why can't a hundreds chart only have odd numbers?
821946
Searching for "hundreds chart" gave me this...\nhttp://math.about.com/library/100.pdf\n\nIf this is what you are talking about, I think the point is that it has every number from 1 to 100.  Why would you want a chart that had only odd numbers?  It would only go up to 99, anyway.
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
621145
With just ! (Factorial), +, -, * and /, it is impossible to get 150 using only 3 5s without the x^2 (Squared) operation (Which means x*x, in case you are in Fifth Grade), or using a '^2' as you say. If you say there is an alternate way, and there really is, then I give up and admit defeat. (I'm sure most others do.) If you are so sure, why don't you post the answer yourself?
979591
Is there some reason it wouldn't be 280?
779526
How are you even sure that such a function exists? the numbers look very random.
662185
Irrational numbers don't have patterns, either.\n\nYou could take pi or e or sqrt(2) out to a hundered digits and stat test it.\n\nI just looked at sqrt(2), sqrt(3), and e.  I couldn't see any pattern to occurances.\n\nAlso, I can't think of any reason there would be more of some number than another.  For years, sqrt(prime) has been used as a source of random numbers.
64745
Your problem is digits dont use large digits to plot small once so use intervals of 0.5 to the next whole number value like 0,0.5,1,1.5,2,2.5 and so on.
957153
They are simply based around counting on your fingers; fives and tens and are designed to be carved on monuments and so are mostly built of straight lines.\nThe problem with them is when you have something on the left of a number that you need to deduct, as with nine, for example.\nThat can be written as 5+1+1+1+1 (VIIII) or more economically as 10 minus one. Except the one comes BEFORE the ten: IX . That rule applies throughout, so IV is 4, XIV is 14, XIX is 19, XXXIV is 34 etc. \n\nOne tip a teacher gave me was to think of ten (X) as five (V) on top of another upside-down five. Works better if demonstrated with your hands!\nTo count to a hundred you will need the standard 1, 5 and 10 symbols; I, V and X, as well as L which is fifty (50) and C which is 100. So, IL is 49 and IC is 99.\n\nIt's really quite logical and you don't need to memorize much.
482701
"mode"\n"display digits"\n"float 12" (or however many you want - 12 is the highest)
1110706
6 - Thousands\n2 - Hundreds\n5 - Tens\n2 - Ones\n\nIf you want to round to the nearest hundreds, then look at the number that comes before it. In this case, it is in the tens spot. That number is a 5. The rule for rounding is if the number is 5 or greater then you round up. If the number is 4 or lower, you round down.  Thus, the answer would be 6300.
234737
because yahoo charts suck\n\nuse bigcharts.com
1176747
It's so that the person you are talking to knows what you are talking about.\n\nWhat would the alternative be? "Twelve and seven hundred and eighty-nine thousandths?" That's a bit cumbersome.\n\nThe decimal point *creates* a division between the numbers, so we need to replicate that division by saying "point." It just makes sense.
112490
Because the numbers after the decimal (to the right of) represent less than one. One times a number is that number. Less than one times a number is only a fraction of that number.\n\nMake sense?
797722
AARGHHH!!  I withdraw my earlier attempt.  It must be dementia, I'm afraid.  It's still asking about 1000! (1000 factorial), of course, but a careful accounting is needed, not only of the effect of all numbers ending in: \n\nA. 1, 2, 3, ... 9 (I did this by a variety of ways, always coming up with a final non-zero digit of 6 for the product of all such sets), \n\nbut also of the following two other types of number, which I regrettably neglected the first time around:\n\nB. 10, 20, 30 ...90 (as often as sets of numbers like that occur), and finally, of types:\n\nC. 100, 200, 300, ... 900 (this set occurs but once).  The final, unique 4-digit number (1000) only adds three zeros at the end and doesn't affect the final non-zero digit.\n\nO.K., here goes:\n\nNumbers of Type A.\n\n1 x 2 x 3 x ... x 9 = 362880; final non-zero digit is 8.\n\nThere are 100 such "basic sets" of type A altogether.  \nMultiplying them together, set after set, we get the \nfollowing repeated cycle of non-zero ending digits \n(top row is # of basic sets used so far; bottom row is the corresponding final digit.)\n\n1   2   3   4   5   6   7   8   9 ...\n8   4   2   6   8   4   2   6   8 ...\n\nNotice that after every four basic sets have been used, we end up with 6 as the final non-zero digit.  Since there are exactly 100 such sets (a multiple of 4), we end up with that same last non-zero digit for the product of ALL numbers of type A.  (But we're not done yet.) \n\nNumbers of Type B.\n\nThere are exactly 10 sets of numbers of Type B, i.e. ending with  exactly one (but not two or three) zeros.\n\nMultiplying just these sets together, we'll have the following analogous results:\n\n1   2   3   4   5   6   7   8   9  10  \n8   4   2   6   8   4   2   6   8     4   \n\nSo, the last non-zero digit in these "10's products" is 4 for this lot.\n\nNumbers of Type C.\n\nWe have just one set of "100's".  For this, the last non-zero digit is 8 (it's just our "basic set" with a large number of zeros following it); only the last non-zero digit is of interest.\n\nSo, to summarize, the different sets we've divided our numbers into, when multiplied together, give us final non-zero digits as follows:\n\nType A:   6\nType B:   4\nType C:   8 \n\nFinally, then, the last non-zero digit of the product of everything is the last digit of:\n\n6 x 4 x 8, or 2.  (PHEW!!)\n\nWell, it's now the value of 2 (as reported by zanti3 from the 100 factorial website at:\n\nhttp://www.justinwhite.com/big-calc/1000.html)\n\n(Finish off this URL with 1000.html, instead of 1000... The Yahoo site would not reproduce the full URL from my draft.  Alternatively, just go to Yahoo or Google with 1000 factorial 1000 - this worked for me.) \n\nBoy; what a problem!  Evidently the jury is still out on the question of my impending dementia.  (See zanti3's kind words, below about my earlier fiasco.) \n\nThanks again to zanti3 for bringing the existence of a (1000 factorial) web site to my attention.\n\nLive long and prosper, everyone.
226140
I'm not sure I understand your question - do you mean adding odd numbers only to odd numbers and even numbers only to even numbers?  For odd I think it would be the numbers that end in (3 and 7) (1 and 9) and (5 and 5) where they add up to 50, and for the evens the numbers ending in (2 and 8) (4 and 6) that add up to 50.
283743
This is 194. For this question, you have to know that 9 is the highest number, and for a difference of 8 to occur for digits it has to be 9-1 = 8, since no other numbers can satisfy this, then the tens digit is 9 and the hundred unit is 1 and then of course the ones digit is 5 less than  the tens unit that is 9-5 = 4 thus the number is 194
194052
I think the contributions thus far have hit upon the reason, but I'm not sure it is all tied together.\n\nChoosing any two-digit number and subtracting its separate digits will yield 81 for any two-digit number that starts with 9, 72 for a choice of 8x (8x = 89, 88, 87, 86, to 80), 63 for a choice of 7x, and so on.\n\n9*9 = 81\n8*9 = 72\n7*9 = 63\n6*9 = 54\netc\n\nIf you look at the corresponding numbers for ALL of these choices you will see the same symbol! So no matter where you start you will always get this answer.\n\nAlso, the symbols change every few refreshes of the page. So when you "Try Again?" you will get a fresh symbol. If you look at them closely you will see this happen as you go back and play again and again.\n\nThis is a very slick game, and even tho I know how it works, I still think it's way kewl.\n\nlol