Good luck to all you warriors out there giving it your all one last time before the summer. May your June SAT scores be well worth all your hard work.

This weekend’s challenge is a bit of a logic question. The prize if you get it: nobody in your testing room will assault your senses with unbridled body odor tomorrow morning. Awesome, right? I know.

If

g,h,j,m,n,p,q,r, andsare all positive integer constants, and (g+h+j)(m+n+p+q)(r+s) is odd, what is the difference between the maximum number of even integers and the minimum number of even integers that could be in {g,h,j,m,n,p,q,r,s}?

Drop your answers in the comments. I’ll post the solution Monday.

UPDATE: We had two correct answers again this week! I’m going to have to start making these harder.

The thing you need to recognize here is that there are certain properties of even and odd numbers that are true no matter what the actual numbers are. (Yes, this is fair game for the SAT to throw your way.)

- even × even = even
- odd × odd = odd
- even × odd = even
- even + even = even
- odd + odd = even
- even + odd = odd

So first: if (*g* + *h* + *j*)(*m* + *n* + *p* + *q*)(*r* + *s*) is odd, then (*g* + *h* + *j*), (*m* + *n* + *p* + *q*), and (*r* + *s*) all have to be odd (Rule 2). In order for that to happen, we’re going to need to use rules 4-6 on the sums within the parentheses.

We’re looking for the minimum *and* the maximum number of even integers that could be used, so let’s examine each separately. (I’m going to opt for clarity over conciseness here, but see the SECOND comment below from StaceyHL for the quick and dirty.)

##### MINIMUM evens

You know this won’t work, but try plugging in nothing but odds (I’ll use 3):

*g*+

*h*+

*j*)(

*m*+

*n*+

*p*+

*q*)(

*r*+

*s*)

The (*g* + *h* + *j*)* *is already odd! (*m* + *n* + *p* + *q*) and (*r* + *s*) are even, though, so we need to find a way to make them odd so that our product will be odd. Note that if I change one value from (*m* + *n* + *p* + *q*) to even (say, from 3 to 2) then I’ll have an odd sum. Same goes for (*r* + *s*). By starting with absolutely no evens, I’m able to see the minimum number of evens by dealing with one sum at a time. Once each sum is odd, the whole expression will be odd.

*g*+

*h*+

*j*)(

*m*+

*n*+

*p*+

*q*)(

*r*+

*s*)

**2**)(3 +

**2**)

So the minimum number of even integers in the set is 2.

##### MAXIMUM evens

Just like we did above, I’m going to start by assuming something I know can’t be true: that all the values are even. Then I’m going to see how few I can change to make my product odd.

*g*+

*h*+

*j*)(

*m*+

*n*+

*p*+

*q*)(

*r*+

*s*)

Oh, that won’t do at all! What is the smallest number of values I’d need to change to make all those sums odd? Looks like I’m going to have to change one in each set of parentheses, right?

*g*+

*h*+

*j*)(

*m*+

*n*+

*p*+

*q*)(

*r*+

*s*)

**3**)(2 + 2 + 2 +

**3**)(2 +

**3**)

So the maximum number of even integers in the set is 6.

##### The big finish

If the maximum number of even integers in the set is 6, and the minimum number of even integers in the set is 2, then the answer to this weekend’s challenge is:

**4**

## Comments (4)

4

@pwnthesat Comments have reappeared.

I’ve tried it with every combination of numbers I can think of, including 1s.

I’ll await the solution….Arrggggg

Let’s see..even x even = even; even x odd = even; but odd x odd = odd. Therefore each () needs to be odd. Even + even = even; even + odd = odd – so there needs to be one odd number in each (). So 3 odds and 6 evens to max the evens. 3 evens and 6 odds to min the evens. Difference of max to min evens is 3. I think 🙂

Bugger. Forgot about odd + odd = even.

() Min even Max even

3 O+O+O E+E+O

4 O+O+O+E E+E+E+O

2 E+O E+O

6 max evens – 2 min evens = 4.

I think 🙂