<p>If the product of six integers is even, what is the largest number of those integers that could be odd?</p>
<p>Thanks</p>
<p>the key to this question is to know that even x odd = even
it means that you need only 1 of the numbers to be even and the whole number will end up being even. So maximum 5 of them could be odd.</p>
<p>Thank you so much. I got it now.
I have 2 more math questions, please help.
Question 1: John has invited 113 guests to his wedding. The wedding hall uses circular tables, each wit a radius of 4 feet. If all of the invited guests are expected to come to the wedding and each guest required 4 feet of space, how many tables will John need at the wedding?
Question 2: Three different medals–gold,silver, and bronze-- are awarded to athletes in two different races in the Sunshine Olympics. If no athlete may win more than one medal, and there are six athletes in total, how many different medal combinations are there?
Thank in advance.</p>
<p>I’m bad at math, so I’m not confirming this answer, rather I’m asking. Is the answer to the first question 16?</p>
<p>Edit: is the answer to the first question 19 tables?</p>
<p>Answer to question #2: 40? 6 nCR 3 = 20 * 2 since there are two races?</p>
<p>Lol for the first question I got 15 tables…I know that’s wrong but I want to see how close I am haha.
Second one I got 720…just did 6<em>5</em>4<em>3</em>2*1 
What are the correct answers?</p>
<p>I got these questions in the SAT prep book, but no answer , no explanation at all.
This book is used in famous SAT prep class, tutor will explain. I do not attend this class.
Please answer with explanation.</p>
<p>These are grid-in questions, not multiple choices.
I typed exactly, except 1 mistyped “each with a radius of 4 feet” instead of “each wit a radius of 4 feet”, I typed “wit”. Sorry.
I do not undestand this :“each guest required 4 feet of space”. Should it be square feet?</p>
<ol>
<li><p>You can fit at most 6 guests per table (space them equally around a table). 113/6 = 18.83, round up to 19.</p></li>
<li><p>Choose two people for gold, choose two people for silver, the bronze medals are uniquely determined. (6C2)(4C2) = 15*6 = 90.</p></li>
</ol>
<p>The medal question is not clear (at least to me): are we to treat gold medals in different events as interchangeable? They say “two different races”. But with no other hint, the only way to tell what the author meant is to work backwards from the answer key, a method that, alas, is not available on the SAT…</p>
<p>Yea, I assumed the two gold medals are indistinguishable. If not, the answer’s just 6! = 720.</p>
<p>How did you get 6 people can fit in one table?</p>
<p>If six people are evenly spaced around the table of radius 4 ft, each person will be at least 4 ft away from any other person at the table. Draw it out and you’ll see.</p>
<p>You’re supposed to use the circumference formula. C = 2pi(4). You get 8pi(25.12) then divide that by 4 ft, so that is 6 people per table. 113/6 = 18.83333 rounds to 19 tables.</p>
<p>Oh thanks a lot, I understand it now :)</p>
<p>For me those other 2 sound like they don’t fit in the SAT. The questions are not clear of what they are asking, and in SAT the questions are always clear. To me it sounds like with that prep book you will gain skills for math but not for SAT. Anyone is free to object this idea of mine 
</p>
<p>For the first question… In SAT (practice and real) I have never seen that you would be required to plug in the value of pi. SAT calculations are always assured to be simple and comfortable to manipulate. </p>
<p>The second question… Again, this one is too convoluted for the real SAT. I am an advocate of training your mind for simplicity because that is what you will have to apply on the real SAT (meaning that if after reading the question you imagine that it is hard, reset and think again; SAT just wants you to apply some simple key rule). I understand this question like that. There are two sets of winners and awards. 3 athletes and 3 medals and another 3 athletes and 3 medals.
So for the first set there can be 3x3=9 combinations and for the second same 3x3=9. 18 in total?</p>
<p>In SAT, every question should evoke one key rule in your mind to solve the problem. After reading those 2 questions I ended up with more questions instead of the key rules.</p>
<p>Now that I think about it, both questions seem a little ambiguous.</p>
<ol>
<li><p>“Each guest required four feet of space.” I interpreted it as, no guest can be 4’ of another guest. If this is the case, spacing six people evenly around a table accomplishes the task, and is optimal. Haphazard intepreted it as taking the circumference of the table and dividing it into 4’ intervals. Fortunately, we both got the same answer…</p></li>
<li><p>This question relies on whether the two gold medals are distinguishable, etc. Clearly, each athlete wins exactly one medal. If anyone can win any medal, the answer’s either 90 or 720 depending on whether the medals are distinguishable or not. If the athletes are already divided into races, each race has 3! = 6 outcomes. 6*6 = 36…</p></li>
</ol>
<p>After rspence insight I see that I combined the outcomes of both races incorrectly. My final answer would be 9x9 = 81</p>
<p>By the way, are these questions for SAT I math or for the subject math?</p>
<p>SAT 1
I absolutely agree with both of you. The questions are poor.
We should just ignore it.</p>
<p>Waitttt!!!
Q1: If we first calculate the number of people can sit in one table, that is
6 people 1table
113 people ? table<br>
→ So we have 18.8883 tables then round up to 19 tables.
BUT IF we work out the circumference of a table (8pi), the space that 113 guests need (113x4) then we have
8pi 1 table
113x4 ? table
→ So we have 17.9845 tables then round up to 18 tables.
It is just the problem of difference of rounded numbers. And can this be available in official SAT test???</p>