Prove "5 divides (11^n)-6" for any positive integer n?
411657
11 ^n means 11*11...--> the unit of the results is always 1. When it is subtracted by 6,the unit is always 5\nSo that 5 can devide it\ntherefore " 5 divedes (11 ^n)-6"
172504
(w^5)^8 = w^(5*8) = w^40
108399
use binomial theorem, it can be applied to any power.if you don't know this theorem, i'll send it to you.
1062229
Technically there are only 4 answers that will work, and they all rely on parentheses to modify the natural precedence rules:\n\n(5+5/5)*5\n(5/5+5)*5\n5*(5+5/5)\n5*(5/5+5)\n\nAs you can see they are all variations on (5 + 1) * 5, where you use 5/5 = 1.  So there is really just one answer.\n\nYou can easily solve problems like this yourself using the link below...\n\nFor example, can you make 24 using 3, 3, 8, 8?  Try it before you cheat... also, note that the solver allows % (mod) as an operator, so you may have to filter out those results.
774052
LetP(n) denote the statement\n1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6\nwhen n=1,P(1)=1(2)(3)/6=1 checks\nlet P(k) be true\ni.e.1^2+2^2+3^2+.......+k^2=k(k+1)(2k+1)/6\nto prove\nP(k+1)=(k+1)(k+2)(2k+3)/6 is true\nP(k+1)=1^2+2^2+3^2+.......+((k+1)^2\n=>[1^2+2^2+3^2+......+k^2]+(k+1)^2\nnow 1^2+2^2+3^2+......+k^2=k(k+1)(2k+1)/6\nsubsttituting\n=k(k+1)(2k+1)/6 +(k+1)^2\n=(k+1)[k(2k+1)+6(k+1)]/6\n=(k+1)(2k^2+7k+6)/6\n=(k+1)(k+2)(2k+3)/6\nhence proved\n\n2.try by your self
67134
1/6^5=1/7776\nI am 100% sure i am right.
531201
the solution is 1. see 6^1,5^2,4^3,3^4,2^5,1^6.
914435
x > 5\n\n- - - - - -s-
861014
13 + (-12) - (-5) is the same as: 13 -12+5 = 6
8869
I made an applet that solves any 24 problem, so I got the answer easily.\n5*(5-1/5)=24
134713
It can be proved with the use of the integral, but you don't have to.\n\n1)The first some is n(n+1)/2\n\nTo prove that you can add all the consecutive numbers twice, but rearraging them, like that: 1+n; 2+(n-1); 3+(n-3)\n...n+1. Then you have n times the number n+1, which means that this sum equals n(n+1). But since you added twice the sum you wanted, to get the desired result, divide by 2.\nThus you obtain n(n+1)/2\n2)Prove (with the use of mathematical induction) that the sum of all cubes of consecutive integers from 1 to n is  \n[n^2*(n+1)^2]/4\n3)Elevate the first sum, n(n+1)/2, to the second power as requested. See that you get \n[n(n+1)/2]^2= [n^2*(n+1)^2]/4\n4)Observe that the results of 2) and 3) are the same.\n\nQ.E.D\n\nGood Luck.
