use prime factorization to reduce 540/600?
914238
540 = 2x2x3x3x3x5\n600 = 2x2x2x3x5x5\nso\n540/600 = 2x2x3x3x3x5 / 2x2x2x3x5x5\n= 3x3 / 2x5 = 9/10\n.........
240962
i think this will work...\n\nint prime=43;\nint isprime=0;\nfor(int i=2 ;i<prime;i++)\n{\n   if((prime%i)==0))\n   {\n       isprime=1;\n       break;\n   }\n}\nif(isprime==1)\n   printf("Not prime");\nelse\n   printf("Prime !");\n\n\n=====\nGot it na??
882012
What you want to do is factor out the same values from the numerator and denominators.  \n\nThe order in which you factor these out does not matter, the key is to continue to factor them out until there is nothing else to factor out.\n\nFor example you can factor out a^8 by turning it into:\n\n36 * (a^8) * (b^4) * 1/2\nover\n27 * (a^8) * a * (b^6) * 1/3\n\nsince (a^8) over (a^8) = 1 you can cancel them out leaving you with:\n\n36 * (b^4) * 1/2\nover \n27 * a * (b^6) * 1/3\n\nwe do the same factoring for b where we can factor out a (b^4):\n\n36 * (b^4) * 1/2\nover \n27 * a * (b^4) * (b^2) * 1/3\n\ncanceling the (b^4)s we get:\n\n36 * 1/2\nover\n27 * a * (b^2) * 1/3\n\nYou can then further reduce the fraction by multiplying out the 36 * 1/2 and the 27 * 1/3:  leaving you with:\n\n18\nover \n9 * a * (b^2)\n\nAs you can see now you can factor out a multiple of 9 leaving you with:\n\n2\nover\na*b^2
807304
1/30 n^5 - 5/8 n^4 + 53/12 n^3 -107/8 n^2 + 371/20 n - 8 works.\n\nI don't think you'll find anything nicer.
735862
PART A:\n\nFactor out the common 2.\n2(x² - 50)\n\nUnfortunately, 50 isn't a perfect square so if you are looking for integer factorization, you wouldn't factor further.   The answer would be:\n---> 2(x² - 50)\n\nNote: if you do use the difference of squares, then you would have sqrt(50) = sqrt(25) * sqrt(2) = 5 sqrt(2).   The final factoring would be:  \n---> 2 ( x + 5 sqrt(2) ) ( x - 5 sqrt(2) ) )\n\n\nPART B:\n\nThe integer factors of 3 are 1 and 3, so start with:\n(3x ... )(x ... )\n\nNow the integer factors of 5 are:\n1, 5 or -1, -5\n\nYou need a negative sign to get -8, so use -1 and -5.\n\nTry them in either order.  In one case you get -16x, the other gives -8x, so that is your answer:\n---> (3x - 5)(x - 1)
522039
Primes from 50 to 100\n53\n59\n61\n67\n71\n73\n79\n83\n89\n97
1121426
I don't know if this is what you are looking for but simplified this fraction would be 24/43 (divided by 3). And you cannot simplify it any further because 43 is a prime number, as far as I remember...
483853
You said "57 + 2 = 59"\nIt should be "57 - 2 = 55", which was what was spent.
165290
Yes - again I agree with Aseem (why is he so much faster than me?) :-)\n\nFirst you have to ask is 7 a prime number ; the answer is yes (see my answer to your other prime number question) so you go to the second test. Can you divide 60 by 7 with no remainder? Answer is no; so 7 is not a prime factor of 60. The prime factors of 60 are 2(2), 3, 5;\n\n60/2 = 30\n30/2 = 15\n15/3 = 5\n5 is a prime so you go no further; \nSo you end up with 2, 2, 3, 5
829703
I did check the above answerers answers, and they are wrong :P\na,b,c,d,e,f = 1,2,5,5,4,6.\nYou could go and solve this by the usual elimination methods, but for something this large its far better to use a calculator or online solver. Google for 'simultaneous equation solver' and you'll find  heaps.\n(Just to confirm above answers are wrong: for those numbers you get 4a + b + 5d + 2e + 3f = 69, not 57.)
387848
Still considered poor. Any score below 600 is in hot water below 620 considered sub prime. Try to boost it over 700 and the world will be yours. ;)
151147
2/3 x 45/60\n\nSteps:\n1)Reduce 45/60(divide by 15)=3/4\n2/3 x 3/4\n2)Multiply across(2 x 3 on the top, 3 x 4 on the bottom)=\n6/12\n3)Reduce(divide by 6)=\n1/2\n\nNOTE:To make it faster you can cross-cancel.\n(the 2 and 4 would become 1 and 2, the 3 and 3 would become 1 and 1) This is not necessary.
997626
http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150\n\nhttp://www.math.sfu.ca/~cbm/aands/subj.htm\n\nhttp://clusterfie.epn.edu.ec/ibernal/html/CURSOS/AbrilAgosto06/Metodos/Libros/Fundamentals%20Of%20Numerical%20Computing%20-%201997%20-Wiley%20%20(By%20Laxxuss).pdf
1079450
FRIST TRY PUTTING THE 512 IN THE SYS 1ST THEN THE 256 THEN THE 128. IN OTHER WOEDS START WITH THE HIGHEST FRIST THEN WORK UR WAY DOWN
777032
Since there are too many numbers to list, here's a program to do it for you:\nhttp://wims.unice.fr/wims/en_tool~algebra~factor.html
713027
Just try out small numbers. 1, 2, 3, 4, 6, 8 all work. Once you start doing anything higher, the second factor becomes smaller again, so the others are 96/8, 96/6, 96/4, 96/3, 96/2, 96/1.\nIn all: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.
452386
7*6*5*4 = 840 if order matters.\n\n7*6*5*4 / 4! = 35 if order does not matter