₦airaland Forum

Hi everyone!
I'm Richard Obinna Agbara. We diagnose and solve math problems here.
This thread is the meeting point for Nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

@kaxmytex, thank alot and happy new year too, do u av any new yr maths question 4 me?

@ Richiez,my question,is how can I become a mathematician,as in having distinctions in the subject.God bless u.

@richiez, what level of math are u talkin abt?

Zikstar4me: @richiez, what level of math are u talkin abt?

Good question.

Zikstar4me: @richiez, what level of math are u talkin abt?

the first poster is not bold enough to ask that kind of question. even those that are studying maths and applied maths in university cannot be bold enough to ask such question

Onaolamipo: the first poster is not bold enough to ask that kind of question. even those that are studying maths and applied maths in university cannot be bold enough to ask such question

.Pride goeth before a fall,if u prove to be too intelligent or wise,u will look miserable b4 a fool.Richliez said,ask him any question on maths,Dumb ass kids came here with unreasonable questions.

Maths gurus pls help me solve the simultaneous equations

x + y =5
x^x + y^y=31

Adol16: Maths gurus pls help me solve the simultaneous equations

x + y =5
x^x + y^y=31

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

@calebsworld, here are a fewtips;
1. you have to develop a great interest for maths, and have that strong desire to become a maths guru.

2. Realize that there's nothing magical about mathematics.

3. make friends with people who appreciate mathematics.

4. practise as many maths questions as possible, starting from the very simple ones to more complex ones. develop the urge to solve them on your own, but you can always seek for assistance whenever you get hooked.

5. do not forget any new thing you have learnt, keep reviewing them.

6. learn to memorize some basic maths formulas as this will build your confidence.

GOOD LUCK

Zikstar4me: @richiez, what level of math are u talkin abt?

both secondary and tertiary, infact anyone

Good thread.

Ok then, please can sm1 explain the pigeonhole principle to me?

Richiez: @calebsworld, here are a fewtips;
1. you have to develop a great interest for maths, and have that strong desire to become a maths guru.

2. Realize that there's nothing magic about mathematics.

3. make friends with people who appreciate mathematics.

4. practise as many maths questions as possible, starting from the very simple ones to more complex ones. develop the urge to solve them on your own, but you can always seek for assistance whenever you get hooked.

5. do not forget any new thing you have learnt, keep reviewing them.

6. learn to memorize some basic maths formulas as this will build your confidence.

GOOD LUCK

thanks boss,that's all I need,God bless u d more.

Youngsage: x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
logx + logy = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

Zikstar4me: Ok then, please can sm1 explain the pigeonhole principle to me?

well funny as the name sounds, the pigeonhole principle is a very important principle in mathematics. it states that if there are x<y and there are x pigeonholes and y pigeons, then at least one pigeonhole must contain more than one pigeon.

this principle is used in solving basic mathematical arguments, for example; if there are only 4 colours of shirts and five students are required to choose any colour of shirt that they prefer, then atleast two students must choose same colour.

wow.....maths is really interesting!

Youngsage: x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
logx + logy = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

Bro pls I don 't understand how you got the fifth equation.

olawalebabs: Good thread.

Calebsworld: thanks boss,that's all I need,God bless u d more.

thanks!

Youngsage: x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
logx + logy = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

Maths is frightening with equations like this

@richiez, was tryin to learn that theorem in a buk, though it is xplained to b easy as u jst said in the buk and d xamples are cheap, Bt the exercises are smtin else

Zikstar4me: @richiez, was tryin to learn that theorem in a buk, though it is xplained to b easy as u jst said in the buk and d xamples are cheap, Bt the exercises are smtin else

maybe you should post some of the exercises here

17 people corespond by mail with one another- each one with all d rest. In their letters only 3 diff topics are discussed. Each pair of corespondents deal with only one of these topics. Prove that there are at least three people who write to each other about the same topic.

Zikstar4me: 17 people corespond by mail with one another- each one with all d rest. In their letters only 3 diff topics are discussed. Each pair of corespondents deal with only one of these topics. Prove that there are at least three people who write to each other about the same topic.

you should note that most pigeon hole problems involves logical reasoning and not calculations.

SOLUTION
no of mail correspondents = 17
no of topics = 3
since the correspondents are in pairs, no of possible pairs = 8 (i.e only 8 pairs from 17)

let pigeons = no of pairs =8
and pigeonholes = no of topics = 3
if we assume that the letters wud be distributed evenly i.e pigeons wud be distributed evenly into the pigeon holes, then atleast all 3 pigeon holes will an extra pigeon
this implies that some pairs will have same topic as other pairs

Yea

U are da bomb @ Richlez,pls can u pls do some explanations and exercises on this assumption topic M.I(Mathematical Induction).thanks.

oh! mathematical induction...dat very interesting topic, i'l do some explanation in d morning, let me av some sleep...i like ur spirit

I dont understand this part pls explain

Richiez: if we assume that the letters wud be distributed evenly i.e pigeons wud be distributed evenly into the pigeon holes, then atleast all 3 pigeon holes will an extra pigeon
this implies that some pairs will have same topic as other pairs

okay what i mean is this; since there are 8 pigeons, the first 3 goes into each 3 hole, it then remains 5, in a similar way, we distribute another 3 to each 3 hole and son till all 8 pigeons get exhausted. by so doing, all 3 holes must have extra pigeons

this means at least 3 pairs must have same topic since each hole represents a topic

These maths looks weird.. Is dis university maths?

Calebsworld: U are da bomb @ Richlez,pls can u pls do some explanations and exercises on this assumption topic M.I(Mathematical Induction).thanks.

Mathematical induction is a method of mathematically proving that a statement is true for all natural numbers i.e positive integers
it usually involves 2 basic steps;
1. showing that the statement is true when the lowest natural number is used e.g n=0 or n=1 (preferably)
2. showing that if the statement is true for any natural number n=k, then it is true for the next natural number n=k+1

let us do some examples;

1. prove by mathematical induction 1+2+3+4+.......+n+(n+1) = n(n+1)/2
SOLUTION
note that n represents the number of elements in the series
step 1---we have to show that the statement is true for n=1
1= 1(1+1)/2
1= 2/2
1=1 therefore since L.H.S = R.H.S the statement is true for n=1
we can even go ahead to show that the statement is true for any other small value of n, say n= 4
1+2+3+4 = 4(4+1)/2
10= 4(5)/2
10= 20/2
10= 10 therefore it is also true for n=4
next is step 2
here we shall consider an arbitrary value of n, say n=k and assume it is true
i.e 1+2+3+4+....+k = k(k+1)/2 ................(1)
if the above statement is true, then the final step is to show that it is also true for n= k+1
1+2+3+4+....+k+(k+1) = (k+1)(k+1+1)/2
1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2 ...............(2)

to show that eqn 2 is true, let us add (k+1) to both sides of eqn 1
1+2+3+4+....+k+(k+1) = k(k+1)/2 + (k+1).............(3)
the job here is to make the R.H.S of eqn 3 look like that of eqn 2
taking l.c.m as 2
1+2+3+4+....+k+(k+1) = [k(k+1)+2(k+1)]/2
since (k+1) is a common factor in the R.H.S we factorize
1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2

This is same as eqn 2 above, hence the statement 1+2+3+4+....n+(n+1) = n(n+1)/2 is true for all natural numbers.

feel free to post more exercises.

marcangelo: These maths looks weird.. Is dis university maths?

yep, but don't be scared, any level of maths is accepted here