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Whats the sqaure root of 1386​

User OmidTahouri
by
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1 Answer

26 votes
26 votes

Answer: "
3√(154) " ;

or, write as: " 37.229 " .
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Explanation:
The digits of "1386" ;

add up to: " (1 + 3 + 8 + 6) " ;

= " (1 + 3) + 8 + 6 " ;

= " (4) + 8 + 6 ;

= " (4 + 8 + 6) " ;

= " (4 + 8) + 6) " ;

= " (12) + 6 "

= "18 " ;

"18" is divisible ("evenly") ; by both: "2" : {" 18 ÷ 2 = 9 ".} ;
and by: "3" : {" 18 ÷ 3 = 6 ".}

Note: If a number is "evenly divisible by both: "2" and "3" ; then said number is divisible by "6" ;
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Other: If the sum of the digits of a number result in a number that is evenly divisible by "9" ; then said number is evenly divisible by "9" ;
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The sum of the digits of "1386" is: "18" ;
and: the sum of the digits of: "18" ; ⇒ " 1 + 8 = 9" .
{and: " 9 ÷ 9 = 1 ".}.
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Note: If number that ends with the last 2 (two) digits; in consecutive order; in which, taken as a "separate number" , is evenly divisible by "4" ; then, the entire said number is evenly divisible by "4".
Note: The number given: "1386 " ; ends with: "86" ;
and: " 86 ÷ 4 = 21.5" ; so we know that:
"1386" is not evenly divisible by "4".
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Note: If a number ends with the digit: "5" ; or "0" ; then the number is evenly divisible by: "5" . The number: "1386" ; does not meant either of these criteria.
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Note: If a number ends with the digit: "0" ; then the number is evenly divisible by: "10" . The number: "1386" ; does not meant this criterion.

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Note: If a number ends with the last 2 (two) digits; in consecutive order; in which, taken as a "separate number" , is evenly divisible by "7" ; then, the entire said number is evenly divisible by "7".
Note: The number given: "1386 " ; ends with: "86" ;
and: " 86 ÷ 7 = 12
(2)/(7) " ; so we know that:
"1386" is not evenly divisible by "7".
_________
Note: If a number ends with the last 2 (two) digits; in consecutive order; in which, taken as a "separate number" , is evenly divisible by "4" ; then, the entire said number is evenly divisible by "4".
Note: The number given: "1386 " ; ends with: "86" ;
and: " 86 ÷ 4 = 21.5" ; so we know that:
"1386" is not evenly divisible by "4".
_________

Note: If a number ends with the last 3 (three) digits; in consecutive order; in which, taken as a "separate number" , is evenly divisible by "8" ; then, the entire said number is evenly divisible by "8".
Note: The number given: "1386" ; ends with: "386" ;
and: " 386 ÷ 8 = 28
(1)/(4) " ; so we know that:
"1386" is not evenly divisible by "8" ;
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Since: we know that "1386" is evenly divisible by "9" ;

and "9" is a perfect square;

{since: " 3² = 3 * 3 = 36 " ; and since: "
\sqrt[]{9} = 3 " }.

Let's see what the other factor of "1386" is when "9" is a factor:
" 1386 ÷ 9 = 154 " ;

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So: √(154) ;

= √9 * √(154) ;

= 3 * √(154) ;

Let's look at the factors of 154:
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⇒ 1 & 154; 2 and 77;

List this factors of "154" as: 1, 2, 77, 154;

→ none of which are "perfect squares" ;

→ So, the answer is: "
3√(154) " ;

or: further calculate with a calculator:
→ "
3 * √(154) =
37.2290209379725698 " ;

→ round to: " 37.229 ".

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Hope this is helpful. Best wishes!
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User Jouell
by
3.3k points