Explain why the product of any three consecutive integers

is divisible by 6

Write out some groups of ‘three consecutive integers’ and try to figure out what they have in common.

since you only want an explanation and not a proof, we could argue as follows

for any two consecutive integers, one has to be even and one has to be odd, so the product of any two has to be divisible by 2
for any 3 consecutive integers, one has to be divisible by 3, (just like for any consecutive four integers, one of them has to be divisible by 4 etc)

any number wich is divisible by 2 AND by 3 is automatically divible by 6

trivial case: if one of the number is zero, then the product of the consecutive integers is also zero, which of course is divisible by 6 with a result of zero.

Since this a problem for you to reason out -- what do you think?

To understand why the product of any three consecutive integers is divisible by 6, let's examine some groups of three consecutive integers:

1, 2, 3
2, 3, 4
3, 4, 5
4, 5, 6
5, 6, 7

Notice that every group of three consecutive integers contains at least one multiple of 2 and one multiple of 3. For example, in the first group (1, 2, 3), 2 is a multiple of 2, and 3 is a multiple of 3. This pattern holds true for every group of three consecutive integers.

If a number is divisible by 2, it means it can be evenly divided by 2 without leaving a remainder. Similarly, if a number is divisible by 3, it means it can be evenly divided by 3 without leaving a remainder.

Since every group of three consecutive integers contains at least one multiple of 2 and one multiple of 3, the product of any three consecutive integers will have both 2 and 3 as factors. This means the product will be divisible by both 2 and 3.

When a number is divisible by both 2 and 3, it is also divisible by their least common multiple, which is 6. Therefore, the product of any three consecutive integers is divisible by 6.