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if x is an integer divisible by 3, then x^2 is divisible by 3. is this true or false? is the converse true or false?
 one year ago
 one year ago
if x is an integer divisible by 3, then x^2 is divisible by 3. is this true or false? is the converse true or false?
 one year ago
 one year ago

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just.chrisBest ResponseYou've already chosen the best response.1
x = N x 3, represents the divisible by 3 integer. x^2 then equals N^2 x 3^2 which having the factors of 3 is therefore divisible by 3. Is this useful?
 one year ago

hana88Best ResponseYou've already chosen the best response.0
what about the converse?
 one year ago

lopusBest ResponseYou've already chosen the best response.0
X=2*3 6/3=2 \[x^{2}=36\] 36/3=12
 one year ago

hana88Best ResponseYou've already chosen the best response.0
that only gives me evidence
 one year ago

just.chrisBest ResponseYou've already chosen the best response.1
I haven't done this stuff in awhile. Do you mean if and integer, x, is divisible by 2 then x^3 is also divisible by 2?
 one year ago

hana88Best ResponseYou've already chosen the best response.0
no if x^2 is divisive by 3, then x is divisible by 3
 one year ago

just.chrisBest ResponseYou've already chosen the best response.1
OK. I am winging it here In order for x^2 to be divisible by a prime, which cannot be further factored then it must also be a factor of x. Is this useful?
 one year ago

lopusBest ResponseYou've already chosen the best response.0
x=n*3 n=1,2,3,4,5,6,7,8............... x^2=3,36,81,144....... divide
 one year ago

hana88Best ResponseYou've already chosen the best response.0
@lopus x is 3,6,9,12,18...
 one year ago

just.chrisBest ResponseYou've already chosen the best response.1
x = 3,6,9,... is correct valus for x but does that indicate that if x^2 is divisible by 3 then so is x? I am asking because I don't know. A question about the question.
 one year ago

lopusBest ResponseYou've already chosen the best response.0
yes, you just divide and see for yourself demostration x=N*3 (N*3)^2 =n^2 * 3^3
 one year ago

just.chrisBest ResponseYou've already chosen the best response.1
So you don't really need the series you just need if x^2 is divisible by 3 then x is because 3 cannot be further factored. And the equations you provided apply in bothe directions. unlike say if x^2 is divisible by 4 then x is not necessarily divisible by for because 4 can be further factored, i.e., 2^2 = 4 is divisible by 4 but x, being 2, is not. Yes?
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
If x^2 is divisible by 3 it can be written in the form 3a where a is an integer. So x^2=3a and x = sqrt3(sqrt a) so the converse is not true.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
3 is a prime number. u should pay attention to that!!!
 one year ago

hana88Best ResponseYou've already chosen the best response.0
um @helder you make no sense
 one year ago

hana88Best ResponseYou've already chosen the best response.0
@Mertsj thanks thats the best answer because it was easy to understand
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
if \(3\mid x^2\) then \(3\mid x\) because 3 is prime.
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
That is not true helder. 315 but 3 does not divide sqrt15
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
So if x^2 = 15, 3 is a factor of x^2 but not a factor of x
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
but 15 is not a perfect square!
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
The converse is: If x^2 is divisible by 3 then x is an integer divisible by 3. So we see that x^2 does not have to be a perfect square. It only has to be divisible by 3.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
so why it is x^2 if it is not a perfect square?
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
x^2 could be 15 or 18 or 21 or 24 or 27 or 12 .
 one year ago

just.chrisBest ResponseYou've already chosen the best response.1
If x^2 does not have to be a perfect square then the converse is not true. However, if x^2 is perfect and divisible by 3 then is the converse true? And would that be the case for any prime? Since primes cannot be factored into smaller integers then I think in order for the square to be factorable by the prime then you must have the prime^2 in order to have the prime as a factor of x^2.
 one year ago
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