Reverse Consecutive Integers Product Calculator

Use the Reverse Consecutive Integers Product Calculator to find consecutive integers, odd integers, or even integers that multiply to a specified product. This is the reverse calculation of the product of consecutive integers.

Okay, from the beginning, let’s review three simple questions:

• Question 1: The product of two consecutive integers is 306, find these integers.
• Question 2: Find three consecutive integers whose product is 120.
• Question 3: The product of four consecutive integers is 840 find the numbers.

Stop and think, how do you solve such problems? The process is simple or complex? Have you thought of other solutions?

Of course, if you are worried about these problems, then congratulations, you have come to the right place. We will introduce three solutions in this article, from common to special, from complex to simple. Are you a little excited? Don’t worry, let’s introduce them one by one.

This is also the most common. Just assume that the first integer is N, then the next consecutive integers are N + 1, N + 2, N + 3, N + 4…

Then the three problems mentioned above can be expressed by equations

N * (N + 1) = 306

N * (N + 1) * (N + 2) = 120

N * (N + 1) * (N + 2) * (N + 3) = 840

Now solve these 3 equations one by one.

Question 1: The product of two consecutive integers is 306, find these integers.

N * (N + 1) = 306

N² + N = 306

N² + N – 306 = 0

(N – 17) * (N + 18) = 0

(N – 17) = 0 or (N + 18)=0

N =17 or N = -18

So the answer to question 1 are 17, 18 or -18, -17.

Question 2: Find three consecutive integers whose product is 120.

N * (N + 1) * (N + 2) = 120

(N² + N) * (N + 2) = 120

N³ + 2 * N² + N² + 2*N = 120

N³ + 3 * N² + 2 * N = 120

N³ + 3 * N² + 2 * N -120 = 0

(N – 4) * (N² + 7 * N + 30) = 0

(N – 4) = 0 or (N² + 7 * N + 30) = 0

N = 4

The answer is out, the first integer is 4, so the second and third integer is 5 and 6.

Question 3: The product of four consecutive integers is 840 find the numbers.

N * (N + 1) * (N + 2) * (N + 3) = 840

(N² + N) * (N + 2) * (N + 3) = 840

(N³ + 2 * N² + N² + 2 * N) * (N + 3) = 840

(N³ + 3 * N² + 2 * N) * (N + 3) = 840

N⁴ + 3 * N³ + 2 * N² + 3 * N³ + 9 * N² + 6 * N = 840

N⁴ + 6 * N³ + 11 * N² + 6 * N = 840

N⁴ + 6 * N³ + 11 * N² + 6 * N – 840 = 0

(N – 4) * (N³ + 10 * N² + 51 * N + 210) = 0

(N – 4) * (N + 7) * (N² + 3 * N + 30) = 0

(N – 4) = 0 or (N + 7) = 0 or (N² + 3 * N + 30) = 0

N = 4 or N = -7

Thus, the answer to third question are 4, 5, 6,7 or -7, -6, -5, -4.

3 problems have been successfully solved, but have you discovered a feature? As the number of consecutive integers increases, the difficulty of solving equations also increases. 2 consecutive integers are 2-dimensional equations, 3 consecutive integers are 3-dimensional equations, and 4 consecutive integers are 4-dimensional equations. If you want to calculate 6 consecutive integers based on a given product, you must solve 6-dimensional equations. It is certain that this cannot be solved by an ordinary person, even if you are a professional, it is not easy too.

Therefore, this method is only suitable for finding 2 or 3 consecutive integers based on the given product. If you want to calculate 4 or more consecutive integers, you must consider the second method below.

Observing the equations of above three questions, do you find any other features? Yes, the coefficients before the Nth power are all 1.

N² + N = 306

N³ + 3 * N² + 2 * N = 120

N⁴ + 6 * N³ + 11 * N² + 6 * N = 840

If we ignore the addition part on the left side of the equation, we can turn the equations into

N² < 306

N³ < 120

N⁴ < 840

Here, we can use the root calculator to calculate the maximum value of N.

N² < 306

N = 17.49

Rounding down, the largest positive integer of N is 17, then 306 / 17 = 18, which happens to be two consecutive integers that meet the requirements, so the answer to first question are 17 and 18.

Another question: why round down instead of up? In fact, it is very simple. If you round up, N = 18, then 18² = 324 is greater than 306, which does not meet the requirements, so rounding down is correct.

Calculate the second equation in the same way

N³ < 120

N = 4.93

Rounding down, the largest integer of N is 4, so which of these 3 consecutive integers is 4?

The third one, 2 * 3 * 4 = 24, does not meet the requirements!

The second one, 3 * 4 * 5 = 60, doesn’t work either!

The first one, 4 * 5 * 6 = 120, the perfect answer!

So the answer to second question are 4, 5, 6.

