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If x is an integer divisible by 3, is x squared divisible by 3?

Q: If x is an integer divisible by 3 then is x2 divisible by 3?

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Yes

Call the unknown integer x. Then, from the problem statement, x - 3/x = 26/3, or:x2 - 3 = 26x/3; or x2 - (26/3)x - 3 = 0x = 9

All numbers of the form 6+12x are, for all integer x.

X is divisible by Y if there is an integer Z such that X = Y*Z that is, Y will go into X without remainder.

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Suppose x = sqrt(3*a) where a is an integer that is not divisible by 3. then x2 = 3*a which is divisible by 3. but x is not even rational and so is not an integer and is certainly not divisible by 3.

Yes

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Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.

Call the unknown integer x. Then, from the problem statement, x - 3/x = 26/3, or:x2 - 3 = 26x/3; or x2 - (26/3)x - 3 = 0x = 9

x2=3x+4 x2-3x-4=0 (x-4)(x+1)=0 x=4 or x=-1 Since you specified a positive integer, the number in question is 4.

All numbers of the form 6+12x are, for all integer x.

No, 250/3 = 83 remainder 1 3 is not a factor of 250 : the prime factors are 2 x 5 x 5 x 5 (2 x 53).

For an integer x to be evenly divisible by another integer y, x/y must also be an integer. Here are some examples: 4/2=2. 15/3=5. 18/6=3. From these examples, you can see that 4 is divisible by 2, 15 is divisible by 3, and 18 is divisible by 6. When two numbers are not evenly divisible by each other, you end up with a remainder, or fractional/decimal answer. Here is an example: 25/2 =12.5

that would be limited to 3 and -3 for values of x

x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)

If we write the problem as 6x-5=x2, then we can write in the form of ax2+bx+c: -x2+6x-5. Then we can use the quadratic equation, x = (-b ± √(b2 - 4ac))/2a, and put in our own values to get the equation x = 3 ± 2. Therefore, x1= 1 and x2=5.