₦airaland Forum

Humphrey77: @ lapacian show it

Humphrey77: if 3 raise to power 1998 is divided by 5 what is the remainder

Mr Calculus: plz guys help me out wit dis......
if 3sin2@=5sin2¥....
find @ & ¥ if(tan@=2tan¥)

benbuks: ......M.O.G...try again o..incorrect.....9ce attempt sha..

benbuks: Compute the derivative of..
*] sqrt [1+sqrt(1+ sqrt(x)) ]

**] sqrt[ (x^2 +1)^2 +sqrt (1 + (x^2 +1)^2 ) ]
nb:"sqrt." Means square root

benbuks: Chairman,@laplacian..
Your boy dey hail o.

benbuks: 1/9=0.111111111. . .
2/9=0.222222222. . .
3/9=0.333333333. . .
4/9=0.444444444. . .
5/9=0.555555555. . .
6/9=0.666666666. . .
7/9=0.777777777. . .
8/9=0.888888888. . .

smurfy: Solve for all odd numbers? Okay.

Here goes...

4-DIGIT NUMBERS

The last digit must be 1, 3 or 5. The first digit must be 4 or 5.
Six scenarios!

(a) first digit is 4, last digit is 1
(b) first digit is 4, last digit is 3
(c) first digit is 4, last digit is 5
(d) first digit 5, last digit 1
(e) first digit 5, last digit 3
(f) first digit 5, last digit...

Ha! We can't use 5 twice, so scenario (e) doesn't count. That means we'll only consider scenarios (a) to (e) only.

(a) gives 1*4*3*1 = 12
(b) gives 1*4*3*1 = 12
(c) gives 1*4*3*1 = 12
(d) gives 1*4*3*1 = 12
(e) gives 1*4*3*1 = 12.

Adding the scenarios gives 5*12 = 60 ways of forming odd numbers greater than 4000 from the digits 0 through 5.

5-DIGIT NUMBERS

We have three scenarios here since we don't have to bother too much about the first digit. We only ensure the digit zero doesn't come first.

(a) ends with digit 1 gives 4*4*3*2*1=96
(b) ends with digit 3 gives 4*4*3*2*1=96
(c) ends with digit 5 gives 4*4*3*2*1=96.

So there are 3*96 = 288 ways of forming odd 5-digit numbers from the digits 0 through 5.

6-DIGIT NUMBERS

Three scenarios. Let me skip a step or two...

(a) 4*4*3*2*1*1 = 96
(b) 4*4*3*2*1*1 = 96
(c) 4*4*3*2*1*1 = 96

So, there are 3*96 = 288 ways of getting 6-digit odd numbers from the digits 0 through 5.

Final answer...

The number of odd numbers greater than 4000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is 60 + 288 + 288 = 636.

Hey, we obtained 684 for even and 636 for odd numbers. Why do we have less ways of forming odd numbers than even numbers?

It's partly because of the silly restriction placed on 4-digit numbers. It has cost the ODD NUMBERS team a whooping 48 points!

Dane17: . 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

Dane17: . 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.

Dane17: . 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.

factorial1: *clears throat*
let's go,
firstly, let the two numbers be 'x' and 'y'. From the qt, it can be infered dhat the sum of this non-zero integers is equal to their product: the mathematical representation is x + y = xy.....(i)
Now, from equation (i), make x the subject of the relation/formulae, the result is x = xy - y.....(ii), also, make y the subject of the relation from eqn (i), the result is y = xy - x....(iii). 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 = xy, which is equal to 2xy - 2x = xy, collecting the like terms, we have 2xy - xy = 2x, xy(2 - 1) = 2x, therefore xy = 2x, divide both side by x, we have xy/x = 2x=x, from there....y = 2. So, subitutute y=2 into any of the above equation. I use eqn (i) which is x + y = xy, x + 2 = 2x, collecting the like terms x - 2x = -2, -x = -2. Note: minus cancelled out, therefor x = 2. With this...I think u are satisfied sir? The two non-zero integers in which their sum is equal to their product is/are ''2'' and ''2''. Thanks

factorial1: The numbers are 2 and 2.