Now calculate the third equation

N⁴ < 840

N = 5.38

Similarly, round down, N = 5, number 5 is the first, second, third or fourth integer of 4 consecutive integers?

The first one, 5 * 6 * 7 * 8 = 1680, much larger than 840!

The second one: 4 * 5 * 6 * 7 = 840, the condition is met, it is the consecutive integers we want, so the answer to the third question are 4, 5, 6, 7.

Obviously, this method will be much more convenient than the first method, but there is still a shortcoming. We must use the root calculator to calculate the Nth root. If there is no Nth root calculator, then we can do nothing? Don’t worry, you forgot, the above said there are three methods. Only two are introduced now. Let’s move on to the third method.

Let’s recall what is a divisor first? The divisor is also called the factor. If integer A is divided by integer B, the quotient is just an integer and there is no remainder, then this integer B is called the divisor of A.

For example, the divisors of 6 are 1, 2, 3, and 6. The divisors of 12 are 1, 2, 3, 4, 6, and 12.

After understanding the divisor, we know that the consecutive integers we seek are all divisors of the product. Next, let’s explain what is the method of minimum divisor? The minimum divisor method is to divide the integer by the smallest divisor, then divide the result by the smallest divisor, and repeat this operation until the result obtained has no other divisors except 1 and itself. Note that the minimum divisor here refers to the smallest positive divisor other than 1, because any number divided by 1 is equal to the number itself, which will cause an infinite loop, so we have to exclude the divisor 1. Finally, group the results to find the answer we want.

Now back to the first question above

306

= 2 * 153

=2 * 3 * 51

=2 * 3 * 3 * 17

17 is no other divisor except 1 and itself. Now, we group these divisors [2, 3, 3, 17], it is obvious that 17 should be a separate group. So one integer we get is 17, then the other number is 2 * 3 * 3 = 18, 17 * 18 = 306, the answer is correct.

The second question

120

= 2 * 60

= 2 * 2 * 30

= 2 * 2 * 2 * 15

= 2 * 2 * 2 * 3 * 5

5 is no other divisor except 1 and itself. Now, group these divisors [2, 2, 2, 3, 5], number 5 should be a single. Because 5 is multiplied by any of them, the result will be too big. So 5 is one of 3 consecutive integers, then the other two integers are 2 * 2 = 4, 2 * 3 = 6.

4, 5, 6 are three consecutive integers, and their product is 120, which meets the requirements.

Third question

840

= 2 * 420

= 2 * 2 * 210

= 2 * 2 * 2 * 105

= 2 * 2 * 2 * 3 * 35

= 2 * 2 * 2 * 3 * 5 * 7

Same as above, 7 is one of 4 consecutive integers, then the result is only 4 cases:

• 4, 5, 6, 7
• 5, 6, 7, 8
• 6, 7, 8, 9
• 7, 8, 9, 10

As long as we verify these four cases one by one, we can easily conclude that 4, 5, 6, 7 are the consecutive integers we want.

Is it much simpler than the above two methods? The only disadvantage is that if the result of the product is large, many steps of calculation and grouping are required. But fortunately, each step is a simple division operation, which can be solved by oral arithmetic.

A little trick about grouping. If the product is relatively small, generally the last divisor is one of the required integers. If the product is relatively large, then the first divisor multiplied by the last divisor may be one of the required integers. Or the multiple of the last divisor is one of the required integers.

After the introduction of the three methods, which one do you think is more suitable for you? If you are still not sure, don’t worry. Next, I will introduce a heavyweight method, which is to directly use our online free calculator to find consecutive integers based on the product. Just enter your conditions and the calculator will immediately give the answer. Suitable for friends who don’t have any professional knowledge. Of course, you can also add some special conditions, such as calculating consecutive even or odd numbers.

Now let’s introduce how to use this product-based consecutive integer calculator. It’s very simple, just 4 steps.

Example 1: Find two consecutive integers whose product is 3906.

First, enter 3906 into the first input box. Second, select 2 consecutive integers. Then click Calculate.

Verify: 62 * 63 = 3906.

Example 2: The product of three consecutive odd integers is 4845. Find these integers.

Enter the number 4845 in the first input box, then select 3 consecutive integers and odd integers type, finally click Calculate.

Verify: 15 * 17 * 19 = 4845.

Example 3: The product of four consecutive even integers is 1488384. What are the integers?

First, enter the number 1488384 into the first input box, then select 4 consecutive integers. Third, select the even integers type. Fourth, click Calculate.

Verify: 32 * 34 * 36 * 38 = 1488384

Well, in summary, the first of the three methods above is preconceived and a more commonly used method, but with a high degree of difficulty, suitable for finding two consecutive integers. The second is more convenient, but requires an auxiliary calculator. If there is a Nth root sign calculator, this method is recommended to solve the problem. The third method, purely manual calculation, has a low degree of difficulty. If you do not have any auxiliary tools, this method is recommended.