Laplacian:
answers mean nothin 2 me without steps...username is adokwu-ondoma

lol... Paying me back in my own coin, eh?

Laplacian: ...questn1; find a number which wen multiplied by 7 and d result added to 1 gives a perfect square
....questn2; List ALL d subgroups of a group of order 7
...questn3; find All integers satisfying; 3x+4y=5
....question4; black king on a8, white king on a6, white pawn on c7...white to move and mate in 1

........for question 3, there is an infinite number of values satisfying 3x + 4y=5, the first being x=-1 and y=2.....when obtaining the other values of x and y, it is observed that x increases by a constant amount of -4 while y increases simultaneously by an amount of 3 i.e: -5 and 5, -9 and 8, -13 and 11, etc

Alpha Maximus: ........for question 3, there is an infinite number of values satisfying 3x + 4y=5, the first being x=-1 and y=2.....when obtaining the other values of x and y, it is observed that x increases by a constant amount of -4 while y increases simultaneously by an amount of 3 i.e: -5 and 5, -9 and 8, -13 and 11, etc

The graph of the function is linear, hence having infinite solutions (unless the domain is restricted) for x and y.

There are 4 different letters addressed to 4 people whose addresses are written on four envelopes. How many ways are there of putting the letters in the envelopes such that
(a) 3 of the letters are in the wrong envelopes and 1 is in the correct envelope
(b) 1 of the letters is in the wrong envelope and 3 are in the correct envelope?

@Laplacian
Visit ur page to accept the game. You're on!

Wow!!!! I cnt believe dat am just seeing dis thread..thumbs up 2 u guys..av bin tinkin ,since we have xo many maths gurus in d house,y dnt u guys put all dis skills in2 AI(Artificial Intellegence)...mayb in d game design aspect...i made some research lately on d various applications of maths in real life,and 3D game design caught my attention,2 design 3D animations a huge amnt of d Knowledge of Maths and physics is highly needed,and u guys have proven ur worth,i really do respect u guys,Doubledx,jackpot,richie,etc u guyz cn give it a shot,wu knowz we kud be having a 3D game made bye our own Naija Brothers and sisters in d tyme beyond,u guyz kud giv it a try and see 4ur selfs..believe u me u all have got wt it takes 2 do it.if u need mor tips u kud visit d programming section of dis forum some dudes will be willing 2 partner wif u guys...this is just my sugestion ,ibeg make una no flog me ooo,am awear am digresing 4rm d main aim of dis thread and am v.sorry abt dat,am only trying 2b nice..ill check bak 2 c ur responses.thanks for reading.

Dekins88: Boss seems richiez is busy these days, he is now one of the moderators in the education sector, I have been following this thread from the inception. Boss it been a long time we've seen your handiwork on mathematics oooooo

Yep that's true! although i'v been following up every post. and would always be here whenever anyone calls on me. THanks to Doubledx and Laplacian for solving Jackpot's question. I was extremely busy but was very happy to know that my Generals are always up to the task.

Alpha maximus and smurfy, I've been watching you guys. keep up the good work.

Laplacian: ...G is a group of order 4, show dat G is abelian...

Soln.
Case 1. Suppose G has an element g of order 3. Then d cyclic subgroup generated by g contains three elements.{g,g^2,g^3=e}, where e is d identity. But the order of every subgroup must divide the order of G, nd dis is a contradiction. So G has no element of order 4.
Case 2. G is nt cyclic. Thus G={e,a,b,c} where e is d identity, nd each of a,b nd c has order 2. Nw let us take a look at d multiplicatn table.
i dnt knw hw to construct a table here. Buh here is a way 4ward. Draw a multiplicatn table.The ist row is (e,a,b,c), ist column (e,a,b,c) , 2nd row is (e,a,b,c), 3rd row is (a,e,c,b), 4th row is (b,c,e,a) nd d 5th row is (c,b,a,e). Nd nw by inspection, we see dat G is abelian. All is well.

rhydex 247:
solution.
Suppose we have 2 subgroups of G defined as H1 and H2. Nw the questn implies H1 intersectn H2 must also be a subgroup of G.
For any element a in H1 there exists a^-1 and H2 is closed. The same holds for H2. So the intersectn will only contain an element c in H1 intersectn H2. If c and c^-1 are in H1 and H2 they must contain e the identity of G thus H1 intersectn H2 cannot be empty. Hence the intersectn of any subgroups of a group G is a subgroup of G.

kudos 2 u...u did a nice wrk...

smurfy: There are 4 different letters addressed to 4 people whose addresses are written on four envelopes. How many ways are there of putting the letters in the envelopes such that
(a) 3 of the letters are in the wrong envelopes and 1 is in the correct envelope
(b) 1 of the letters is in the wrong envelope and 3 are in the correct envelope?

.......here we go:
(A) let A, B , C and D be the letters and let a, b , c and d be their respective envelopes
When correctly arranged we have;
Aa, Bb, Cc, Dd
When only one is in the right envelope and the rest are in the wrong envelopes, one of them can't be touched while the remaining three letters can be arranged in 3! =6 ways, since there are only 3 envelopes into which the 3 letters can be put , there are 6/3=2 ways this can be done when one of the letters is still in the right envelope; now, since during the course of these arrangements the four letters at some point in time will be put into their appropriate envelopes, the total number of ways this can be done is 2*4=8 ways
(B) this is an invalid question seeing as it isn't possible for only one of them to be in the wrong envelope unless we are assuming its own envelope has been discarded and it shares the same envelope with the letter matching the inhabited envelope meaning there would be two letters in one envelope

Alpha Maximus: ........for question 3, there is an infinite number of values satisfying 3x + 4y=5, the first being x=-1 and y=2.....when obtaining the other values of x and y, it is observed that x increases by a constant amount of -4 while y increases simultaneously by an amount of 3 i.e: -5 and 5, -9 and 8, -13 and 11, etc

...very correct...

