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Find all positive integers n such that n, n + 2, and n + 4 are all primes.

All prime numbers past 3 are of one of the form 6n+1 and 6n-1.

All integers are of one of this forms:
\(\begin{cases}6n-2 & \Rightarrow& 2·(3n-1) \\ \mathbf{6n -1} \\6n & \Rightarrow & 2·3·n \\ \mathbf{6n+1} \\ 6n+2 & \Rightarrow& 2·(3n+1) \\ 6n+3 & \Rightarrow & 3·(2n+1)\end{cases}\)

Note that all other than 6n-1 and 6n+1 can be expressed as a product of two integers bigger than 1.

So a prime number cannot be of any form other than \(\mathbf{6n\pm 1}\)

(That doesn't mean that all numbers of the form \(\mathbf{6n\pm 1}\) are prime)

So \(n=6n-1\), and \(n+2 = 6n+2\) , we see \(n+4 = 6n+3\) is always divisible by 3, so there is no more solution, if \(n > 3\).

The only solution is \(n=3\), \(n+2=5\), \(n+4= 7\)

This isn’t the correct answer Hugo! You didn’t compare your result to the criteria that defines the problem. 

If you did compare you would see this:

If 75 eggs are counted in threes, then 0 remain; it should have 2

If 75 eggs are counted in fives, then 0 remain; it should have 4

If 75 eggs are counted in sixes, then 3 remain; it should have 5

And the most obvious one:

If 75 eggs are counted in sevens, then 5 remain; it should have 0

Here’s a hint: Reread the question, and notice that the remainders are all one less than the divisor, except for seven.

It’s not likely you will know what to do with this hint. So here are few more hints.

What do you get by taking the LCM of the divisors that have a remainder other than zero?

What happens when 1 is subtracted from this LCM?

Is the result divisible by 7?

If not, then what about  2*(LCM - 1)?

If x^2 = 16. what is the sum of all possible values of x?

Hello Guest!

 \(x^2=16\)            | take the root

 \(x=\pm\sqrt{16}\)       | calculate the root

 \(x_1=4\\ x_2=-4\)

 \(x_1+x_2=4+(-4)=0\)

 \(The\ sum\ of\ all\ possible\ values\ of\ x\ is\ 0\ .\)

  !

What is the slope of the line determined by any two solutions to the equation $\frac{2}{x}+\frac{3}{y} = 0$? Express your answer as a common fraction.

Hello Rayjplayz!

    \(\large \frac{2}{x}+\frac{3}{y}=0\)               | common denominator

     \(\large\frac{2y+3x}{xy}=0\)                | \(\times xy\)

    \(2y+3x=0\)                 | \(-3x\)

              \(2y=-3x\)           | \(:2\)

               \(\large y=-\frac{3}{2}x\)

\( The\ slope\ of\ the\ equation\ is\ {\color{blue}-\frac{3}{2}}.\)

  !

If
\(x^2 + \dfrac{1}{x^2} = 171\),
then what are the possible values of
\(x - \dfrac{1}{x}\)?

\(\begin{array}{|rcll|} \hline \left(x - \dfrac{1}{x}\right)^2 &=& x^2-2*x*\dfrac{1}{x}+\dfrac{1}{x^2} \\\\ &=& x^2+\dfrac{1}{x^2}-\dfrac{2x}{x} \\\\ &=& x^2+\dfrac{1}{x^2}-2 \quad | \quad \mathbf{x^2 + \dfrac{1}{x^2} = 171} \\\\ &=& 171-2 \\\\ &=& \mathbf{169} \\ \hline x - \dfrac{1}{x} &=& \pm \sqrt{169} \\\\ \mathbf{x - \dfrac{1}{x}} &=& \mathbf{\pm 13} \\ \hline \end{array}\)

CORRECTION... 

When you solve for the hypotenuse you'll have the RADIUS not the diameter. 

You end up with A = 3.1416 • [sqrt(50)]²  =  3.1416 • 50  =  157.08 units²  

Sorry for the clumsy mistake in my original answer above.  This is what happens when I try to do it in my head, instead of actually drawing the diagram on paper.   

_.

A circle has a chord of length 10, and the distance from the center of the circle to the chord is 5. What is the area of the circle?   

Draw the circle and the chord.  These don't have to be to scale. 

Draw a perpendicular from the center of the circle to the chord. 

Draw a diameter from the center of the circle to the end of the chord. 

