short but sweet question:

Question

anonymous · Answer

find an integer n such that:

$$n^2 < 33127 < (n+1)^2$$

find a small integer m such that:

$$(n+m)^2 - 33127 $$

is a perfect square

hence write 33127 in the form$$ pq$$ where p and q are integers greater than 1.

experimentx · Answer

n = floor(sqrt(33127))

experimentx · Answer

(m+n)^2 = m^2 + 2mn + n^2 - 33127 => 33127 = 4 mn

experimentx · Answer

Oo ... something went wrong LOL

anonymous · Answer

its a non calculator, i am very interested in seeing if there is any nice way of getting the first part, when i did it i used a little trick, but its not much

experimentx · Answer

7 am and haven't slept ... i'll book mark this thread. LOL

anonymous · Answer

haha fair enough, i am also very tired, heading to bed now. night!

experimentx · Answer

GN

anonymous · Answer

for first part take sqrt of 33127 = 182.0082...
so just try n=182 first
which satisfy our euation so n = 182

anonymous · Answer

yeah, although i wouldn't have a calculator to sqrt, i suppose i could use a numerical method

i saw that 180^2 = 32400

33127 - 32400 = 727

so we are looking for t such that

360t + t^2 = 727

and t = 2 gets us close

so n = 182

experimentx · Answer

thats cheating ... LOL

anonymous · Answer

haha

anonymous · Answer

try the other parts if you want

anonymous · Answer

the only clever trick in this question really is the pq bit

anonymous · Answer

i felt so stupid when i realised what to do, i had been sitting there for  a while lol