Yo Smurfy check out my solution on the 'envelope' question and comment

rhydex 247:
Soln.
Case 1. Suppose G has an element g of order 3. Then d cyclic subgroup generated by g contains three elements.{g,g^2,g^3=e}, where e is d identity. But the order of every subgroup must divide the order of G, nd dis is a contradiction. So G has no element of order 4.
Case 2. G is nt cyclic. Thus G={e,a,b,c} where e is d identity, nd each of a,b nd c has order 2. Nw let us take a look at d multiplicatn table.
i dnt knw hw to construct a table here. Buh here is a way 4ward. Draw a multiplicatn table.The ist row is (e,a,b,c), ist column (e,a,b,c) , 2nd row is (e,a,b,c), 3rd row is (a,e,c,b), 4th row is (b,c,e,a) nd d 5th row is (c,b,a,e). Nd nw by inspection, we see dat G is abelian. All is well.

precise...thumbs up...

Alpha Maximus: Yo Smurfy check out my solution on the 'envelope' question and comment

Nice try, but both answers are wrong. The answer to (b) is 0. Remember the question says 'how many ways'. Try again on the first one.

smurfy:
Nice try, but both answers are wrong. The answer to (b) is 0. Remember the question says 'how many ways'. Try again on the first one.

....so technically I was correct for the second

Yeah, (un-)technically
You missed something on the first one, and I ain't giving any hint

lets keep solving problem!!!

smurfy: Yeah, (un-)technically
You missed something on the first one, and I ain't giving any hint

......
these are the only ways of scrambling three letters and putting one of them in the right place:
Aa Bd Cb Dc
Aa Bc Cd Db
Bb Ac Cd Da
Bb Ad Ca Dc
Cc Ad Ba Db
Cc Ab Bd Da
Dd Ab Bc Ca
Dd Ac Bc Cb...........that's 8 different formats? what did I miss in this question!!! somebody help!!

@Maximus
Correct! I missed out on something earlier. Good job!

...@Smurfy...all d pieces are alignd vertically on a straight line...how do i make it rectangular?...

M47h3m471c5.1ncr353 y0ur 1Q.17.3nh4nc35.your.cr3471v17y..57u6y m0r3 0f 17.bu7.d0n7.b3.20b355355.w17h.17

benbuks: M47h3m471c5.1ncr353 y0ur 1Q.17.3nh4nc35.your.cr3471v17y..57u6y m0r3 0f 17.bu7.d0n7.b3.20b355355.w17h.17

Mathematics increases your IQ...it enhances your creativity...study more of it but dont be obsessed with it...

Laplacian: ...@Smurfy...all d pieces are alignd vertically on a straight line...how do i make it rectangular?...

Change ur phone's image quality settings to high or something like that. That's what I did to my operamini.

The math general in the house,

Nobody has done justice to the third question posed by mikebis on page 57

I believe dat he extracted the question from a textbook called bouday

u^2/v + v^2/u =12

1/v + 1/u = 1/3.

smurfy:
Change ur phone's image quality settings to high or something like that. That's what I did to my operamini.

dat's been done succefully...how do i make a move?

Laplacian:
dat's been done succefully...how do i make a move?

Now click on the piece u want to move, wait a moment, click on where u want to put it, wait a moment then click Submit.
Note: You can play multiple games at the same time.

d citizen:
The math general in the house,

Nobody has done justice to the third question posed by mikebis on page 57

I believe dat he extracted the question from a textbook called bouday

u^2/v + v^2/u =12

1/v + 1/u = 1/3.

Reduced to x + y = 1/3 (1)
(1/x)^2y + (1/y)^2x = 12 (2). [x = 1/u, y = 1/v].
I'm no maths general though. I'm just here to learn from the Greats...

smurfy: @Maximus
Correct! I missed out on something earlier. Good job!

so I got it right?

d citizen:
The math general in the house,

Nobody has done justice to the third question posed by mikebis on page 57

I believe dat he extracted the question from a textbook called bouday

u^2/v + v^2/u =12

1/v + 1/u = 1/3.

i guess nobody attemptd it 'cos its straightforwrd...multipy both eqns by uv to get, u^3+v^3=12uv,
u+v=uv/3, recall dat
u^3+v^3=(u+v)^3-3uv(u+v), d first eqn bcoms
(u+v)^3-3uv(u+v)=12uv, substitut 4 u+v frm eqn1
(uv)^3/27-(uv)^2=12uv or
(uv)^2-27uv-324=0 or
uv=0 or 36 or -9 so dat from
u+v=uv/3
u+v=0 or 12 or -3, solvin d quadratics givs, u=6,-4.85 and
v=6, 1.85

d citizen:
The math general in the house,

Nobody has done justice to the third question posed by mikebis on page 57

I believe dat he extracted the question from a textbook called bouday

u^2/v + v^2/u =12

1/v + 1/u = 1/3.

i guess nobody attemptd it 'cos its straightforwrd...multipy both eqns by uv to get, u^3+v^3=12uv,
u+v=uv/3, recall dat
u^3+v^3=(u+v)^3-3uv(u+v), d first eqn bcoms
(u+v)^3-3uv(u+v)=12uv, substitut 4 u+v frm eqn1
(uv)^3/27-(uv)^2=12uv or
(uv)^2-27uv-324=0 or
uv=0 or 36 or -9 so dat from
u+v=uv/3
u+v=0 or 12 or -3, solvin d quadratics givs, u=6,-4.85 and
v=6, 1.85