there are two men waiting to go upstairs with an escalator. one takes steps three times faster then the ohter. they move step by step and begin to move at the same time. when the faster one finishes the esclator he takes 75 steps and meanwhile the slow one took 50 steps. so the question is how many steps are there in he esclator.
120528
This was the best question I have seen on this site.  (That isn't saying much, so let me say it this way:  This is an excellent question!)\n\nThe answer is 100 steps. (Not considering the hidden steps since it's constantly revolving).\n\nFirst we have to consider the ACTUAL speed of each man since the escalator is moving, too.\n\n(Fast Speed) + (Escalator Speed) = (Fast Adj. Speed)\n(Slow Speed) + (Escalator Speed) = (Slow Adj. Speed)\n\nAnalyzing that if the fast man had been able to climb the entire time that the slow man did, he would have gone 150 steps, we can see that he finished in half the time it took the slow man (because 75/150 = 1/2).  Therefore we also have:\n\n2 (Fast Time) = (Slow Time)\n\nWe also use the simple equation defining speed (s=d/t) to determine that:\n\n(Fast Adjusted Speed) = (Distance) / (Fast Time)\n(Slow Adjusted Speed) = (Distance) / (Slow Time)\n\nwhere (Distance) is equal to the actual distance traveled (eg. the distance from the top of the escalator to the bottom of the escalator).\n\nInserting the few givens and abbreviating the variables, we have, respectively:\n\nEq 1) 3 + ES = FAS\nEq 2) 1 + ES = SAS\nEq 3) 2 FT = ST\nEq 4) FAS = D/FT\nEq 5) SAS = D/ST\n\nThis looks like 6 variables with only 5 equations, but it is still actually solvable.\nUsing Eq.3 to simplify Eq.4 and Eq.5:\n\nEq 4) FAS = D/FT\nEq 5) SAS = D/2FT\n\nand we can reduce our number of variables by simply saying that X = D/FT, hence,\n\nEq 4) FAS = X\nEq 5) SAS = X/2\n\nSo now we use these to analyze Eq.1 and Eq.2:\n\nEq 1) 3 + E = X\nEq 2) 1 + E = X/2\n\nTwo variables, two equations.  E turns out to be 1.  This means that the escalator moves at the same speed as the slow man (remember that we assigned his abitrary, but proportional speed to be 1 in the equations).  If he and the escalator are both the same speed, and he contributes 50 steps to the journey, then the escalator must also have contributed 50 steps, totalling 100 steps.\n\nOr guess and check works pretty well, too.
384128
1st person shakes with 7, next with 6 more etc\ntotal is 7+6+5+4+3+2+1=7*8/2=28 handshakes.\nfor n people there will be n(n-1)/2 handshakes.
493187
1st step: OK we need to know how many times  faster the car went on the way to the service station than on the return trip:\n49/16=3.0625 times faster on the way up than the way back. \n\n2nd step: find the how much time it took on the way up and the way back. \n\nNow comes the Algebra:\nif we let X stand for the way up, the equation for the round trip would be:\n X+ 3.0625X= 4\n4.0625X=4\nX=4/4.0625\nX=.985 \n\nso the time up = .985 hours and the time back = 4 -.985, or 3.015 hours.\n\nso how far is the station? well, it's 49km/h x .985 h, (using the data for the trip to the gas station) or:\n\n48.265 km\n\nok we're DONE!!\n\nPS: I have a sneaking suspicion that if you look at the problem again, you might find it said 48 km/h on the way up, which would work out to be that the trip was EXACTLY 3 times faster on the way up. ok just a hunch. But from what you gave me, that is the answer.
150645
1>INPUT 1\n2>STORE 1 in A\n3>STORE 1 in B\n4>ADD A to B\n5>STORE A\n6>DISPLAY A\n\nThat's total of 6 steps. If you have 2 different numbers than you will have one more step of INPUT 'X' so it will be total 7 steps in that case.
