Care to try a math problem that may be a little more difficult than it at first appears?

About 6/7×300, give or take 1. I don’t care enough to figure it exactly.

Your answer is very close, but it was fun to find a method for solving exactly.

Well we know that there are 300 even numbers between 1 and 600, right? So although it might take a few minutes, I think I would just count up by sevens until I got to 600, then eliminate every one that was even and however many of those there were I would subtract from the 300.

SPOILER ALERT: (So, 77 is clearly the 11th one, so 98 is the 14th. So 98×3 = 294, so that’s 14×3 = 42, so there are 258.)

Here is a way of looking at it, The total number of even numbers is 600/2=300. Even numbers divisible by 7 are divisible by both 7 and 2, and are therefore divisible by 7×2=14. There are 600/14 = 42 (using whole number division). of them, Even numbers that are not divisible by 7 account for the remaining even numbers, of which there must be 300 – 42 = 258.

Suppose we want to generalize this problem. There is something that we left out. Consider the following. Of the numbers from 1 to 600, how many of the numbers divisible by 10 are not divisible by 15?

A simple generalization of this problem turns out to be trickier than it may at first appear. Suppose we want find all numbers up to 600 that are divisible by 10 but not divisible by 15.

Start as before, by finding how many numbers are divisible by 10 = 600/10=60.

A number will be divisible by both 10 and 15 if and only if it is divisible by the least common multiple of 10 and 15, lcm(10,15) = 30. lcm is what we find when we want the least common den decomnator, We then, similar to what we did before, compute 60 – 600/30 = 60 – 20 =40.

In general the number of values from 1 to n that are divisible by a and not divisible by b is:
a/b – a/lcm(a,b), using integer division.