Submitted by Heirio on Sat, 08/05/2017 - 20:02
Forums:
So the hints (I looked because I was knee deep in algebra and didn't know what to do) said that you take (k1 - 1)...(k4 - 1) = 175 and write it as it's factors. 1, 5, 5, 7. The numbers that k1 could hence be is 12. These numbers are 8,-6,6,-4,2,0,26,-24,36,-34,176,-174.
It then said that via the k1k2k3k4 = 576, k1 could only be factors of 1,2 AND 3. This was said to eliminate 5 values, but it doesn't. The values it eliminates are 0,8,-4,26,34,2,176.
Am I reading something wrong, am I not seeing something?
Elimination
$k_1$ can have factors of 1, 2 and 3 (and can have up to $2^6$ and $3^2$). Hence $k_1=8=2^3$ is not eliminated, nor are $k_1=2$, $k_1=-4=-2^2$. $k_1$ does not have to have both factors of 2 and 3!
So it can have a factor of 2
So it can have a factor of 2^something but it doesn't necessarily have to have a factor of 3^something? Okay, I think the "and" was confusing me. Wording in STEP is usually very specific (if ... then for example), so I was caught out.
8, -6, 6, -4, 2, 0, 26, -24, 36, -34, 176, -174.
So the resulting numbers should be a factor of 576, right? Hence 176, -174 are also out, as are -34, 26, and 0?
Which results in 5 numbers being cut?
I think you're reading something wrong
Hello !
I think there is a mistake in what you wrote :
From the " ... = 175 " equality you can get the 12 possible values for k1 you said.
However, since " ... = 576 ", then k1 must divide 576.
576 = $2^6$ * $3^2$
Therefore, 8; -6; 6;-4; 2; -24; 36 are possible values for k(1). The 5 values eliminated for k(1) are 0; 26; -34; 176 and -174 (which, you can check on a calculator, don't divide 576).
Maybe your mistake was to consider (k(1) - 1) and not k(1).
Anyway, good luck for the assignments!
My mistakes were the
My mistakes were the misunderstanding of the word "and", which in terms of the actual STEP questions are usually very specific, but apparently not so in the hints. There was that and the fact that I forgot that they had to be a factor of 576.
Cross posts!
I think your mistake was that you thought that $k_1$ had to have factors of both $2$ and $3$, where as $k_1$ could be anything of the form $2^n \times 3^m$ where $n \le 6$ and $m \le 2$.
Wording
STEP papers go through a rigorous quality control process (several people work through them in advance and make sure that there is nothing that could be possibly misleading etc.). Our hints and solutions do not go through as many quality checks (if we did we would still be working on the foundation assignments!), which is why we are always very grateful for people pointing out mistakes!