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Stupid C3 question.

Prove,using direct arguement, the a four digit number, formed by writing down two digits, and then reversing them, is divisible by 11.

So,for example, 2112, or 3993.

Reply 1

Necro Defain

Hippysnake

Prove,using direct arguement, the a four digit number, formed by writing down two digits, and then reversing them, is divisible by 11.

So,for example, 2112, or 3993.

Let x and a represent these numbers then the number is axxa = a(10^3)+x(10^2)+x(10^1)+a = a(10^3+1)+x(10+100) jaja you get the idea :smile:

Reply 2

Hippysnake

One more- For any pair of numbers, x and y, 2(x^2 + y^2) is the sum of two squarea.

Reply 3

Necro Defain

Hippysnake

One more- For any pair of numbers, x and y, 2(x^2 + y^2) is the sum of two squarea.

Clearly 2(x^2+y^2) = 2x^2 + 2y^2 = (x*sqrt(2))^2 + (y*sqrt(2))^2 ... fairly obvious from there I think :biggrin:

Reply 4

Hippysnake

Necro Defain

Clearly 2(x^2+y^2) = 2x^2 + 2y^2 = (x*sqrt(2))^2 + (y*sqrt(2))^2 ... fairly obvious from there I think :biggrin:

Is that not it? That's the answer right?

EDIT: How can 2x^2 be written as (Root2 x X^2)? It doesn't work.

the answer is 5.

Hippysnake

Is that not it? That's the answer right?

EDIT: How can 2x^2 be written as (Root2 x X^2)? It doesn't work.

It's (root2 x X)^2, kinda hard to write that out but Idk how to use latex well

Reply 7

Hippysnake

Necro Defain

It's (root2 x X)^2, kinda hard to write that out but Idk how to use latex well

If you expand that though do you not get something completelt different? And neg to the other guy for being a ****.

Reply 8

Necro Defain

Hippysnake

If you expand that though do you not get something completelt different? And neg to the other guy for being a ****.

Nope... (root(2)*x)^2 = (root(2))^2 * (x)^2 = 2x^2 :smile:

Reply 9

stevencarrwork

Any number is divisible by 11 if the 1st, 3rd, 5th digits add up to the sum of the 2nd, 4th, 6th digits etc, or the difference is a multiple of 11.

Ie 15323 is divisible by 11 because 1+3+3 = 5+2

Obviously the rule can be extended to any size integer.

Reply 10

stevencarrwork

Hippysnake

One more- For any pair of numbers, x and y, 2(x^2 + y^2) is the sum of two squarea.

So for 1 and 1, 4 is the sum of two squares?

I think they have to be different numbers.

I usually stick in a pair of numbers to see if I can see a pattern.

3 and 2 gives us 26, which is 5^2 + 1^2. There is a pattern there.

Reply 11

meatball893

Necro Defain

Clearly 2(x^2+y^2) = 2x^2 + 2y^2 = (x*sqrt(2))^2 + (y*sqrt(2))^2 ... fairly obvious from there I think :biggrin:

A square is usually a square of an integer...

OP:
I wish I could be more help, but I can't without giving it away. Just experiment writing 2(x²+y²) in different ways.

Reply 12

Hippysnake

meatball893

A square is usually a square of an integer...

OP:
I wish I could be more help, but I can't without giving it away. Just experiment writing 2(x²+y²) in different ways.

Don't have a clue....