Maths Quiz

[Krazzy Warrior]

MATHS QUIZ

One friend of mine asked me a question and told me to solve..question is based upon Simultaneous Linear Equation..Solve if if u think u r damn good enough..but i think u can't..and don't forget to show the process but first of all u have to solve which is impossible..

Note: We all guys (me and my friend) study in class IX and i m damn sure there are many senior in this forum..so just check this problem...

Here is the question:-

Find a two digit no. whose sum is equal to 10 and when 18 is subtracted from the no. both the digit of the no. become the same....

Look damn easy but do then u will understand...One more question but that I will post afterward...

U can even post ur question too...

[trublu]

The answer is 73.
7+3=10
73-18=55.

This is what u wanted,right?

Answers is 73

yes trublu i am agree with you.

thanks for your answer

[srinivasdevulapally]

Let the no. be xy

x + y = 10

[(10x + y) - 18 ]/11 = p {Where p is an integer since the two digits are equal it should be divisible by 11}

putting y = 10 - x we have

[(10x + 10 - x) - 18]/11 = p

x = ( 11p+8 )/9

Substituting p = 5 (no other value of p solves the equation so that x lies b/n 0 and 10)
x = 7; y = 3 hence 73 is the no.

[Krazzy Warrior]

U guys rocks....
U can even post ur question too...

Here is another but this is easy but still..

x+1/x = 2

find value of x^999999919191

damn easy...

Originally Posted by srinivasdevulapally

Let the no. be xy

x + y = 10

[(10x + y) - 18 ]/11 = p {Where p is an integer since the two digits are equal it should be divisible by 11}

putting y = 10 - x we have

[(10x + 10 - x) - 18]/11 = p

x = ( 11p+8 )/9

Substituting p = 5 (no other value of p solves the equation so that x lies b/n 0 and 10)
x = 7; y = 3 hence 73 is the no.

the bold letters proves me that u r excellent in maths...

[srinivasdevulapally]

x=1 simply solves the equation and hence answer is (1)^999999919191 = 1

But if u want the solution here it is

x + 1/x = 2

(x^2+1)/x = 2

x^2 - 2x + 1 = 0

(x-1)^2 = 0

Hence x = 1

and (1)^999999919191 = 1

That's all!!!!

[Krazzy Warrior]

You are awesome...Give me some question and I will try to solve...

Q) A lotus from the surface is 2 cm high.whose root is fixed to the surface in a sea.A blow of wind moves the lotus 18 cm away..find the depth of the sea..

[srinivasdevulapally]

Let the depth of the sea be x cm

(x+2)^2 = x^2 + 18^2

x^2 + 4x + 4 = x^2 + 324

4x = 320

x= 80 cm ---------> depth of the sea

Is it really a sea or a road side pond?

Can u find out what is the last digit(units digit) of the number obtained after this operation?

(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ...(so on) ... + (100!)^(100!)

Quite simple if u get the logic... try....

[Krazzy Warrior]

Originally Posted by srinivasdevulapally

(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ...(so on) ... + (100!)^(100!)

Quite simple if u get the logic... try....

what is meant by ! sign

[srinivasdevulapally]

Factorial
eg: 3! means 3*2*1
4! means 4*3*2*1

[Krazzy Warrior]

huh! what a question...answer plz..

and after this plz post any other question...let me and u rocks in maths..

it is "0" or "1" or I donot know..

[srinivasdevulapally]

Its takes time to understand if u don't get the logic but very simple if u are through

1!= 1
2!= 2
3!= 6
4!= 24
5!= 120

From 5! every no. ends with a zero...
Any no ending with zero raised to any power gives a number which has the last digit as zero.

So all the other no.s from 5! onwards will end in zero so we should calculate only the last digit for the following

(1!)^(1!) + (2!)^(2!) + (3!)^(3!) +(4!)^(4!)

(1!)^(1!) = 1

(2!)^(2!) = 2^2 = 4

(3!)^(3!) = 6^6 =... ends with 6 (since 6 raised to any power gives a no. whose last digit is always 6

(4!)^(4!) = 24^24 = 4^24 * 6^24 =....ends with 6

{since 6^24 ends with 6 and

4 raised to the odd power gives no. ending wit 4
4 raised to the even power gives no. ending with 6
eg:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256 and so on...
4^24 ends with 6

6*6 ends with 6}

hence the last digit will be the addition of these last digits
1+4+6+6 = 17 hence ends with 7

Now a simple question:

There are hens and cows in a farm. When counted they add up to 30 but when their legs are counted they added up to 100. Find the no. of cows and hens in the farm

[dreamcatcher]

^^ old trick..

Same questions

last digit 1!+2!+3!+4!+5!+6!...100!

last digit 2^12345

[slugger]

Originally Posted by srinivasdevulapally

There are hens and cows in a farm. When counted they add up to 30 but when their legs are counted they added up to 100. Find the no. of cows and hens in the farm

Cows = 20
Hens = 10

[dreamcatcher]

4x+2y=100
x+y=30

x=y-30
4(y-30)+2y=100

120-2y=100
y=10

x+10=30

x=20

So 20 cows and 10 hens

[The_Devil_Himself]

Fuall.

:<

[Faun]

^^no horses for ya n00b

[shady_inc]

Prove that 0.999999....... = 1

[nvidia]

Check this out -
i = √-1.
i² = -1.
i² = i * i
-1 = √-1 * √-1
-1 = √(-1*-1)
-1 = √1
-1 = 1
Why does this happen?

[dreamcatcher]

i^2=i*i cannot be part of the equation.

[nvidia]

Why not?

[Kl@w-24]

Hmm... Is this one valid?

