<p>From blue book, pg. 473</p>
<li>If 18sqrt(18) = Rsqrt(t), where R and t are positive integers and R > t, which of the following could be the value of rt?</li>
</ol>
<p>(A) 18
(B) 36
(C) 108
(D) 162
(E) 324</p>
<p>Is there a quick way to do this one?</p>
<p>Break up the number in the square root:
18sqrt(18) = 18<em>sqrt(9</em>2) = 18<em>3</em>sqrt(2) = 54sqrt(2)
R = 54; t = 2</p>
<p>Check the stated conditions:
"R and t are positive integers and R > t"? Yup, it fits</p>
<p>So rt = 108</p>
<p>The answer is
C. 108</p>
<p>Given:
18sqrt(18) = Rsqrt(t)</p>
<p>sqrt 18 = 3sqrt2</p>
<p>18(3)sqrt(2) = Rsqrt(t)</p>
<p>18*3= 54
so,</p>
<p>54sqrt(2) = Rsqrt(t)
R=54 t=2
R>t? yes.</p>
<p>Tanman you beat me to it...ha!
R<em>t = 54</em>2
= 108</p>
<p>Thanks, gentlemen. Would you like to solve another one. I know you would. Here goes</p>
<p>There are 5 cards: a b c d e</p>
<p>How many 5 card arrangements can be made if card c cannot be on either end</p>
<p>72.
Any more?</p>
<p>There are 5 cards: a b c d e</p>
<p>How many 5 card arrangements can be made if card c cannot be on either end</p>
<p>c cannot be at either end, so for slot a there are 4 possibilites (all the cards except c)</p>
<p>now go to the other end (slot e) - there are 3 possibilites because one of them was used up on the first slot</p>
<p>now go back to slot b - there are 3 possibilities - all 5 cards minus the 2 that are already placed at either end
2 possibilites for slot c
1 possibility for slot d
so there are 4 x 3 x 2 x 1 x 3 = 72 different arrangements.</p>
<p>yesss. thanks guys. I finally figured it out. And your explanation confirms it. Much love.</p>
<p>4<em>3</em>3<em>2</em>1=72</p>
<p>There are 4 possibilities for the first end, then three for the other end, then 3 for one of the middles, then 2 for the next middle, then 1 for the final middle (idk if that makes sense....lol)</p>
<p>Edit: man you people are fast...</p>
<p>Here is a straight foward explanation; This is a simple permutation problem...</p>
<p>5 cards arranged in random order (no repetition):
5! = 5<em>4</em>3<em>2</em>1 = 120 total combinations</p>
<p>Subtract the number of combinations possible with C at the end.
That value = 4! because C is fixed and there are 4 cards left to be arranged in random order.</p>
<p>4<em>3</em>2<em>1 = 24
Multiply this by 2 to account for the other end,
24</em>2 = 48
120 total - 48 with C at the end
120 - 48 = 72</p>
<p>That is so cool.</p>
<p>Anything else? I love helping people understand math.</p>
<p>haha... not at this time. but stay tuned...</p>
<p>Ask something other than probability please. Next challenge question..........</p>
<p>If it takes 4 men 8 hours to dig a hole, how many hours will it take 2 men to dig a half of a hole?</p>
<p>i may be wrong but is it half an hour?</p>
<p>If it takes 4 men, 8 hours to dig 1 hole - THEN::</p>
<p>That means it takes 1 man 2 hours to dig 1/16 of a hole.</p>
<p>This means it takes 2 men 2 hours to dig 1/8 of a hole.</p>
<p>This means it takes 2 men 16 hours to dig 1 complete hole.</p>
<p>Then, it's obvious that it takes 2 men 8 hours to dig 1/2 a hole.</p>
<p>hmm... i dont understand your reasoning eternity... could you perhaps make it a bit clearer.</p>
<p>I'm sorry, but this was a smartass question. There's really no such thing as a half of a hole. A hole is a hole. :)</p>
<p>Yeah but a hole can still be quantified. </p>
<p>You can have a hole the size of a lake or a hole the size of a small house.</p>
<p>In any case, I'm sure my prior answer is correct.</p>
<p>Don't have time to explain right now but I'll try and get to it later.</p>
<p>here was my reasoning.. </p>
<p>If it takes 4 men, 8 hours to dig 1 hole - THEN::</p>
<p>That means it takes 1 man 2 hours to dig a hole.</p>
<p>This means it takes two men 1 hour to dig a hole. </p>
<p>This means it takes two men half hour to dig half a hole. </p>
<p>im not good at math so i'd appreciate it if someone could tell me why my logic is flawed.</p>