congrats JARYEH our Maths genius

Goodluck Contestants.

Laplacian:
thank u very much bro....

Why are d even numbers more dan d odd numbers?....**:ojust thinking:/**

. 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

Laplacian:
i do not claim dat my solution is right but u hav decisively justified my approach...using my approach u correctly predicted d four numbers, but d odd and even numbers are not equal because my FIRST OBSERVATION is not satisfied; 0,2(even) and 1(odd)...
Try it d numbers 0,1,2,3 and c

. 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

Laplacian: @smurfy
2 ansa ur ques i'll first mak a few observations

1.) d numbers 0,1,2,3,4,5 wen used 2 form numbers greater dan 4000 will contain equal number of even numbers as odd numbers because the last digits can each be in 3 ways; 0,2,4(even) and 1,3,5(odd)

2.)if u are to arrange a set of n numbers and u do not desire 0 to
start d digit, then first arrange d numbers i.e nPr
Then keep zero constant as d first number and arrang d rest i.e
(n-1)P(r-1)
If now u take their differenc
i.e nPr-(n-1)P(r-1)
Then the result is d number of ways of arrangin d numbers without zero appearin as first digit

Now Ur Solution

ARRANGE IN SIX DIGITS
number of ways of arrangin 0,1,2,3,4,5 in six digit witout zero appearin first is
=6P6-5P5=600ways

ARRANGE IN FIVE DIGITS

6P5-5P4=600ways

ARRANGE IN FOUR DIGITS

in dis case 0,1,2,3 MUST not occur as first digit, so number of ways of arrangement
=6P4-4*(5P3)=360-4*60
=120ways

in summary, number of ways of formin digit wit 0,1,2,3,4,5 so dat d number formed is greater than 4000=600+600+120=1,320ways
finally if d numbers formed must be EVEN, then number of ways
=1320/2=660ways

. 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.

smurfy:

Beautifully correct! Wow! I love your answer. Now, for the sake of others, let me explain the steps below...

. Thanks *smiling*

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.

smurfy:

The numbers are integers, so only 2 and 2 qualify as answers.

. Ya I forgot . U are right.

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

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

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

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)

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.