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Mathematics 19 Online
OpenStudy (anonymous):

Find the smallest value of k, when 264k is a perfect square

imqwerty (imqwerty):

okay 1st write the prime factorization of 264 @Love.Study :)

OpenStudy (anonymous):

I did

OpenStudy (anonymous):

@imqwerty

imqwerty (imqwerty):

okay what did you get

OpenStudy (anonymous):

264

OpenStudy (anonymous):

@imqwerty

imqwerty (imqwerty):

i mean what is the prime factorization of 264 for example prime factorization of 12 is-> \(12=2 \times 2 \times 3\) :) like this

OpenStudy (annie12m):

cant you just divide by 4 because a square as 4 sides and its suppose to be all even so take 264 and divide 4 to get your answer

imqwerty (imqwerty):

@annie12m we have to multiply and k prolly has to be a positive integer so that this problem becomes legit

OpenStudy (anonymous):

Oh, 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 3 multiplied by 11 @imqwerty

imqwerty (imqwerty):

its almost correct you just gotta remove one 2

imqwerty (imqwerty):

correct prime factorization of 264 is this- \(264=2 \times 2 \times 2 \times 3 \times 11\)

imqwerty (imqwerty):

we can write it like this- \(264=2^3 \times 3 \times 11\)

OpenStudy (anonymous):

Yes I wrote that @imqwerty

imqwerty (imqwerty):

you wrote one extra 2 http://prntscr.com/9m0y67

OpenStudy (anonymous):

I wrote it accidentally in the message, in the copy I've written three 2, one three and one 11 @imqwerty

imqwerty (imqwerty):

to be a prime the power of each term in the prime factorization must be even but here we see that-> \(2^{\color{red}3} \times 3^{\color{red}1} \times 11^{\color{red}1} \) so all powers are odd so k must be a number which on multiplication with 264 makes all those powers even

imqwerty (imqwerty):

its alright :)

imqwerty (imqwerty):

to make all powers as even we go step by step- 1st we gotta make \(2^3\) to have an even power so we just multiply it by \(\color{red} 2\) to get this-> \(2^3 \times 2=2^4\) 2nd we make \(3^1\) to have even power so we just multiply it by \(\color{red}3\) to get this- \(3^1 \times \color{red}3=3^3\) similarly we multiply \(11^1\) by \(\color{red}{11}\)

imqwerty (imqwerty):

so overall we are multiply k and k is nothing but the extra numbers multiplied to make the powers even :) so \(k= 2 \times 3 \times 11 = 66\)

OpenStudy (anonymous):

Thank you @imqwerty

imqwerty (imqwerty):

np :)

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