₦airaland Forum

LORDDICE:
u Also a madmatician ?

MathsChic:
Problem 2
(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(-√2+√11+√13)
=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-((√2+√11)²-(√13)²))((√2-√11)²-(√13)²)
=-(2+11+2√22-13)(2+11-2√22-13)
=-(2√22)(-2√22)
=4.22
=88

ladokuntlad:
(2x-1)²(x-1)²(2x+7)pie

-1 and not +1

Laplacian:
can u write out one or two of his other questns?

Karmanaut:
Laplacian, welcome back.

all4naija:
Please, show how you arrive at that. I don't think that is the correct answer.

all4naija:
How is that correct when the radius R is 2x ² - 3x + 1 ?

ladokuntlad:
taaah
u ar too big for dat question naaa
leave it for babies

attempt odas

stlaibrowne:
(2x+1)(x+1)(2x+1)(x+1)(2x+7)pie

masperano:
I am not comfortable with the 'log cancels log' thingy, albeit i know what you mean. I thought they should be a better explanation for that

Karmanaut:
Yeah, a programmable calculator.
Looks like this:
https://upload.wikimedia.org/wikipedia/commons/thumb/7/7d/Voyage200.jpg/1200px-Voyage200.jpg
Costs $223(Approximately #45, 000) on Amazon.

Karmanaut:
Yeah, a programmable calculator.
Looks like this:
https://upload.wikimedia.org/wikipedia/commons/thumb/7/7d/Voyage200.jpg/1200px-Voyage200.jpg
Costs $223(Approximately #45, 000) on Amazon.

MathsChic:
pi

ladokuntlad:
The integer n is the smallest positive multiple of 15 such that every digit of n is either 0 or 2. Compute n/15

MathsChic:
Okay, for integer n which is the smallest positive multiple of 15 such, and has only 0 or 2 as digits, we need to look at possible solutions.
Firstly though, we can assume multiplicant m, to give m*15 = n.

In order for multiplicant m to give integer n, we can see that multiplicant has to end with an even number, i.e. set of (2,4,6,. Termination in an odd number will give integer n which has a '5' in its digits.

Secondly, n has to be divisible by both 5 & 3. This means that n (which is made up of only 0 & 2) can only end in a '0' and not a '2'.

So, we have *****0 as the n we are looking for. This makes '20' available as possible number. Obviously not the answer, anyways.

Again, number in tenth position cannot be a '0', hence n/5 will produce another number which will end in a '0' and will not be divisible by '3'.
So, we can only have ****20.

Next, we can attempt 220. 220/5 = 44 which is not divisible by 3.

So, we move on with ***220. We have next, 2220 or 2020. 2020/5 = 404 which is not divisible by 3.

However, 2220/5 = 444 which is divisible by 3.

Hence, smallest n is 2220.

stlaibrowne:
Question: the result when a number is divided by 2 is the same result when that number is divided by 4 what is the number?

stlaibrowne:
Question: the result when a number is divided by 2 is the same result when that number is divided by 4 what is the number?

Karmanaut:
0.

ladokuntlad:
I can only think of zero

stlaibrowne:
Give a logic for your answer

ladokuntlad:
Briony takes a standard pack of 52 cards and throws some cards away. However she makes sure she keeps all four aces among the remaining cards. She then selects four cards at random from these remaining cards. If the probability of her selecting the four aces is 1/1001, how many cards did she throw away?

Laplacian:
can u write out one or two of his other questns?

Each of the interior angles of a heptagon (a seven-sided polygon) is obtuse and the number of degrees in each angle is a multiple of 9. No two angles are equal. Find in degrees the sum of the largest two angles in the heptagon.

Let abcd be a four digit number with the following properties: a + b + c + d = ab = c x d, where ab denotes the two-digit number 10a + b. What is the remainder when the largest number with these properties is divided by 1000?

ladokuntlad:
clap for ursef joor.
good girl
so answer= 148



Wat course did u study bak in skul?

MathsChic:
Let number of cards Briony now has be x.
Probability of selecting four aces in a row in pack is

(4/x).(3/(x-1)).(2/(x-2)).(1/(x-3)) = 1/1001
24/(x.(x-1).(x-2).(x-3))=1/1001

x.(x-1).(x-2).(x-3)=24.1001
x=14 is a positive solution. Others are negative and complex.

Hence Briony has 14 cards. She had 52 before and so must have thrown away 52-14 = 38 cards.

Not too sure about answer sha

MathsChic:
Hahaha
It's quite an easy question though. I studied Elect Engr.

ladokuntlad:
Now is see.
dat means u are limited to applied maths
because i av some pure maths questions here which is my field.
Or can u attempt d maths Olympiad advance level?

ladokuntlad:
jst a little tweaking needed

ladokuntlad:
Now is see.
dat means u are limited to applied maths
because i av some pure maths questions here which is my field.
Or can u attempt d maths Olympiad advance level?

MathsChic:
What's wrong with the solution?

stlaibrowne:
Drop them

ladokuntlad:
Well there is currently no light
So i cant upload for now

I bet most of us will spend weeks getting solution to some of them.

PS: the ones av been uploading are secondary level olympiad. Bt as u av made mention of me dropping advance levek, i will gladly do dat