What is The product of any 5 consecutive integers is always divisible by?

Answer 1

Short answer: The product (n) of any 5 consecutive numbers will always be divisible by 120 and its factors.

Long explanation:

n(n+1)(n+2)(n+3)(n+4)

One of these numbers will always be divisible by 5 since they are consecutive. Between two integers that are divisible by 5, 5(n+1) and 5n there are only 4 integers that aren't divisible by 5.

5(n+1)-5n-1 (the -1 is there because we'd be counting one of these two numbers as well)

5n+5-5n-1 (removing the brackets)

5n-5n+5-1 (regrouping the numbers)

0+4 (calculating both groups)

4 (our end result)

As a rule we can state that the amount of integers between two multiples of a is a-1.

a(n+1)-an-1

an+a-an-1

an-an+a-1

a-1

From this it follows that a list of n consecutive integers must always be divisible by any number that is smaller than or equal to n.

So, in our list we have:

• 1 integer that is divisible by 5
• 1 integer that is divisible by 4 (and by 2, but that's not important for what we want to do)
• 1 integer that is divisible by 3
• 1 integer that is divisible by 2, but not by 4. This is very important!

Now, to be find the easiest (which is not necessarily the smallest) number that divides our final product we multiply all of these numbers. It may seem a bit redundant to multiply by 2 while we also multiply by 4, and it is, but we can always make our final answer smaller if we need to.

5*4*3*2=120

Now, every integer that is divisible by 120 is also divisible by any of the prime factors of 120. To get the prime factors of a number you divide it by integers, going up and starting from 2. If the division yields an integer as well you write the divisor (the number you divided by) down as a prime factor and the result of the division as our new dividend (number you divide by). I'll show you how to do it with 120.

120/2=60, which is an integer, so we can add 2 to our list of prime factors.

Prime factors: 2

now we'll start again at 2 and use the result from this division, which would be 60, as our new dividend.

60/2=30, which is an integer, so we can add 2 again to our list of prime factors.

Prime factors: 2,2

30/2=15 (continuing the list)

Prime factors: 2,2,2

15/2=7,5 which is not an integer, so we keep 15 as our dividend and make 3 our new divisor.

15/3=5 which is an integer, so 3 is added to our list of prime factors.

Prime factors: 2,2,2,3

Now we have 5 as our final number, which is a Prime number, so we can add 5 to our list as well (since it is only divisible by 5) and get our complete list of prime factors. Note that prime factoring can be much faster if you only divide by prime numbers, since for example dividing by 4,9 or 10 would be useless since if a number would be divisible by any of those you would have divided it by 2,3 or 5 before.

Prime factors: 2,2,2,3,5

So, the product will be divisible by all of the prime factors 2,2,2,3 and 5 and all multiples. A list of those would be 2,3,4,5,6,8,10,12,15,20,24,30,40,60 and 120.