• M5

    The part in bold is very very confusing, can someone explain what the part in bold has to do with the problem??
    "Any multiple of 6 is even, since 6 is even. Thus we can look just at the term that is the multiple of 5. A multiple of 5 is even if the integer multiplied to 5 is even, so the cases where this is true are 0 × 5, 2 × 5, 4 × 5 and 6 × 5. We can also add an even number less than 5 to this base term. There are three ways to add an even (0, 2 and 4, and not anything bigger, because that would make another 5.)

    The ways are:

    There are 10 ways in total. "-Solution

  • MOD

    @neatlobster Great question! Think about the grid box corresponding to a number N. The box has a purple number (N mod 5) and a yellow number (N mod 6).

    We want the yellow number to be even. Write N as 6a + b, where b is N mod 6. For instance, if N is 19, we'd let a = 3 and b = 1. But because any multiple of 6 is even, this implies that 6a is always even. And we want the yellow number, which is b, to be even. So, 6a + b = N must be even since both 6a and b are even. (Basically, the whole point of the bolded sentence is to show that N is even)

    From there, we want the purple number (N mod 5) to be even. BUT we already established that N is even! Therefore, N = 5*(even number) + (even number). (If N = 5*(odd number) + (even number), N would be odd, which can't be true)

    Hope this explanation helped! 🙂