703605
n^5 - n\n= n(n^4-1)\n= n(n^2-1)(n^2+1)\n= n(n-1)(n+1)(n^2+1)\n\nWe must prove that one of these four factors is even; that one is a multiple of 3; and one is a multiple of 5.  The product will be multiple of 2*3*5 = 60.\n\nThree of the factors are consecutive numbers (n-1, n, n+1) as factors.  One of these must be a multiple of 3 and one of them must be even.  Therefore the result is a multiple of 6.\n\nIf one of n-1, n or n+1 is a multiple of 5, the proof is over.  Assume then that none of them is.  Then either n=5i+2 or n=5i+3 for some integer i.  The factor (n^2 + 1) then becomes\n(5i+2)^2 + 1\n= 25i^2 + 20i + 4 +1\n= 25i^2 +20i + 5\n= 5 * (5i^2 + 4i + 1)\n\nor\n\n(5i+3)^2 + 1\n= 25i^2 + 30i + 9 +1\n= 25i^2 +30i + 10\n= 5 * (5i^2 + 6i + 2)\n\nBoth of these are multiples of 5 - therefore, if none of (n-1), n or (n+1) is a multiple of 5, then n^2+1 is.\n\nQED.\n\nJust saw the hint to go for induction, let's try it!\n\nfor n=2 we have n^5 - n = 32-2 = 30 OK.\n\nAssume it works for n, i.e. n^5-n is a multiple of 30, then\n(n+1)^5 - (n+1)\n= n^5 + 5n^4 + 10 n^3 + 10 n^2 + 5n + 1 - n + 1\n= (n^5 - n) + 5n(n^3 + 2n^2 + 2n + 1)\n= (multiple of 30) + 5n(n+1)(n^2 + n + 1).\nn(n+1) is certainly even.  Is it a multiple of 3? if not this means n=3i+1 for some i.  Therefore the third factor equals (3i+1)^2 + 3i + 1 + 1 = 9i^2 + 9i + 3 is a multiple of 3.  So that big ugly thing is a multiple of 5, 2 and 3 - and thus of 30.\n\nWorks too for negatives too, since (-n)^5 - (-n) = - (n^5-n).\n\nEDIT: Want a third proof???\n\nIf n= r mod a (meaning r is the remainder of n/a) then n^5-n = (r^5 -r) mod a\n\nSetting a=2:\nIf r=0 then n^5-n = 0^5-0 mod 2 = 0 mod 2\nIf r=1 then n^5-n = 1^5-1 mod 2 = 0 mod 2\nHence n^5-n is a multiple of 2 in all cases\n\nFor a=3\nIf r=0 then n^5-n = 0^5-0 mod 3 = 0 mod 3\nIf r=1 then n^5-n = 1^5-1 mod 3 = 0 mod 3\nIf r=2 then n^5-n = 2^5-2 mod 3 = 30 mod 3 = 0 mod 3\nHence n^5-n is a multiple of 3 in all cases\n\nFor a=5\nIf r=0 then n^5-n = 0^5-0 mod 5 = 0 mod 5\nIf r=1 then n^5-n = 1^5-1 mod 5 = 0 mod 5\nIf r=2 then n^5-n = 2^5-2 mod 5 = 30 mod 5 = 0 mod 5\nIf r=3 then n^5-n = 3^5-3 mod 5 = 240 mod 5 = 0 mod 5\nIf r=4 then n^5-n = 4^5-4 mod 5 = 1020 mod 5 = 0 mod 5\nHence n^5-n is a multiple of 5 in all cases.\n\nBeing a multiple of 2, 3 and 5, n^5-n is therefore a multiple of 30.
432827
(n-1)!/(n+1)! = (n-1)!/((n+1)n(n-1)!)\n= 1/((n+1)n)\n= 1/(n^2+n)
320291
6p^6 + 24p^5 + 18p^3 = 6p^3(p^3 + 4p^2 + 3)\n\nIt doesn't break down any farther than that.
814830
HI!\nFor the first: 5 times six equals 30.\n\nSecond: x to the yth power.\n\nWhen in doubt...don't be afraid to ask! :) :)
1132135
I think you need to rephrase this question.  I'm going to solve this problem:\n\n6^(2x)>5^(x+2)\n\nFirst, logarithm both sides and simplify.\n\n2x*ln6>(x+2)*ln5\n\nln5 and ln6 are both positive, so you don't have to worry about changing your sign anytime in the future.  Now just get all of the x's on one side and factor.\n\nx*(2ln6-ln5)>2ln5\n\n2ln6-ln5>0, so no worries in solving for x.\n\nx>(2ln5)/(2ln6-ln5)\n\nor \n\nx>1.631
701385
√605 = √(121*5) = 11√5