what is the greatest number of times you would regroup when multiplying a 3 digit factor by a 2 digit factor?
970698
(a + b)(c + d + e)\n= a(c + d + e) + b(c + d + e)\n= ac + ad + ae + bc + bd + be\n\nOR\n\n(c + d + e)(a + b)\n= c(a + b) + d(a + b) + e(a + b)\n= etc.
761643
How many at a time?  You didn't say.\n\nThe number is an expression called  nCr, which is \nn!/((n-r)!r!), but it's pretty simple to think of it this way:\n\nI'll suppose you want them 3 at a time.  Then, for the first one, with 6 letters to choose from, write 6*5*4 on top (three factors because we want 3 letters) and 1*2*3 on the bottom\n6C3 = 6*5*4/(1*2*3)\n        = 20\n\nIf there had been 10 letters and we wanted to know the number of combinations 4 at a time, we would put four factors on top and bottom:\n10C4 = 10*9*8*7/(1*2*3*4)\n          = 210
155698
2/3...  multiply 1/3 by 2.
691350
3)4320\n(4!/2!)*(6!/2!)
53474
If you break it up it is easier to do in your head\n2 x 1000\n2000\n+\n2 x 200\n400\n2400\n+\n2 x 20\n40\n2440\n+\n2 x 6\n12\n\n2452
97133
7,248,000,000.\n\nLook at the digit of the place you want to round to.\n\n7,248,193,675.\n------^\nLook to the immediate right. If this digit is 4 or less, keep the digit of the place you want to round to the same and make everything else a zero. If this digit is 5 or more,   make the digit of the place you are rouding to 1 greater, and then make everything else a zero. It is like seeing whicho ne the number is closer to.
492512
6 - after two re-counts.,  Very clever!\n\nI vote more, also!\n\nFun!
133768
Assuming each digit is used no more than once in any given integer, this one would be fairly simple (and perhaps time-consuming) to draw out on paper:\n\n1. All integers have to end with the 3. For single-digit integers you of course have:\n\n3\n\n2. For two-digit integers, you have:\n\n43\n63\n83\n\n3. Three-digit integers:\n\n403\n603\n803\n463\n643\n483\n843\n683\n863\n\n4. Four-digit integers:\n\n4063\n4083\n4603\n4803\n6043\n6083\n6403\n6803\n8043\n8063\n8403\n8603\n8463\n8643\n4683\n4863\n6483\n6843\n\nI count 31.
670692
Good question! Easy to state but unless you are familiar with this type of problem (which I was not) it is tricky. \n\nYou really want to know? The answer is:\n\n573,147,844,013,817,084,100\n\nThis is how I started. It's easy to prove it's > 2^50\n\n100/2 = 50  So you can take 2 steps at a time anywhere from 0 to 50 times. 0 would be that you climbed the 100 steps 1 at a time. If we let r equal the number of times (from 1 to 50) that you take 2 steps at a time, and for the sake of simplicity we will assume that we can only take 2 steps either from the 0th step or from an even step, then for each value of r the number of distinct ways of doing it are given by the binomial coefficient nCr = n!/(n-r)!r!. So 50C1 = 50; 50C2= 1225; 50C3 = 19,600 etc..  \nNote: in reality there are 98 ways you can combine 1 two-step with 98 1-steps so 50C1 = 50 is too low. However,let's continue. \nWe have to sum up all of these coefficient values.\nBut this is not hard to do because the sum of nCr for r= 1 to n is 2^n (2 raised to the power of n). I am not going to prove this; that's already been done. but you get the idea from the following expansions:\n\n2^3 = (1+1)^3 = 1+3+3+1 =8\n2^4 = (1+1)^4 = 1+4+6+4+1 =16\n2^5 = (1+1)^5 = 1+5+10+10+5+1 =32 and so on...\n\nWhen we use (1+1)^n as a specific case of (a+b)^n we just get the binomial coefficients and we that they sum to a power of 2. So we know the answer is greater than 2^50.\nYou will be closer if you add 1+98C1+96C2+94C3+...+54C48+52C48+50C49+1\n\nBut there is still a problem..because you have to subtract some because of starting on an odd step..\n\nAnyways, I realized it was problem in restricted partition compositions so I went to this site and they had a calculator and I used it. That's how I got the answer lol!\n\nhttp://www.btinternet.com/~se16/js/partitions.htm
295676
My dear sweet Franklin. You must simplify your fraactions. That would be 1 in every 248,000
372976
yes u can\nt^2(s^4-4)is the factored form\nt^2 is the highest common factor\n\n\n75x^2 - 147y^2\n3 is the common factor\n3(25x^2-49y^2)\n3(5x^2-7y^2)(5x^2+7y^2) is the factored form of 75x^2 - 147y^2
24732
Maybe you will get a mixture of them until some of them ofcourse will reeact. It depends onthe method and the sequence of addition. It is a very bad idea anyways to do so!!
115942
Break each number into prime factors (factors that are prime numbers).\n12=2*2*3\n16=2*2*2*2\nThen, counting each duplicate factor once, multiply those factors together.\n2*2*3*2*2=48
757625
1) \n\n1+2+3+4+5+....n +(n-1)+(n-2)+...+3+2+1...regrouping\n\n=2(1+2+3+4+5+....(n-1)) +n...regrouping the term in parenthesis\n\n=2( (1+n-1)+(2+n-2)+(3+n-3)...)+n\n=2(n+n+n+...(n-1) terms)+n\n=2(n)(n-1)/2 +n\n=n^2-n+n\n=n^2\n\n\n2)\n 1=10^0 and 11=(10^1+10^0) and 111=(10^2+10^1+10^0)...\n\n(10^n+10^(n-1)+....10^2+10^1+10^0) times itself\n=(10^0+10^2+10^3+....10^n)^2\n=10^2n+2*10^(2n-1)+.....\n\n\n3)\n\nn^2-n=n(n-1)...one of them must be even and is therefore divisible by two.\n\nn^3-n=n(n^2-1)=n(n+1)(n-1)=(n-1)(n)(n+1) or three consecutive integers.\n\nIf the product is three consecutive integers, then one of them must be divisible by three and therefore the product is divisible by three.
1170732
6 SIX\n\nLAST THREE ARE CONSECUTIVE
939247
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
819558
Theres 7 choices for the first digit, 6 more for the second .. 2 for the sixth. So 7*6*5*4*3*2 = 5040.