441266
Int(sqr(2y))\n\nIf you double the blocks, that gives you an 'almost' square since the diagonal of the incline made by the blocks would be 45 deg.  So take the square root of that to get an estimate of the number of rows (steps).  Take the integer of that number to get actual total steps.\n\nint(sqr(2*1)) = 1\nint(sqr(2*2)) = 2\nint(sqr(2*3)) = 2\nint(sqr(2*4)) = 2\nint(sqr(2*5)) = 3\nint(sqr(2*6)) = 3\n\nthis gives a close answer but apparently not full proof.  if you follow this idea further you may find a better equation.\n\nThe problem is when you double the number of blocks, you end up with extra row of blocks beyond a true square.\n\nInt(sqr(2y - Int(sqr(2y))))\n\n1 = 1\n2 = 1\n3 = 2\n4 = 2\n5 = 2\n6 = 3\n\nso this second equation looks good.  basically, I'm subtracting out the extra row.  I don't know if you are allowed to use the Integer function in your calculation but works good for programing.\n\n         1\n        11\n       111\n      1111\n     11111\n    111111\n   1111111\n  11111111\n 111111111\n1111111111
470985
1st contact with floor\nS1 = 8\n\n2nd contact with floor\nS2 = 8 + 2*3/4*8\n\n3rd contact with floor\nS3 = 8 + 2*8*3/4 + 2*8*(3/4)²\n----------------------- ----------------------- ----------------------- \nnth contact with floor\nSn = 8 + 2*8*3/4 + 2*8*(3/4)² + ..... + 2*8*(3/4)^(n - 1)\n= 2*8  + 2*8*3/4 + 2*8*(3/4)² + ..... + 2*8*(3/4)^(n - 1) - 8\n= 2*8(1 - (3/4)^n)/(1 - 3/4) - 8\n= 64(1 - (3/4)^n) - 8\n\neg After 10th bounce\n\nS10 = 64(1 - (3/4)^10) - 8\n≈ 52.4m\n\nThe limiting distance is 64(1 - 0) - 8 ie 56 m
368733
add 4, divide by 4\nadd 3, divide by 3\nadd 2, divide by 2\nthe answer is 3
520743
t = d/r\n\nAndrei: t = 50/7.7 = 6.49 sec\nBrother: t = 45/6.5 = 6.92 sec\n\nSo Andrei wins\n\nBrother runs 42.21 m (6.49 x 6.5) so he still had 2.79m to go\n\n(rounded answers ok I hope)
240586
a) That means the number has to start with 5\nso you only need to choose the second and third digits.\nThere are 4 choices for the second and 3 choices for the third\n4*3 = 12\nb) The first digit must be 1 or 2\n2 choices for the first 1\n4 choices for the second \n3 choices for the third\n2*4*3 = 24\nEDIT: I read the part b incorrectly. This is the answer if no repetitions are allowed.\nIf repetitions are allowed, then you still have 5 choices for the second and third digits.\n2*5*5 = 50
366261
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour
462322
maximum number of steps = N\n\nsince ther r N arrows, N circles u cannot make more than N steps.
185784
How many ways to pick 5 from 20 if order isn't important?\nAnd, how many ways to pick 2 from 7 if order isn't important?\n\nFor the second question, you have... 1,2 1,3...1,7 or 6 that include 1. Then 2,3 2,4... 2,7 which is 5 more. Then 3,4 3,5...3,7 which is 4 more. Keep doing this and you will see the answer is 6+5+4+3+2+1 = 21\n\nFor the first part 5 out of 20, note that the total number of arrangements of 20 things is 20! which is 20 x 19 x 18 x17 ..... x 2 x 1. But for every arrangement that you take the first 5 of, there will be 15! ways to arrange the rest.\nAnd for those first 5, you can get them in any order which means that 1 2 4 3 5 is the same for you as 2 3 1 4 5. So, you need to divide the 20! by both 15! and 5! which looks like this-  20!/(15! x 5!) and equals 15,504.\n\nNow for each pick from the cable houses (15,504 possible) you have 21 different ways to pick a no cable set. Therefore, multiply 15,504 by 21 to get your final answer. 325,584
210639
Zeno ignored the fact that the infinitely many smaller steps take infinitely less amounts of time to do.\n\nAs the earlier answerer pointed out, sum the power series and you get a finite value for the time taken...\n\nLike a lot of these logical tricks, there's something being forgotten or ignored to make it look impossible.
351485
As the elevator moves up with an increase in the weight of the student from 849N to 942N, there was a net force upward acting on the student. This net force equals 942-849=93N. If the scale read 849 originally when the elevator was at rest then the mass of the student must have been 849/9.8=86.6kg. Now solve for the acceleration using the formula F=ma, where F is the net force, m the mass of the student, and a the acceleration.\n\nF=ma\n93=86.6a\na=93/86.6\n  =1.07m/s^2\n\nTo bring the reading back to 849, will need a net downward force of 942-849=93N. m remains the same. With the same net force as in the above, but in the downward direction, the acceleration must be the same in magnitude, i.e, 1.07m/s^2. But this time it is deceleration because the elevator would be slowing down, the net force being in the downward direction.\n\nThe elevator decelerates as it approaches the 74th floor because of the lower reading of 774N. The net force is 849-774=75N downward.\n\nF=ma\n75=86.6a\n a=75/86.6\n   =0.86m/s^2.
91142
With the right number of steps forward and enough time you will eventually reach the edge and fall. The more steps you take the closer you get to the edge but never quite reaching it because your distance to the edge is merely halved with each step taken. So, the greater the number of steps forward, the closer to the edge.\nThe instant you make step number infinity you will have reached the edge. The problem is when and how many steps will that take. This is one eccentricity of infinity.
284387
7.\n\nThe last attendant got 1.  1/2(1)+1/2 = 1.\nThe second attendant got 2.  1/2(3)+1/2 = 2.\nThe first one got 4.  1/2(7)+1/2 = 4.
73776
One correction to Duane G's response. When you multiply the bottom by sqrt(3) you get sqrt(3) * sqrt(3) = 3, not 1. Therefore your final answer would be 2 * sqrt(6).