Question

can u help me with this
Given that 198 = 2 * 3^2 * 11
find
the smallest integer, k , such that 198k is a perfect square?

Answer 1

@SmoothMath @tiaph @robtobey

Answer 2

22

Answer 3

how?

Answer 4

if you have a perfect square then all its prime factors must have an even power. Correct?

Answer 5

yes

Answer 6

typo I meant multiply by 2

Answer 7

Well, 3 already has an even power, but 2 and 11 do not, but if you multiply by 2 then the 2 gets an even power. correct?

Answer 8

can u show the working, if u dont mind?

Answer 9

\[2*3^2*11\]
to be a square just get all the powers to be even so multiply by 2 to get
\[2^2*3^2*11\]
and multiply by 11 to get
\[2^2*3^2*11^2\]

Since you multiplied by both 2 and 11 in total you multiplied by 2*11=22.

Answer 10

2*3^2*11
2*9*11
\sqrt 2*9*11
3* sqrt 2*11
in order to be square it should be multplied by 22

Answer 11

thanks :)