Question

Find two 4 digit natural numbers that multiply to 4 to the power of 8 + 6 to the power of 8 + 9 to the power of 8

Answer 1

@kidspence did you see answer to last problem?

Answer 2

Yeah I'm confused though

Answer 3

what step?

Answer 4

i cant understand it properly

Answer 5

All.. :$ LOL

Answer 6

Find two 4 digit natural numbers that multiply to equal 4 to the power of 8 + 6 to the power of 8 + 9 to the power of 8

Answer 7

oh this one i don't know. i meant last one

Answer 8

I know the last one I'm confused on

Answer 9

Find the smallest positive whole number that when added to: 2001 x 2002 x 2004 x 2005 - 2003 x 2003 you get a square number

Answer 10

Yup

Answer 11

that one yes?

Answer 12

ok let me start again. to make life easier put
\[2003=x\]

Answer 13

so
\[2001=x-2\]
\[2002=x-1\]
\[2004=x+1\]
\[2005=x+2\] clear so far?

Answer 14

YUp

Answer 15

now some algebra.
we get
\[(x-2)(x-1)(x+1)(x+2)-x^2\]

Answer 16

\[(x-2)(x+2)(x-1)(x+1)-x^2\]
\[(x^2-4)(x^2-1)-x^2\]
\[x^4-5x^2+4-x^2\]
\[x^4-6x^2+4\]

Answer 17

that is what you have if x was 2003

Answer 18

and this is not a perfect square, but if you add 5 you get
\[x^4-6x^2+9\] and that is a perfect square. it is
\[(x^2-3)^2\]

Answer 19

so if you add 5 to that big mess above you get
\[(2003^2-3)^2\] a perfect square

Answer 20

we can check with a calculator and see that this is right, but the algebra does all the work for you

Answer 21

yup, they are equal look here where i typed them both in together
http://www.wolframalpha.com/input/?i=%282003^2-3%29^2%2C2001*2002*2004*2005-2003^2%2B5