sputnik Posted December 19, 2005 Share Posted December 19, 2005 320, the numbers are arranged alphabetically. Good one - I thought it might be a famous phone number at first. We should start a new thread called "Solve That Math Problem" - the geek version of "Name That Tune". Quote Link to comment Share on other sites More sharing options...
Ray Garrison Posted December 19, 2005 Share Posted December 19, 2005 Okay, here's another. What's the biggest number you can write using only three digits? Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 19, 2005 Share Posted December 19, 2005 Okay, here's another. What's the biggest number you can write using only three digits? Hah. I can do it using two digits: 1/0 = infinity Quote Link to comment Share on other sites More sharing options...
Ray Garrison Posted December 19, 2005 Share Posted December 19, 2005 Actually, 1/0 is undefined. Let's change that to "What is the largest integer you can write using only three digits?" Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 19, 2005 Share Posted December 19, 2005 I'm sure the answer isn't 999. If exponents are allowed then 9^99 (9 to the 99th power). Quote Link to comment Share on other sites More sharing options...
DrWho Posted December 19, 2005 Share Posted December 19, 2005 but what about (9^9)^9 ? [] I think I can do better though....in fact I know I can. My number is in binary: 111 Since representations with binary are trivial, the first digit represents 2^infinity. The next digit is 2^(infinity - 1) and the last digit is 2^(infinity - 2). So the value of "111" is really big, lol. (replace infinity with any number you want). Ok Ok, that's cheap I know. How bout something like 1/(1-.999999999999999999999999)? -------------------- Since we're being nerdy...one of my favorite "flaws" in math can be shown with such a simple model. there is a property of multiplication where A / B * B = A. then there is the property of long division, which is hard to show in this message screen, but we all probably know how to divide 1 by 9, which yields .111111111111111 repeating forever If we long divide 1 by 9 and then multiply by 9, we get .111111111111 repeating forever times nice, which yields .999999999999999 repeating forever. If we divide 1 by 9 and then multiply by 9, we get the answer 1. This means that either .9999999999 repeating forever = 1, or that there is some fundamental flaw with one of the basic concepts. My theory is that math doesn't differentiate between touching points and the points .99999999 repeating forever and the point 1 are both touching each other. If this is true (which arguably it can be shown to be true), then it is impossible to have any function other than y=x to continue for infinity. (for example, y=x^2 is limited by the point density of the x-values, so at a very large x-value, the next y-point will be infinitely far way and therefore the line will cease to continue). It also means that no other function besides y=x is a continuous line - and when you get into the presuppositions of non-euclidian geometry this is indirectly an assumption being made...and which is where other systems of math come into play (which are more accurate at the extremes, but not so much in the every day life regions). Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 19, 2005 Share Posted December 19, 2005 but what about (9^9)^9 ? [] 9^99 > (9^9)^9 we should go with 9^(9^9) Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 19, 2005 Share Posted December 19, 2005 ......Since we're being nerdy...one of my favorite "flaws" in math can be shown with such a simple model. there is a property of multiplication where A / B * B = A. then there is the property of long division, which is hard to show in this message screen, but we all probably know how to divide 1 by 9, which yields .111111111111111 repeating forever If we long divide 1 by 9 and then multiply by 9, we get .111111111111 repeating forever times nice, which yields .999999999999999 repeating forever. If we divide 1 by 9 and then multiply by 9, we get the answer 1. This means that either .9999999999 repeating forever = 1, or that there is some fundamental flaw with one of the basic concepts. My theory is that math doesn't differentiate between touching points and the points .99999999 repeating forever and the point 1 are both touching each other. If this is true (which arguably it can be shown to be true), then it is impossible to have any function other than y=x to continue for infinity. (for example, y=x^2 is limited by the point density of the x-values, so at a very large x-value, the next y-point will be infinitely far way and therefore the line will cease to continue). It also means that no other function besides y=x is a continuous line - and when you get into the presuppositions of non-euclidian geometry this is indirectly an assumption being made...and which is where other systems of math come into play (which are more accurate at the extremes, but not so much in the every day life regions). The "flaw" is not in the math but rather in our ability to express the value either on paper or in a calculator register. Go to a museum and find a slide rule and you'll see the folly of the .99999999999 stuff. y=x^2 (a parabola) doesn't have the limits you describe, x and y can extend forever. You're implying that there is an asymptote out at a "very large x-value" but that's not the case. Quote Link to comment Share on other sites More sharing options...
