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Education / Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by Dane17: 4:33pm On Nov 22, 2013 |
congrats JARYEH our Maths genius |
Education / Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by Dane17: 10:35am On Nov 22, 2013 |
Goodluck Contestants. |
Education / Re: Nairaland Mathematics Clinic by Dane17: 4:09pm On Nov 21, 2013 |
Laplacian:. I think it's cos u can't start wit zero. if it start with an odd numba dere are more even numba form and vice versa( i.e when dere are equal odd and even numba). cos d numba is keep constant and dere will be n odd and n-1 even to occupy d last position or vise versa |
Education / Re: Nairaland Mathematics Clinic by Dane17: 3:19pm On Nov 21, 2013 |
Laplacian:. I'm not saying I'm right either I just stated an observation. for 0,1,2,3 u get 10 even (102,120,130,132,210,230,302,310,312,320) and 8 odd (103,123,201,203,213,231,301,321). ur oda formula should be correct as 4P3-3P2=18 |
Education / Re: Nairaland Mathematics Clinic by Dane17: 2:38pm On Nov 21, 2013 |
Laplacian: @smurfy. d 6 and 5 digit will not have d same numba of odd and even numba. let's use ur method for 0,1,2. from ur form numba of 3 digit will be 3P3-2P2=4 and numba of even numba would be 4/2=2. But dere are 3 even(210,102,120) and one odd (201 numba. |
Education / Re: Nairaland Mathematics Clinic by Dane17: 1:36pm On Nov 21, 2013 |
smurfy:. Thanks *smiling* |
Education / Re: Nairaland Mathematics Clinic by Dane17: 1:28pm On Nov 21, 2013 |
since y= x/x-1 and x and y are integers and non zero then x=2 and so is y. since 2 is d only case that x-1 is a factor of x. |
Education / Re: Nairaland Mathematics Clinic by Dane17: 1:23pm On Nov 21, 2013 |
smurfy:. Ya I forgot . U are right. |
Education / Re: Nairaland Mathematics Clinic by Dane17: 1:16pm On Nov 21, 2013 |
smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]. (6 digit ending with 0 can be arranged in(cbai)) 5×4×3×2=120 ;(6 digit ending with 2 or 4 cbai) 4×4×3×2=96 each. total six digit =96+96+120=312. total five digit numba is also 312. (4 digit ending with 0 and 2 cbai) 2×4×3= 24 each ; (4 digit ending with 4 cbai) 1×4×3=12 ; total 4 digit= 24+24+12=60. Total numba greater dan 4000 dat can be formed is 60+312+312=684 |
Education / Re: Nairaland Mathematics Clinic by Dane17: 12:37pm On Nov 21, 2013 |
factorial1: Brov, u aint told to do try and error, laplacian asked for a proof, and I've done that. #Cheers. u manipulated . I don't get dis part (Therefore, add eqn (ii) and (iii)...(i.e x + y) together and equate it with their product(i.e xy), x + y = xy, xy - x + xy - x y = xy, which is equal to 2xy - 2x = xy,). how did u arrive at 2xy-2x=xy. Finally d answer to his question is not just 2 and 2 . they are many values like 3 and 1.5 which gives 4.5 . look at my previous solving carefully and u will understand |
Education / Re: Nairaland Mathematics Clinic by Dane17: 12:21pm On Nov 21, 2013 |
smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]. In solving ur question do I assume that 023514 is different from 23514 |
Education / Re: Nairaland Mathematics Clinic by Dane17: 12:09pm On Nov 21, 2013 |
every value of x has its y value which is (x/x-1) for d equation to be true provided x =/= 1(provided x is not equal to 1) |
Education / Re: Nairaland Mathematics Clinic by Dane17: 12:05pm On Nov 21, 2013 |
Laplacian: ...d sum of two non-zero integers is equal to their product...find d numbers..... x+y=xy ; x=xy-y ; x=y(x-1) ; y=x/x-1 ; therefore if x=2 ; y= 2/1 ; if x=3 ;y=3/2 if x=4 ; y=4/3 and so on. |
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