-1 = √-1 * √-1

[dreamcatcher]

i^2=i*i

here both sides should cancel out. The same way we can prove 1=2, 1=4, blah balh. Its not feasible.

[srinivasdevulapally]

Originally Posted by dreamcatcher

4x+2y=100
x+y=30

x=y-30
4(y-30)+2y=100

120-2y=100
y=10

x+10=30

x=20

So 20 cows and 10 hens

Dont just go for equations, just use ur brain

For cows or hens, there will be 2 legs in common. There are 100 legs in total and there are 30 in number. Hence 30 x 2 = 60 legs are common.

But the extra 40 (100 - 60) legs are nothing but cows' legs.

Since each cow has 2 extra legs,

40/2 = 20 cows

rest (30-20) 10 are hens...That's all !!!

Originally Posted by nvidia

-1 = √1
-1 = 1
Why does this happen?

√1 can be (-1)^2 or (1)^2 so u cannot say that √1 is 1

[Krazzy Warrior]

Its takes time to understand if u don't get the logic but very simple if u are through

1!= 1
2!= 2
3!= 6
4!= 24
5!= 120

From 5! every no. ends with a zero...
Any no ending with zero raised to any power gives a number which has the last digit as zero.

So all the other no.s from 5! onwards will end in zero so we should calculate only the last digit for the following

(1!)^(1!) + (2!)^(2!) + (3!)^(3!) +(4!)^(4!)

(1!)^(1!) = 1

(2!)^(2!) = 2^2 = 4

(3!)^(3!) = 6^6 =... ends with 6 (since 6 raised to any power gives a no. whose last digit is always 6

(4!)^(4!) = 24^24 = 4^24 * 6^24 =....ends with 6

{since 6^24 ends with 6 and

4 raised to the odd power gives no. ending wit 4
4 raised to the even power gives no. ending with 6
eg:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256 and so on...
4^24 ends with 6

6*6 ends with 6}

hence the last digit will be the addition of these last digits
1+4+6+6 = 17 hence ends with 7

I more trick(i got that last night):-

(1)^1 + (2)^2.....+(n)^n --->the unit digit will be same as 1+2+.....+n...and to find 1+2+.....+n we apply this formula :- n*(n+1)/2

so for ur question 100*(100+1)/2--> 100*101/2 = 5050 the last digit is zero..therefor the unit digit of (1!)^(1!)....+(100!)^(100!) will be zero....

A question (just u have to explain..you will be shock):-

Three friends went to a restaurant and had a bill of Rs. 30 ...each decided to pay Rs.10...the owner was impressed by the waiter who served this friend and return Rs. 10 to the waiter telling him to return Rs. 2 to each person (2*3=6) and to keep Rs. 4 with u (waiter).. (6+4=10)...As Rs. 2 is returned so each person paid Rs.8 (10-2) so 8*3=24 (all three together) and Rs. 4 from waiter make 24+4=28 ..Where is this Rs. 2 gone ?

Originally Posted by shady_inc

Prove that 0.999999....... = 1

totally wrong question...

[shady_inc]

Originally Posted by Krazzy Warrior

totally wrong question...

Nope..it's true.
0.999..... means 0.9999....forever [till infinity] and this value is exactly equal to 1.I just want someone who can prove it.

[thewisecrab]

@nvidia

-1 = √-1 * √-1
-1 = √(-1*-1)

This is wrong.

√-1 * √-1 = -1

This is correct

@krazzy warrior
looks like your concept still needs more shaping up
Regarding that 0.9999999999999 = 1, it is not wrong (simple explanation would mean rounding up of decimal places to 1)
But here is a better method I found on a blog

First we set:

x=0.999999999…… (infinitely recurring)

Multiplying both sides by 10, we have,

10x=9.999999999….. (infinitely recurring)

subtracting the first equation from the second one,

10x - x = 9.999999999…… - 0.999999999…….

Therefore,

9x = 9

We divide both sides by 9 to get,

x = 1

so do we have, from the first statement,

1 = .999999999….. ?

Also see this link which I found on the same blog
http://mathforum.org/library/drmath/view/55746.html

[The_Devil_Himself]

Originally Posted by Krazzy Warrior

Three friends went to a restaurant and had a bill of Rs. 30 ...each decided to pay Rs.10...the owner was impressed by the waiter who served this friend and return Rs. 10 to the waiter telling him to return Rs. 2 to each person (2*3=6) and to keep Rs. 4 with u (waiter).. (6+4=10)...As Rs. 2 is returned so each person paid Rs.8 (10-2) so 8*3=24 (all three together) and Rs. 4 from waiter make 24+4=28 ..Where is this Rs. 2 gone ?

Which 2 rupees you twat?
final:friends paid Rs30 in bills off which Rs20 went to the owner(he returned Rs10),Rs4 as tip to the waiter,and the rest ^ rupees weer distributed amongst friends.I have no Idea what 2 rupees you are blabbing about.

24+4 is wrong as the waiters tip is included in that 24 already.

[ico]

Originally Posted by shady_inc

Prove that 0.999999....... = 1

Hah easy.....Any 8th standard guy who has learnt to change Recurring decimals to fractions would have solved this.

Let x = 0.999......

10x = 9.999......
9x = 10x - x = 9.999..... - 0.9999 = 9
9x = 9
x = 9/9 = 1

And srinivasdevulapally, Krazzy is still in 9th (if I'm not wrong) and he hasn't learnt about factorials.

Edit: lol thewisecrab solved this already,

[Pathik]

Originally Posted by nvidia

Check this out -
i = √-1.
i² = -1.
i² = i * i
-1 = √-1 * √-1
-1 = √(-1*-1)
-1 = √1
-1 = 1
Why does this happen?

It can be i*i or even -i * -i

and √1 can be -1 or 1.

Consider both cases then.