Interesting Math Game!
 

Found an Interesting math game on the internet and thought I should share it

Assume the following is a square

[B][SIZE="5"]
? 1 ?
? ? ?
4 9 2[/SIZE][/B]

You can use numbers from 1 to 9 in such a way so that the rows and columns both of this square adds to the same number, also 2 diagonals also must add to the same number.

Give it a try

[QUOTE=mukt;286285]Found an Interesting math game on the internet and thought I should share it

Assume the following is a square

[B][SIZE="5"]
? 1 ?
? ? ?
4 9 2[/SIZE][/B]

You can use numbers from 1 to 9 in such a way so that the rows and columns both of this square adds to the same number, also 2 diagonals also must add to the same number.

Give it a try[/QUOTE]

this goes as

[B]8 1 6
3 5 7
4 9 2[/B]

Challenge
 

Two friends went to buy a wrist watch which cost them 50 Rs. Each contributed 25. When they left the shop, shopkeeper suddenly remembered that price of watch was 45 so, he sent his servant with 5 Rs to return the boys. Servant was sharp so, he thought to eat toffees of 2 Rs. He came to to boys afterward and handed over them 3 Rs. Since they had contributed equal money, they divided it and took 1.5 Rs each. It means both contributed 23.5 Rs each.

Now 23.5+23.5= 47
2 Rs were taken by servant for toffees which means 47+2=49

Problem is: Where is rest 1 Rs?

[QUOTE=Fortune;286309]Two friends went to buy a wrist watch which cost them 50 Rs. Each contributed 25. When they left the shop, shopkeeper suddenly remembered that price of watch was 45 so, he sent his servant with 5 Rs to return the boys. Servant was sharp so, he thought to eat toffees of 2 Rs. He came to to boys afterward and handed over them 3 Rs. Since they had contributed equal money, they divided it and took 1.5 Rs each. It means both contributed 23.5 Rs each.

Now 23.5+23.5= 47
2 Rs were taken by servant for toffees which means 47+2=49

Problem is: Where is rest 1 Rs?[/QUOTE]

[FONT="Comic Sans MS"]you see, the 2 rupees which was spend by the servant was also come from that 50 Rs, and you have to subtract it from 23.5, which in this case you are neglecting...and adding up the spending of two friends with the spending of servant...

each friend spend
25+25=50
2.5 need to be subtracted to get real spending of frien[/FONT]ds

[QUOTE=Rixwan;286316][FONT="Comic Sans MS"]you see, the 2 rupees which was spend by the servant was also come from that 50 Rs, and you have to subtract it from 23.5, which in this case you are neglecting...and adding up the spending of two friends with the spending of servant...

each friend spend
25+25=50
2.5 need to be subtracted to get real spending of frien[/FONT]ds[/QUOTE]

I m just asking spending of friends, but also adding 2 Rs of the servant. :D

[QUOTE=Fortune;286309]

2 Rs were taken by servant for toffees which means 47+2=49

Problem is: Where is rest 1 Rs?[/QUOTE]

You are adding Rs. 2 in 47 but Rs.2 needs to be subtracted from Rs. 47

The cost went 47 when servant returned Rs.3. And if he also gave Rs. 2 back to the two friends then 47-2=45 (which is the original price of watch).

Yet Another mathematical Riddle!
 

Assume the following Equation

x = y

then let's multiply both sides by x

we get

x^2 = xy.

now let's subtract y^2 from both sides

x^2 - y^2 = xy - y^2

Now, we know that x^2 - y^2 = (x+y)(x-y)
so
(x+y)(x-y) = xy - y^2

now taking y common
(x+y)(x-y) = y(x-y)

now divide both sides by (x-y) we get
(x+y)(x-y)/(x-y) = y(x-y)/(x-y)

from the above, we get
(x+y) = y

we know that x = y, so y = x. thus putting value of "y" as "x"
x+x = x
2x = x
divide x by both sides
2x/x = x/x
we get
2 = 1 .. lol

[B]The above is valid algebra but there's a logical error/flaw, The riddle is to find the error :D [/B]

[QUOTE=mukt;286377]Assume the following Equation

x = y

then let's multiply both sides by x

we get

x^2 = xy.

now let's subtract y^2 from both sides

x^2 - y^2 = xy - y^2

Now, we know that x^2 - y^2 = (x+y)(x-y)
so
(x+y)(x-y) = xy - y^2

now taking y common
(x+y)(x-y) = y(x-y)

now divide both sides by (x-y) we get
(x+y)(x-y)/(x-y) = y(x-y)/(x-y)

from the above, we get
(x+y) = y

we know that x = y, so y = x. thus putting value of "y" as "x"
x+x = x
2x = x
divide x by both sides
2x/x = x/x
we get
2 = 1 .. lol

[B]The above is valid algebra but there's a logical error/flaw, The riddle is to find the error :D [/B][/QUOTE]

How can you put as value (in this case, x = y) when you started your procedure with the same equation....

[QUOTE=mukt;286377]Assume the following Equation
x = y
then let's multiply both sides by x
we get
x^2 = xy.
now let's subtract y^2 from both sides
x^2 - y^2 = xy - y^2
Now, we know that x^2 - y^2 = (x+y)(x-y)
so
(x+y)(x-y) = xy - y^2
now taking y common
(x+y)(x-y) = y(x-y)
now divide both sides by (x-y) we get
(x+y)(x-y)/(x-y) = y(x-y)/(x-y)
from the above, we get
(x+y) = y
we know that x = y, so y = x. thus putting value of "y" as "x"
x+x = x
2x = x
divide x by both sides
2x/x = x/x
we get
2 = 1 .. lol
[B]The above is valid algebra but there's a logical error/flaw, The riddle is to find the error :D [/B][/QUOTE]

At the end, there is something wrong

2x=x
means 2x-x=0
so, x=0 that means y=0 because x=y

[QUOTE=mukt;286377]

now divide both sides by (x-y) we get
(x+y)(x-y)/(x-y) = y(x-y)/(x-y)

[/QUOTE]

In mathematics, you absolutely cannot divide any figure by 0 (zero)

but as you said x = y therefore x-y=0.
Therefore dividing by x-y is illegal operation in mathematics.