Does there  is  integer  x   s.t   10x  modulo 1(21),5x  modulo 2(6),4x modulo  1(7)?
460848
no there doesnt. If you solve the congruences one at a time and use your solution for the previous congruence in the next one, you will eventually get the colution 1 congruent to 0 (mod7) which is obviously impossible
103600
usually it's 4, but if u work with modular classes (modulo n) it's a bit diffrent. for example: if u have the class Z4, that means u work with {0,1,2,3) so in that class, 2+2=0, because 4 is considered 0\nif u work in Z3 with {0,1,2} then 2+2=1. because 4 it's one more then 3.
822221
Using mods...\n\nn^3 + n - 4 = 0 mod 7\n\nn^3 + n = 4 mod 7\n\nn(n^2+1) = 4 mod 7\n\nIf n=0 mod 7 then n(n^2+1) = 0 mod 7, doesn't work.\nIf n=1 mod 7 then n(n^2+1) = 1(1^2+1) mod 7 = 2 mod 7, doesn't work.\nIf n=2 mod 7 then n(n^2+1) = 2(4+1) mod 7 = 3 mod 7, doesn't work.\nIf n=3 mod 7 then n(n^2+1) = 3(9+1) mod 7 = 2 mod 7, doesn't work.\nIf n=4 mod 7 then n(n^2+1) = 4(16+1) mod 7 = 5 mod 7, doesn't work.\nIf n=5 mod 7 then n(n^2+1) = 5(25+1) mod 7 = 130 mod 7 = 4 mod 7 - YESSSS!\nIf n=6 mod 7 then n(n^2+1) = 6(36+1) mod 7 = 222 mod 7 = 5 mod 7, doesn't work.\n\nSo, n must be equal to 7m+5 for some integer m.  This gives us:\nn^3 + n - 4 =\n(7m+5)^3 + 7m+5 - 4 =\n7^3m^3 + 3*7^2m^2*5 + 3*7m*5^2 + 5^3 + 7m + 1 = \na whole bunch of multiples of 7 + 5^3 + 1 = \nmultiples of 7, plus 126 =\nmultiples of 7, plus 18*7.
759151
a. This is true. \nThe example is mod n group over integers.\nm mod n is the remainder whe m is devided by n.\nThis mod n group will have exactly n number of elements.\n\neg. mod 3 = {0,1,2} and  mod 5 = {0,1,2,3,4,5}\n\nb. This is false .\n\nThe example are mod 3 and the set {i, -1, 1}\nThey dont have same group structure, also both have different operations for group structure. One had remainders while other has complex number multiplication, addtion etc.\n\nc. This is true.\nThis is by definition of subgroups.
1093659
Or is it this:\n300+20+4\n400+60+7\n100+10+1\n800+90\n60+8\n1000+600+50+4
527322
1+1-21+30-5\n2-21+30-5\n-19+30-5\n11-5\n6
1059114
3x^2 - 9x + 6 - 3(x - 1)(x - 2)\n3x^2 - 9x + 6 - 3(x^2 - 2x - x + 2)\n3x^2 - 9x + 6 - 3(x^2 - 3x + 2)\n3x^2 - 9x + 6 - 3x^2 + 9x - 6\n\nEverything gets cancelled out, so it doesn't matter if i put 10 in for x or not.\n\n3(10)^2 - 9(10) + 6\n3(100) - 90 + 6\n300 - 90 + 6\n210 + 6\n216
786009
-2(x-3) + 4(x+7)\n-2x+6   +4x+28\n2x+34
838764
- 7 - 4x < 13\n\n-7 - 4x + 7 < 13 + 7\n\n- 4x < 20\n\n- 4x/- 4 > 20 /- 4\n\nx > - 5\n\nThe answer is x > - 5\n\n- - - - -s-
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
729005
a.xa+xb\nb.21x^3+24x^2y\nc.3x(8x+5)+2(8x+5)\n=24x^2+15x+16x+10\n=24x^2+31x+10\nd.4x(2x+1)-6(2x+1)\n=8x^2+4x-12x-6\n=8x^2-8x-6\ne.8x(5x^2+7x+3)+2(5x^2+7x+3)\n=40x^3+56x^2+24x+10x^2+14x+6\n=40x^3+66x^2+38x+6\nf.x^2-1\ng.x(x^2+2x+4)-2(x^2+2x+4)\n=x^3+2x^2+4x-2x^2-4x-8\n=x^3-8
823097
Can't prove it as n cannot = + or -1 and k be an interger.\nIt appears to work with other odd integers. Change your question then it may be possible to prove it.
866528
Consecutive even integers would be represented by x and x+2.  So the equation for your problem is x = 10 + .5(x+2) and you solve that equation.\n\nGood luck !
543381
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.
875039
The first five are simple\nP(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5) \n5 + 20 + 60 + 120 + 120 = 325\n\nSix and seven are a little more complicated\n6!/2! + 7!/2!2!\n360 + 1260 = 1620\n\nEight is special because you limit the maximum.\n\nList of bad values\n9x,xxx,xxx\n7x,xxx,xxx\n69,xxx,xxx\n67,xxx,xxx\n66,xxx,xxx\n65,9xx,xxx\n65,7xx,xxx\n65,6xx,xxx\n\n8!/3!2! - (2 * 7!/3!2!) - (3 * 6!/3!) - (3 * 5!/3!)\n3360 - 840 - 360 - 60 = 2100\n\nNow add the results.\n325 + 1620 + 2100 = 4045
419172
(0, 1)\n(1, 1.1)\n(2, 1.21)\n(3, 1.331)\n(4, 1.4641)
175474
1+2+3+4+5+6+7+8+9+10+..........+....