Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

Agnishom
2013-04-17 00:29:01

Oh yeah! I missed the question totally

bobbym
2013-04-17 00:18:56

I am getting 6 with a>b.

13 with order counting.

I counted them up.

{0,432}
{3,363}
{12,300}
{27,243}
{48,192}
{75,147}
{108,108}
{147,75}
{192,48}
{243,27}
{300,12}
{363,3}
{432,0}

Oh, sorry it already got answered.

Nehushtan
2013-04-17 00:17:13

Agnishom wrote:

Since, √432 = 12√3, therefore the pairs are 0  + (12√3)^2, (√3)^2 + (11√3)^2, (2√3)^2 + (9√3)^2, and so on. So a total of 12 ordered pairs.

Wont there actually be 13 ordered pairs (because of the 0)?

Agnishom
2013-04-17 00:06:16

How many ordered pairs of non-negative integers (a,b) are there such that √a+√b=√432 ?


Since, √432 = 12√3, therefore the pairs are 0  + (12√3)^2, (√3)^2 + (11√3)^2, (2√3)^2 + (9√3)^2, and so on. So a total of 12 ordered pairs.

Why is this wrong?

Board footer

Powered by FluxBB