<li>Sam had 20 real coins and 21 fake coins. A fake coin weights 1 gram less than a real one.Sam lost 1 coin. He has a balance scales that shows the weight difference , but he can only use it only once. Is it possible to find out which coin he lost-real or fake?</li>
<li>Two cars- Old Junky and New Sporty-started to move from point A to point B at the same time. After driving ond third of the way Old Junky stopped and started moving only when reaching B New Sporty turned back and continued to drive towards A. Which car will come first- Old Junky to B or New Sporty to A? Assume that each car is moving with the constant speed.</li>
<li>Santa has three kinds of canides in his bga. He knows that if he takes out any 100 candies , he will definitely have all three kindas of candies. What is the maximum number of candies in his bag? </li>
<li>There are N points on the plane. The distance between any two of them is greater than d. Find the maximum number of pairs of points thatt are exactly at distance d.</li>
</ol>
<p>I tried the first one with the coins. </p>
<p>If you put 20 coins on each side of the scale:
Assuming the lost coin is fake, then every combination of coins that could be on either side,will always give an even difference.</p>
<p>Assuming the lost coin is real, then every combination of coins that could be on either side,will always give an odd difference.</p>
<p>your not answering the question...</p>
<p>Yes, it is possible</p>
<p>Aren't we suppose to say how it is possible?</p>
<p>true.......</p>
<p>A3) 133 (I'm not too sure, though)</p>
<p>show work =)</p>
<p>Penn253, is hello right? I cant figure it out.</p>
<p>I think number four is 0, the wording seems like a trick question.</p>
<p>Unless I misinterpreted the problem, number 2 has insufficient information.</p>
<p>The question is equivalent to asking, which takes longer: O.J. driving 2/3 distance from A to B or N.S. driving full distance from B to A. But since the problem doesn't tell you the relative speed of O.J. and N.S., it's impossible to determine the answer.</p>
<p>(O.J. = Old Junky, N.S. = New Sporty)</p>
<p>3) 148, or 49 of two and 50 of one. He can take out 49 of each one, or 50 of one and 49 of the second, and the 99th or 100th must be of the third kind. (I think)</p>
<p>1.) Yes, Real = x, fake = x - 1: 20x + 21x - 21 = 41x - 21. If he loses a real one, he will have 40x - 21 which are odd no matter what and wouldn't balance. If he loses a fake one, he will have 40x - 20 which is even no matter what and would balance so yes it is possible.
2.) If they are moving at the same speed, then Old Junky would win because it only has to go 2/3 of the distance in the time New Sporty has to go the whole distance. Your wording is quite awkward so you may have left out part of it.
3.) 148, because if he took all of any two kinds, the total has to be 99 or below.
4.) If I understand it it is 0 because "The distance between any two of them is greater than d."</p>
<p>
But you didn't say O.J. and N.S. are moving at the same speed. You only said that each of them is moving with a constant speed:</p>
<p>
</p>
<p>
</p>
<p>But you're assuming that the real coin's weight "x" is an integer, which you never stated in the original problem. And even if "x" were an integer, your method doesn't work, because you'd need to balance up to (40 choose 20)/2 times to see whether a set of 20 coins balances the other 20 coins.</p>
<p>I'd stick to inhaven's reasoning, which is valid and can be proved algebraically.</p>
<p>haha, Penn253 went on the Columbia board and stole the answers from post number 5, see the link below:
<a href="http://talk.collegeconfidential.com/showthread.php?t=33690%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?t=33690</a></p>
<p>He stole it ten minutes after it was posted.</p>
<p>I guess Penn253 is too brilliant to solve these easy problems on his own.</p>
<p>i think he probably asked because he wanted our imputs and wasn't trying to pass those answers off as his own. but that's just how it seems to me</p>