Laplacian: ...d sum of two non-zero integers is equal to their product...find d numbers....

smurfy: Here goes...

The last digit can be arranged in 3 ways, since it must end with 1, 3 or 5. Each of the first two digits can be arranged in 5 ways, since each digit can be used more than once. The answer therefore is 5 * 5 * 3 = 75 ways.

ameer!:
Prove dat
cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise.

akpos4uall: Well, I don't know what you mean by "its because of its negative sign" but let me explain
The equation is n(n - 1) = 99750156
Using the approximation as explained earlier
Sqrt of 99750156 = 9987.5
Round up and down to get 9987 & 9988
These two numbers are n and (n-1)
n is greater than n-1
n is the unknown in the equation
Hence my reason for choosing n = 9988 as the solution i.e 9988 (9988 - 1) = 99750156

This is just one of the solutions to the equation. To get the other, use the same procedure but this time,
Sqrt of 99750156 = - 9987.5
Round up and down to get - 9987 & - 9988
The same way, n is the bigger of the two.
Hence n = - 9987
(-9987)*(-9987 - 1) = 99750156

This means the solution to the equation
n^2 - n - 99750156 = 0 is
n = - 9987 or 9988
This also means if we wanted to factorize, we'll get
( n + 9987 ) ( n - 9988 ) = 0

We can even generalize that an equation that is factorizable in the format;
n^2 - n - z = 0
will be ( n + a ) ( n - b ) = 0
Where a is the rounded down value of the sqrt of z while b is the rounded up value of the sqrt of z
Similarly
n^2 + n - z = 0
will be (n - a ) ( n + b ) = 0

echibuzor: I aint gonna lie, you lost me from the first line...

benbuks: Find the anti-derivative of (cosx +1)^-2

smurfy: Thanks Laplacian. Thanks for this method.

Tell me, assuming you got something like dC/C = 1/2(dr/r), what would you have said about P%?

ameer!:
Prove dat
cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise.

smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.

Mbahchiboy: THANKU VERY MUCH SIR 4 D HIGHLIGHT.....BUT PLZ COMPLETE D NO 2 I DON PAUSE

Mbahchiboy: plz u guys should help solve dis using "d citizen" approach.i tried doing it but i dont know my mistake.
(1)if 4^x=8x find x.
(2) find x if (4/x)^x=2x

Alpha Maximus: Hmmm, nice approach bruv but here's a shorter approach:
(I) Pr(sum of 11) : for this to occur the first two numbers must have a sum of at least 5 and at most 10(11-6 and the 2 dice can't sum up to 11 since there's no die bearing 0) and from sample space there, are 27 outcomes out of 36
Thus a probability of 27/36, then the third number must make the previous sum add up to 11 and there is a 1/6 probability of this occurence since the last number is from the last die(singular of dice)
Thus,pr(sum of 11 from 3 dice)=27/36 * 1/6=27/216(it wasn't meant to be this lengthy if not for explanation's sake)
(II) Pr(sum of 12 from a single throw of 3 dice)
Same method above: the first 2 dice must have a sum of at least 6(12-6) and at most 11, from sample space there are 25 outcomes out of 36 possible outcomes satisfying this designation...then the last die must show a number which makes the sum 12 and there is a 1/6 chance for this to occur, therefore:
Pr(sum of 12 from single throw of 3 dice)=25/36 * 1/6 =25/216...this method saves mathematicians fro the strenuous approach of having to mentally try out all possible combinations such as 1,4,6...4,1,6 ...etc, like you did in your solution

Laplacian: ...a two digit number is divisble by 4 prove that d unit digit MUST be even...

Laplacian: ...a two digit number is divisible by 4, if d unit digit is also divisible by 4, show that d tens digit MUST be even...