Why is there only one representation of 100 as the sum of different powers of 2 (check details)?
439138
1=2^0\n2=2^1\n3 =2^0 + 2^1\n4 =2^2\n5 =2^2 + 2^0\n6 =2^2 + 2^1\n7 =2^2 + 2^1 + 2^0\n8 =2^3\n9 =2^3 + 2^0\n10=2^3 + 2^1\n11=2^3 + 2^1 + 2^0\n12=2^3 + 2^2\n13=2^3 + 2^2 + 2^0\n14=2^3 + 2^2 + 2^1\n15=2^3 + 2^2 + 2^1 + 2^0\n16=2^4\n17=2^4 + 2^0\n18=2^4 + 2^1\n19=2^4 + 2^1 + 2^0\n20=2^4 + 2^2\n21=2^4 + 2^2 + 2^0\n22=2^4 + 2^2 + 2^1\n23=2^4 + 2^2 + 2^1 + 2^0\n24=2^4 + 2^3 \n25=2^4 + 2^3 + 2^0\n26=2^4 + 2^3 + 2^1\n27=2^4 + 2^3 + 2^1 + 2^0\n28=2^4 + 2^3 + 2^2\n29=2^4 + 2^3 + 2^2 + 2^0\n30=2^4 + 2^3 + 2^2 + 2^1\n31=2^4 + 2^3 + 2^2 + 2^1 + 2^0\n32=2^5
167875
You can do it on Google.  Do a search on: 2^900\nThe ^ means "to the power of" and it is the shift of 6.\nIt will only give you an approximate answer because the exact answer has hundreds of digits.  It would take you many hours to "show the work" and you would probably make many mistakes because there are so many digits to copy.
601075
check out this website:\n\nhttp://www.wehirealiens.com/resultsall.asp\n\nI'll boycott with you. Anyone who says one or two people can't make a difference is wrong. It starts with one, then two, then ten, then a hundred....keep adding zeros and it starts to add up.
657261
isn't it to add pairs of mumbers like (1+100) (2+99) (3+98).......(50+51)\n\nall the answers are 101, and there are 50 answers\n\nso 101 x 50 = 5050
533127
2*2*2*5*5 or 2to the third power times 5 to the second power
1071317
There are exactly 343 combinations.  I can list them all, but others have already cluttered this thread quite well on their own.  ^.^\n\nYou can think of the solution set as a series of tiered combinations.  These would start with those sets beginning with a $50 bill, which can contain up to 2 x $50,  2 x $20, 5 x $10, etc. which I believe someone has started already in this thread.  Or group them as pairs, triplets, etc. with the equation 50j+20k+10l+5m+1n=100, which may be a bit easier for calculations.  Or a convergent series, where A=(ways to form $5), B=(ways to form $10, including A-sets), etc.  With any method, find the different combinations for each set, then add them all up.\n\nDoing this is still tedious and prone to mistakes, and thus probably not much of an improvement over manually listing all of the combinations one by one.  It can also quickly become an insane exercise for larger values (for example, add the $2 bill, or find change for $200, or even worse).  So instead of trying to think this problem through mathematically, I found it much easier to think it through as a programming problem.  It lends itself very readily to a simple recursive subroutine using the following pseudocode:\n\nmake an empty stack\nset total combinations to 0\ncall CheckCurrency\nprint total combinations found\n\nrecursive subroutine CheckCurrency\n- if total currency in stack = 100\n- - add 1 to the total combinations\n- - (optional: display stack contents)\n- - end recursion\n- if total currency in stack > 100\n- - end recursion\n- if total currency in stack < 100, then\n- - for each valid denomination (50,20,10,5,1)\n- - - if the denomination < = smallest denomination in the stack\n- - - - push it to the stack\n- - - - call CheckCurrency\n- - - - pop it from the stack\nend of subroutine\n\nThis required only a few minutes to design, another few minutes to code, another few minutes to debug, less than 2 seconds to run, and has the wonderful bonus of being able to display every single combination if desired.  It would also be trivial to modify it so that, for example, you can find all combinations if you also include the $2 bill.  (The answer is:  4562 combinations.)  Or to find all combinations totalling $200.  (The answer is: 2930 combinations.)\n\nThe line that checks if the denomination is less than or equal to the smallest denomination currently in the stack is present to avoid a common mistake:  We want the total combinations, not total permutations.  Otherwise, the stack "50 20 20 10" would be counted differently than the stacks "50 20 10 20" and "50 10 20 20", and even "10 20 20 50".  The total permutations is much, much greater, especially if you consider each bill as unique.  Five $20 dollar bills alone have 5!=120 permutations because, after all, they are all completely different $20 bills.  Imagine the number of permutations of 100 x $1 bills, and everything in between.  o.O
359204
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers
1117885
23\n\nAny number exactly divisible by three can be represented just using the three. Any number that yields a remainder of 1 when divided by three and is at least 13 can be represented as 13 plus some number of 3s. Any number that yields a remainder of two when divided by three and is at least 26 can be represented as 13+13+some number of threes. Because all numbers must have a remainder of either 0, 1, or 2 when divided by 3, all numbers at least 26 can be represented with one of these schemes. It is easily seen that adding the ability to include more than two 13s, or any number of sixes, adds no additional power to the system. Therefore, the greatest number not representable would be the largest number less than 26 and having a remainder of 2 when divided by three. This number is 23.
