what is Armstrong's number and what are the values ?
351410
An Armstrong number is a narcissistic number.  One which if you raise each of its digits to the power of the number of digits in the number, and then add them together will yield the number istelf.\n\nExample:\n153 is a 3-digit Armstring number because:\n1^3 + 5^3 + 3^3 = 153\n\n1634 is a 4-digit Armstriong number because:\n1^4 + 6^4 + 3^4 +4^4 = 1634
520658
#24,#99,#16,or #20\nWho knows?
603507
52\n\naequeosalinocalcalinoceraceoaluminosocupreovitriolic
929256
760,320  \n\nWhy the hell would you need to know that???
211094
http://en.wikipedia.org/wiki/7_%28number%29\n\nread down the page
698821
The symbol "^" means raised to the power of ___ , in other words, it shows what exponent is used.\n\nIt doesn't matter what the initial amount is, the increase of 1% will result in the same amount of increase after a certain number of repetitions... which can be multiplied by the initial amount to determine the net amount...and you must add 1 to the rate of increase because you are retaining the initial amount.\nso after 1 cycle you will have: \n1.01 X 33,000 = 33,330\n\n2 cycles will result in:\n1.01 X 1.01 = 1.01^2 = 1.0201 X 33,000 = 33,663.33\n\n3 cycles will result in:\n1.01 X 1.01 X 1.01 = 1.01^3 = 1.030301 = 33,999.933\n\nPV = Present Value\nFV = Future Value\nR = Rate expressed as a decimal, 1% = .01\nn = number of cycles\n\nFV = PV (1+R)^n\n\nFor values of n greater than four or five, a scientific calculator is a big advantage!
1006003
Armstrong 74 games started\nJordan 78 games started\nPippen 81 games started\nGrant 77 games started\nCartwright 63 starts\nPaxson 8 games started
120160
http://mathworld.wolfram.com/ComplexNumber.html\n\nfind definitions of absolute value (||), power and product.\nIt should be straightforward from there.
537505
292\n490\n274\n472\n454\n436\n418 \n238
829046
SrSe\nK2O\nCa3N2\nCoI3\n\nthe numbers should be subscripts
597363
100\n\nhttp://en.wikipedia.org/wiki/Names_of_large_numbers
942278
As the hint says, every number is some power of two multiplied by an odd number, or 2^j * s, where s is odd.\nNow, lets say you have written every number in the set T like that. What possible values for s have you come up with? They all have to be <= 2n-1, since the whole number is <= 2n. That means there are precisely n possibilities: 1, 3, .., 2n-1.\nHowever, you have chosen n+1 numbers. So two of those numbers must have the same 's' part.\nBut then if the numbers were 2^a * s and 2^b * s, one of those numbers is a multiple of the other - assuming a>b, (2^a * s)/(2^b * s) = 2^(a-b) (and similarly for a<=b).
768317
The letter o is a number
303516
Each element x of S_n can be written as a product of disjoint cycles, e.g. x=(132)(45). Since the order of a cycle is the number of elements it permutes, the order of x is the least common multiple of the orders of each of its cycles.\nNow, the order of x should be 3000 = 2^3*3*5^3. To make the number n of elements permuted by x minimal, we have to find the set of coprime numbers d_i, i=1..k with d_1*...*d_k = 3000 that makes n=d_1+d_2+d_3 minimal. All possible choices are listed below :\n1. (k=3) = 3, 2^3, 5^3   \n2. (k=2) : 3*2^3, 5^3    \n3. (k=2) : 3, 2^3*5^3    \n4. (k=2) : 2^3, 3*5^3    \n5. (k=1) : 3*2^3*5^3     \nObviously, the first choice where k is maximal (= the number of different prime factors of 3000) minimizes n to the value 3+2^3+5^3=136.\n\njust in case you want to know... :\nQuite generally, the minimal n for an element x with order d is given by n = p_1^e_1+...+p_k^e_k where the p_i's are different prime numbers with d = p_1^e_1*...*p_k^e_k.
753355
125.6\n\nhttp://www.webelements.com/webelements/elements/text/Ag/ionz.html
392075
By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions.\n\n1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O2, O3, P4, S8, and aluminum metal all have an oxidation number of 0.\n\n2. The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of sodium in the Na+ ion is +1, for example, and the oxidation number of chlorine in the Cl- ion is -1.\n\n3. The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH4, NH3, H2O, and HCl.\n\n4. The oxidation number of hydrogen is -1 when it is combined with a metal as in. LiH, NaH, CaH2, and LiAlH4.\n\n5. The metals in Group IA form compounds (such as Li3N and Na2S) in which the metal atom has an oxidation number of +1.\n\n6. The elements in Group IIA form compounds (such as Mg3N2 and CaCO3) in which the metal atom has a +2 oxidation number.\n\n7. Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O2, O3, H2O2, and the O22- ion.\n\n8. The elements in Group VIIA often form compounds (such as AlF3, HCl, and ZnBr2) in which the nonmetal has a -1 oxidation number.\n\n9. The sum of the oxidation numbers in a neutral compound is zero.\n\nH2O: 2(+1) + (-2) = 0\n\n10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of the sulfur atom in the SO42- ion must be +6, for example, because the sum of the oxidation numbers of the atoms in this ion must equal -2.\n\nSO42-: (+6) + 4(-2) = -2\n\n11. Elements toward the bottom left corner of the periodic table are more likely to have positive oxidation numbers than those toward the upper right corner of the table. Sulfur has a positive oxidation number in SO2, for example, because it is below oxygen in the periodic table.\n\nSO2: (+4) + 2(-2) = 0
878412
Na = +1 , O = -2\n\noxidation number Cl at NaClO = 0-1+2 = +1\n                              at NaCl = 0-1 = -1\nreduction can be defined as the reduction of oxidation number.\nfrom +1 to -1 (proved)