Ray Garrison Posted December 19, 2005 Share Posted December 19, 2005 Actually, the infinately repeating decimal .999 is *EXACTLY* equal to one, as can be easily demonstrated... (Here the underscore represents the repeating decimal) Let a = 0.999 multiply both sides by 10 10a = 9.999 Subtract a from both sides 10a - a = 9.999 - 0.999 9a = 9.0 (the infinately repeating string of .999 are equal, and when subtracted from each other yield 0) thus a = 1 And the answer to the other question was indeed nine to the (nine to the ninth). That equates to 9 to the 387,420,489 which is a, uh, rather large number. Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 19, 2005 Share Posted December 19, 2005 Here is a moderately challenging problem: From the diagram below we can see that the number of blocks on the perimeter, 16, exceeds the number of blocks on the inside, 8. How many rectangles exist for which the number of blocks on the perimeter are equal to the number of blocks on the inside? I just worked through this one with my nephew. The question asks for the number rectangles made up of blocks, so you're dealing with whole numbers. Quote Link to comment Share on other sites More sharing options...
Jay481985 Posted December 20, 2005 Share Posted December 20, 2005 i been studying for finals all week but im delirious so I think 15? Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 20, 2005 Share Posted December 20, 2005 Maybe, we should reshape the question. There are only two such rectangles. What are the dimensions (X number of blocks by Y number blocks) of each one? Quote Link to comment Share on other sites More sharing options...
Ray Garrison Posted December 20, 2005 Share Posted December 20, 2005 5 X 12 6 X 8 Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 20, 2005 Share Posted December 20, 2005 Nice cipher'n there Ray. [*] Solution: If the rectangle measures a x b, the area of the rectangle is ab and the area of the inner rectangle is ab - (a-2)(b-2). The area (or number of blocks) inside is to equal the area (or number of blocks) on the perimeter: (a-2)(b-2) = ab - (a-2)(b-2) ab - 4a - 4b + 8 = 0 ab - 4a - 4b + 16 = 8 (a-4)(b-4) = 8Since a and b are both whole, the only factor pairs are 1 × 8 and 2 × 4. The only two solutions are rectangles measuring 5 × 12 and 6 × 8. Quote Link to comment Share on other sites More sharing options...
Ray Garrison Posted December 20, 2005 Share Posted December 20, 2005 Okay, another one. Three guys check into a hotel for the night - they're going to share a room. The hotel manager charges $30 for the room (This was quite a long time ago...). Each guy chips in $10. Manager takes the $30, guys go to their rooms. Short time later, manager realizes he overcharged - room should have been $25. Manager calls bellhop over, says "Take this $5 and give it back to the guys." Bellhop realizes $5 doesn't divide by 3 evenly, figures he'll make some money. Bellhop pockets $2, gives each guy back a dollar. Now, each guy paid $9. That's $27. The bellhop has $2. Where'd the missing dollar go? Quote Link to comment Share on other sites More sharing options...
ancientdude Posted December 20, 2005 Share Posted December 20, 2005 Wouldnt each guy only have payed 8.333333, seeing as 30-5=25/3, the missing dollar is simply a language problem. Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 20, 2005 Share Posted December 20, 2005 A better question might be "What are those three guys doing in that cheap hotel room?" Wait, I don't want to know. As far as the "missing" dollar, there isn't one. It's probably better to use double entry accounting for this than math but the trick is in the sly wording of the question. Each of the three "fellows" has one dollar, the bell hop has two, and the manager has 25 in his till. The moral of this story is - don't let Ray ever break a fifty dollar bill for you. Quote Link to comment Share on other sites More sharing options...
Ray Garrison Posted December 20, 2005 Share Posted December 20, 2005 The total is $7.97? Here ya go, ma'am... oh, wait, while you've got the till open, can you take that $10 and this $10 and give me a $20? thanks... Quote Link to comment Share on other sites More sharing options...
customsteve01 Posted December 20, 2005 Share Posted December 20, 2005 Nope the first ten has to pay the $7.97 and you get $2.03 in change and you can keep your other ten and get out of the store. Nope not going to get me on that one. Steve Quote Link to comment Share on other sites More sharing options...
sputnik Posted December 21, 2005 Share Posted December 21, 2005 The total is $7.97? I love the look on the cashiers face when I hand over $13.22 and they figure out the right change. When I was in high school I worked at a McDonalds and an old guy would pull that stuff on me all the time. Now it's my turn. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.