How do you simplify $sqrt (1242)$ ?

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Answer 1 · 2016-04-17T15:40:58

$3sqrt(138)$

First decompose 1242 in prime factors:

It ends in 2 so it is dividible by 2:

$1242/2=621$

The sum of the digits of 621 is 9, therefore it is multiple o $3^2$ :

$621/9=69$

#The sum of digits is multiple of 3, so it is divible by 3 again:

$69/3=23$

23 is prime.

$sqrt(1242)=sqrt(2*3^2*3*23)=3sqrt(2*3*23)=3sqrt(138)$

How do you simplify sqrt (1242)sqrt (1242)?