Thread: Calling all math geeks!!!!!!!!

Dagnabbit, I like math but remainders and probability are the two areas in which I'm terrible for whatever reason.

Right now I'm thinking there has to be a number, x, that can be divided by all those dividers and leave a remainder of 2 except for one. For example, 32 might have been a good choice, because if you subtract 2 and take the 30, you can divide that by 3, 5, 6 and 10. Unfortunately, you can't divide it by 11 or 17, but it sounds to me like the question only wants one answer, not two, right?

Yes, one answer. And I'd give you that answer, but then you'd do like my bf did and solve it backwards because you already know the answer.......which doesn't help me.

Ugh

1+ 1= 2

I was thinking about that, asking for the answer so I could solve it backwards and then maybe you could use the explanation to apply similar math to similar questions.

Do you have a study guide that explains the math portion? I was just looking at some online, but all the remainder explanations cover examples that are different from this.

But it's probably just something silly easy that we're overlooking anyway.

Lol Sidney

Tex - in the book (Princeton Review) it discusses how when you have an unknown variable you can use the "plugging in" method as opposed to writing some sort of complex algebraic equation. This would include choosing a number that makes sense (even if the other numbers are even, odd if the other numbers are odd, factors of the numbers, etc) and using that number to solve the question. Then take whatever answer you get and fill that in to all the answer choices and see which one gives you the number you chose for x. But because of the "remainder" in this one, it just doesn't make sense. Remainders are the elementary form of decimals and fractions, which is why most of us haven't dealt with them since grade school. I don't think the remainder concept plays a vital role in this question. And of course an integer is any whole number whether positive or negative.

*grumble grumble* MUST FIGURE THIS OUT!

I don't think the remainder is important in this particular question either. I'm still thinking there must be a number that can be divided by all those except for one. But you can spend an awful lot of time trying to think of, definitely more than two minutes.

Oh, it's on like mother f'n Donkey Kong... it annoys me that I can't think of the answer, so I'm going to do what I can to get it right, even if it's the only thing I'll then have time for all day

It annoys me that you can't think of the answer too. LOL!! No really, my bf spent over an hr last night substituting in numbers until he found the right answer. And while I appreciate the effort, there is a more logical way to do this problem....I just don't know what it is.

One trick might have been to multiply some of those numbers by each other (e.g. 3 * 17 or 6 * 11) and then try to divide by another one of those numbers to verify, but I can't come up with anything. Grrr...

Now I'm going to ponder if there might be a simple equation involved...