what is the remainder left after dividing?
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
1105851
8....it's half.\n\nEach reduce down to 1/2.
545599
subract 32 then divide by 1.8 (or mulitply by 5/9 --- same thing)
275765
you need to know the relation that the greeks found thousands years ago Δ=δπ+υ. this is the identity of the incomplete division Δ is the number you want to divide,\nδ is the divider, π is the quotient and υ is the remainder.\nyou just can´t round up or down the remainder
235381
in dividing fractions, you just multiply them by its reciprocal.\n\nso \n\n[(x+5) / (16x)]  * [ (4x ^2) / [(3x -2)] \n\nperform the operations: \n\n[ (x+5) / 4 ] * [ x / (3x-2) ]\n\nthe answer is:\n\n[x(x+5)] / [ 4(3x-2) ]
823408
Your ratio is 1:8.15          You fid this out by starting at 13:106 and work yoiur way down to 1. \n\nyou divide both sides by 13 to get your answer.
716901
You can use polynomial long division to find the value of k. Be very careful! For the first expression, after dividing the numerator by the denominator a remainder of  8 - 4k  is obtained. Set this equal to zero to find the desired value of k.\n\nTo review polynomial long division, see:\n\nhttp://en.wikipedia.org/wiki/Polynomial_long_division
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.
110638
use a calculator, or do it the old fashion way. first u divide 13 into 80 which euals 6 with a remainder of 2 then u bring the 9 down and 13 goes into 29 2 times. the answer in 62 with a remainder of 3or 62 3/13
357131
It is a problem in long division with a remainder. You need to simply divide the whole integers (everything before the decimal point) and keep the remainder as the inches.  Thus your problem becomes 20 feet and 6 inches, you can check your answer by reversing the problem.  In 20 feet there are 20 times 12 inches, or 240 inches with 6.138 inches left over.  Thus the answer is 20 feet and 6 inches (rounding to whole inches as in your example). \n\n-LOVE YOUR NIEGHBOR AS YOURSELF
545361
u can't get a remainder only a decimal
230715
We know that 2^10=1024.1024/11 leaves 1.Let x=2^10.Since 11 is a factor of (x-1) which itself is a factor of (x^n-1) (to prove this put x=1 in x^n-1).Now 2^(10*n)-1 divides 11.\n        Its double 2^((10*n)+1)-2 exactly divides 11.\n        Hence remainder is 2 in the given question.
190233
When you divide by 3, the remainder must be 0, 1 or 2. \nThen the remainder is 1 , since -5 = 3x(-2)+1\n\nThere are many incorrect answers here, so I will explain clearly.\n\nThe definition of the division of the integer  is :\n\nIf a = b*q + r   (0 ≤ r < |b| ),  then a ÷ b = q remainder r\nIt means the remainder must be non-negative integer.\n\nWell -5 = 3 * (-2) + 1,  then the remainder must be 1.\n\n-2 is wrong !!!!\n\nsource : I am a maths prof.
670503
The answer is x2+2x-4 with a remainder of 11/(x+1)
915778
1) 6.4/100\n2) 100|6.400..since 100 cannot divide into 6, nor 64, you have to move the decimal point over until you reach a whole number that is divisible by the denominator (100). \n3) 0.0640\n\nI hope this clarified the problem for you!
810284
i got 2x^2-x-3 R -4\n\nfirst divide 2x+5 into 4x^3+8x^2-11x-19 and you get 2x^2 then you subtract and your left with -2x^2-11x-19 and then you divide that by 2x+5 and you get -x add that to 2x^2 and you get 2x^2 - x then you subtract from your original and get -6x-19then divide 2x+5 into that and you get -3 ((2x^2-x-3)) multiply that by 2x+5 and you subtract from -6x-19 and you get -4 you cant divide 2x+5 into that so -4 is your remainder
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6