Can you find out the integers n with the proprety that n^3+n-4 is divided by 7?
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.
924341
(6 + 7) * (4-2) = 26\n(14-5) / 3 = 3\n27 / (4+5) - 1 = 2\n6 * (7+2) - 1 = 53
1169872
(7/(3x)) / (x/3)\n(7/(3x)) * (3/x)\n21/(3x^2)\n7/x^2\n\nBut if you meant\n\n((7/3)x)/(x/3)\n((7x)/3)*(3/x)\n(21x)/(3x)\nANS : 7\n\n-------------------------------\n\n2 1/3 - 3/4\n2 1/3 = (7/3)\n\n(7/3) - (3/4)\nMultiply everything by 12\n(28 - 9)/12\n(19/12) or 1 (7/12)
961314
easy the number is found as such \n\n21 = 3*7\n\n27 = 3^3 \n\n33 = 3 *11\n\n55 = 5 *11\n\nIn order that a number must be divisble by these, it must contain \nall factors. if you look for the samllest avoid repetition \n\n3^3 , 5, 11  =1385\n\nand you add 5 \n\nso the number is 1390
928756
m^6-n^6=(m^3+n^3)(m^3-n^3)\n(m^2-n^2)(m^4+m^2n^2+n^4)\nm^4-n^4=(m^2+n^2)(m^2-n^2)\nplugging in\nm^6-n^6/m^4-n^4\n=(m^4+m^2n^2+n^4)/(m^2+n^2)
403642
just remeber, if you want to know if any number is divisible by 4 just check to see if the last two digets are divisible by 4... that should greatly help in solving this problem\n\nNow if you would please, stop asking all of your homework questions
287547
Condition: The other number is different from 7.\n\nLet n = the other number\n\nSolution:\n\n7 / n = n / 7\n\nn^2 = 49\n\nn^2 - 49 = 0\n\n(n + 7)(n - 7) = 0\n\nTherefore:\n\nn + 7 = 0...........or...........n - 7 = 0\nn = -7..............or...........n = 7 (reject)\n\nTherefore, the other number is -7 ANSWER
423594
(3x^3-7x^2+x-20)/(x-3) = 3x^2+2x+7 with a remainder of 1\n         \n         3x^2+2x+7 \n     _______________\nx-3 | 3x^3-7x^2+ x-20\n      -3x^3+9x^2\n    __________\n          2x^2+ x\n         -2x^2+6x\n         ________\n              7x - 20\n             -7x + 21\n             ________\n                   1\n\nI tried to show what this would look like as a division problem but it keeps jumbling up.
786009
-2(x-3) + 4(x+7)\n-2x+6   +4x+28\n2x+34
598513
n + (n+1) + (n+2) = 102\n3n + 3 = 102\n3n = 99\nn = 33\nn+1 = 34\nn+2 = 35\nsecond integer = 34
884048
n+n+2+n+4=201\n3n+6=201\nadding -6\n3n=195\ndividing by 3\nn=65\nthe numbersare 65,67 and 69
195989
Just by inspection, if n=3 the expression is divisible by 7. A more thorough examination shows, though, that n = 3m works for all integers m>0, i.e. n=3,6,9,12,15,18,21,... will all be divisible by 7.\n\nUPDATE: n can not be any real number because the term "divisible" in this context has a precise mathematical meaning, i.e. when you divide by 7 you will get a natural number as the result. The use of the variable n usually suggests too that n belongs to the set of naturals.
357472
Use the formula S(n) = k x (1- r^n+1)/(1-r) where k=6, r=3 and n = 17....the answer is 1,162,261,464
495174
if u meant {4/(q^2+2q+1)}- {q/7(q+1)}\nthe answer is\n\n(28-q^2-q)/{7*(1+q)^2}
473264
all factorials greater than 6! divide evenly since they contain 7\nso add 1! to 6! and divide by 7, find remainder\n=1!+2!+3!+4!+5!+6!\n=1+2+6+24+120+720\n=873/7\n=126 R 5\n\nanswer is 5
793566
10 - 7 = 3\n\ntherefore 3 x 4 (for the 1/4) = 12\n the number is 12
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too