Finding Integers

Question

If n is an integer greater than 5, then find three consecutive integers n, n+1 and n+2 such that these are divisible by 5, 6 and 7 respectively.

Answer 1

Hello Luke,

 

I'm certain that there's a very rigorous mathematical method for solving this, but I'm not much of a mathematician. Seeing as nobody else has volunteered to answer this problem yet, I'll give it a go. Hopefully, my solution, as unrigorous as will be, proves helpful to you.

 

So, here's what we know:

 

     1) n is greater than 5

     2) n must be divisible by 5

     3) n + 1 must be divisible by 6

     4) n + 2 must be divisible by 7

 

Let's work with 2).

 

If n is divisible by 5, then we're looking at all numbers that end in either 5 or 0 (10, 15, 20, 25, 30, etc).

 

Now, let's move on to 3).

 

If n + 1 is divisible by 6, then n + 1 must be even. Going off of what we learned earlier, that rules out all numbers for n that end in 0, because n + 1 would then end in 1, making it odd. Odd numbers aren't divisible by 6. So, we're down to the set of numbers that end in 5 (15, 25, 35, 45, 55, etc).

 

Well, if n ends in 5, then n + 1 ends in 6, and n + 2 ends in 7.

 

What numbers end in 6 and are divisible by 6? Answer: 6, 36, 66, 96, 126, etc.

 

What numbers end in 7 and are divisible by 7? Answer: 7, 77, 147, 217, etc.

 

Another way to write this is:

 

     n + 1 = 6 + 30i, where i is an integer

     n + 2 = 7 + 70j, where j is an integer

 

And the lowest common multiple of 30 and 70 is... yep. 210. With i = 7 and j = 3, we have:

 

     n + 1 = 6 + 210

     n + 2 = 7 + 210

 

So, the numbers 215, 216, and 217 are each divisible by 5, 6, and 7, respectively.

 

Extra credit: using the same constraints as above, can you find a different set of numbers that are divisible by 5, 6, and 7? How many different sets can you find?