + 405 works out to

even + even + even + odd = odd - If N is odd, then the expression 2N^3 + 27N^2 + 171N + 405 works out to

even + odd + odd + odd = odd

So, regardless of the parity of N, the expression 2N^3 + 27N^2 + 171N + 405 always gives an odd integer. And 5 times an odd integer always yields a number whose last digit is 5.

## Comments

Hello, this is problem I have:

Prove that a number L cannot simultaneously be the product of four consecutive numbers and the sum of four consecutive numbers.