# Maths Quiz

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#### Krazzy Warrior

##### "Aal Izz Well"
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...

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#### trublu

##### Ta da !
7+3=10
73-18=55.

This is what u wanted,right?

M

#### moverspacker

##### Guest

yes trublu i am agree with you.

#### srinivasdevulapally

##### The Masterminded
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.

OP

#### Krazzy Warrior

##### "Aal Izz Well"
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...

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...

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#### srinivasdevulapally

##### The Masterminded
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!!!!

OP

#### Krazzy Warrior

##### "Aal Izz Well"
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..

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#### srinivasdevulapally

##### The Masterminded
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....

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OP

#### Krazzy Warrior

##### "Aal Izz Well"
(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ...(so on) ... + (100!)^(100!)

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

what is meant by ! sign

#### srinivasdevulapally

##### The Masterminded
Factorial
eg: 3! means 3*2*1
4! means 4*3*2*1

OP

#### Krazzy Warrior

##### "Aal Izz Well"

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

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

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#### srinivasdevulapally

##### The Masterminded
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

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#### dreamcatcher

##### Rockin g33k
^^ old trick..

Same questions

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

last digit 2^12345

#### slugger

##### Banned
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

##### Rockin g33k
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

Fuall.

:<

#### Faun

##### Wahahaha~!
Staff member
^^no horses for ya n00b

##### Pee into the Wind...
Prove that 0.999999....... = 1

#### nvidia

##### -----ATi-----
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

##### Rockin g33k
i^2=i*i cannot be part of the equation.

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