740336
1. 101\n2.  There are 2 ones.  The first and last entries of the row.\n3.  Each entry is the sum of the 2 entries above and to     each side of it.\n4.  If the 2'nd entry in a row is a prime, it divides each of the other entries in the row (except for the last entry in the row which is always a 1)\n\nCheck out http://ptri1.tripod.com/ for all manner of kewl sh|t about Pascal's Triangle.  Also Fibonacci Series and other fun things ☺\n\n\nDoug\n\nEDIT:  Give Peter D the points.  He needs 'em worse than I do ☺
496936
the first can be shown by induction.\n\nthe 2nd is.. pretty obvious\n\nI couldn't do the 2nd one with Binomial Theorem... btw, what is Binomial theorem? but it is very easy to see 99^50+100^50<101^50\n\nu can write 99^50+100^50<101^50 as 0.99^50+1<1.01^50 and use a calculator, or\n\nTo prove this is true, show 101^50-(99^50)>100^50\n\nWith Binomial expansion\n101^50=(100+1)^50=100^50+50(100^49) +a +b +c +d +...\n99^50=(100-1)^50=100^50-50(100^49) +a -b +c -d +....\n\nwith a,b,c,d,e,f... being some positive real numbers.\n\nu can see that the difference is (50+50)100^49 +2b +2d +2f +... =(100)100^49 +2b +2d +2f +... =100^50 +2b +2d +2f +...> 100^50\n\nI'm not sure if this has anything to do with the Binomial theorem u mentioned.. but.. it has Binomial expansion
1144806
The set {1, 2, 3, ...,} has been closed under multiplication, because any product belongs to that set.\n24 and 25 belong to {1, 2, 3, ...,}. 24 x 25 = 600 as well.\n\nThe identity of the multiplication is 1, because n x 1 = 1 x n = n.\n\nAll numbers n have the inverse 1/n except n=0.\n\nAll the precious answers have no sense.
280758
Solution: 2 to the 5th power \nor written out it looks like this: \n2 X 2 X 2 X 2 X 2 = 32 outcomes\n\nBasically, you place what each position's outcomes can be as the base number. In this case it is 2 because every position can only be either a T or an F exclusively. There are 5 questions. That is why you raise it to the 5th power.\n\nIt is hard to show how to work this one with the The counting principle or the tree diagram method with the limitations of this text processor. I wish I had a whiteboard to show you. LOL\n\nSo that you can see it better, I have a link to a video that covers this topic. It is free to watch on the net - http://www.rogerbacon.org/~jgutzwiller/mathVideos/fundamental%20counting%20principle.asx\n\nOne method you can use is on the video at time mark 2:40. It is a tree diagram.\n\nAnother method you can use is the fundamental counting principle which is at time mark 5:50.
338816
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
940226
Computers run on base-2 arithmetic, since a bit in the computer can only be on or off (positive or negative charge) and thus has only two distinct states.  Since this is true, the amount of memory is also expressed in powers of 2, rather than in decimal (base 10) numbers.\n\n1024 is a power of 2 (2 raised to the 10th power):\n\n2 x 2 = 4 (two of the 2s)\n4 x 2 = 8 (#3)\n8 x 2 = 16 (#4)\n16 x 2 = 32 (#5)\n32 x 2 = 64 (#6)\n64 x 2 = 128 (#7)\n128 x 2 = 256 (#8)\n256 x 2 = 512 (#9)\n512 x 2 = 1024 (#10)
110013
Well, all numbers would have to be represented equally, right?  If there are 27 combination I assume that 1, 1, 1 is included in these combintations, not just every combination that includes all 3 numbers...  However, if all three numbers are represented equally, each number should appear 27 times.  So the answer should be 27*9, or 243.
3940
1. It's one multiplied by ten to the twenty-third power.\n  -Otherwise ten to the twenty-third power.\n\n2. It is also an example of multiplicative identity.\n\nSorry if that is not what you are looking for.
171045
2^16-1 = 65535\n\nThe reason is that you have the summation of powers of two from 0 to 15. Looking at a smaller example, we can see that a 2 bit register (0th and first powers of two) containing all 1's would sum to three 2^0+2^1=3 which is one less than 2^n where n is the number of bits in the register (therefore in this case 2^n = 2^2 = 4).