Now you have a right triangle, with one leg is half the chord = 5 

The other leg is the distance from the center of the circle to the chord = 5  

Use Pythagoras' Theorem to solve for the hypotenuse

and you'll have the diameter of the circle = sqrt(50)  

Half the diameter is the radius and the area of the circle is calculated  A = π r² 

A = 3.1416 • [sqrt(50) / 2]²   =  3.1416 • 50/4  =  3.1416 • 12.5  =  39.27 square units  

_.

Try these:     (1,1)     (1,-1)     (-1,1)

If   s(t(x)) = s(3x-8)=s(1) 

then it stands to reason that   3x-8=1

solve for that and then think about what you have to do next.

Please no one else finish this

so say t is total eggs

t<=150

t is odd

2 more than multiple of 3, or 1 less than multiple of 3

3 more than multiple of 4, or 1 less than of 4

4 more than of 5, or 1 less

5 more than of 6, or 1 less

Multiple of 7

 multiples of 7 to 150

7,14,21,28,35,42,49,56,63,70,77,84,91,98,105,112,119,126,133,140,147

Two more than multiples of 3

21-2 = 19, no

35-2=33, huh. 

49-2 = 47, no

63-2=61, no

77-2 = 75, huh.

91-2 = 89, no

105-2=103, no

119-2 = 117, huh.

Rest are no's.

35, 77, 119

35-3= 32, huh.         

77-3 = 75, huh.

119-3 = 116, huh.

35-4 = 31, no.

77 - 4 = 73, no

119-4 = 115, yes.

119.

Just to check, 119-5 = 114,/6 = 19, yes, and it's a multiple of 7,

Apologies for my prior mistake and thanks for bringing this to my attention, please check to ensure correctness.

What have you done so far?

In the future, please post the AoPS question you are most stuck with.

Funny I was crunching on these earlier today...

For the first... I'll give a hint

By triangle inequality we have

\(\begin{align*} n + 18 &> 25, \\ n + 25 &> 18, \\ 18 + 25 &> n, \end{align*}\)

Also, maybe search these up.

Typically AoPS questions are asked very often, and are answered.

BTW are you in 2387? The one with Hokaj?

ok 

f(n) includes n^2+1 is odd and n/2 is even. For how many integers n from 1 to 100 does f(f(...f(n)...)) =1 for some number of applications of f?

can u please use a latex calculator for the other one?

I'll give hints if you make it readable...

You are welcome. I like to help people who I think are serious about learning.

Just use a latex calculator

(x - 2)² + y²  =  2   defines a circle whose center is (2,0) and whose radius = sqrt(2)

(x - 2)² + y²  >=  2  defines that circle and its exterior 

I can't interpret the question.

A.  Has all sides equal   --->   square = yes     rectangle = no

B.  Has all angles equal to 180^o   --->   square = no     rectangle = no

C.  Has adjacent sides unequal in lenght   --->   square = no     rectangle = maybe

D.  Has sum of all interior angles equal to 360^o   --->   square = yes     rectangle = yes

I can do this; but I use coordinate geometry.

Place this diagram on a coordinate axis:  Place Q at the origin, QR on the positive x-axis and QP on the

positive y-axis.

Q = (0,0)    P = (0,20)    R = (a,0)             [use a variable like 'a' because we don't know exactly where it is]

Since  RT : TS  =  1 : 2,  RT will be one-third of SR.

Since  SR = 20,  RT = 6 2/3.

Coordinates of T = (a, 6 2/3)     S = (a,20)

Equation of QS:  slope  =  (20 - 0) / (a - 0)  =  20/a     y-intercept = 0     --->   y =  (20/a)·x

Equation of PT:  slope  =  ( 6 2/3 - 20) / (a - 0)

            [multiplying the numerator and the denominator of the slope by 3]

                           slope  =  (20 - 60) / (3a)   =  -40/(3a)     y-intercept = 20

     --->     y  =  [ -40/(3a) ]·x + 20

Finding where these two lines intersect:  (20/a)·x  =  [ -40/(3a) ]·x + 20

   [multiply each term by 3a]                            60x  =  -40x + 60a

                                                                       100x  =  60a

                                                                             x  =  (3/5)a

Substituting this value of x into the equation:  y =  (20/a)·x     --->     y  =  (20/a)·(3/5)a  =  12

If the formula for height from the top of a building 150 feet high is:  H(t)  =  40t - 10t²,

then the formula for height from the ground is:  H*(t)  =  40t - 10t² + 150.

Solving H* to find the length of time it takes to reach an altitude of 180 feet:

     H*(t)  =  180     --->     40t - 10t² + 150  =  180

                                         -10t² + 40t - 30  =  0

    dividing by -10:                      t² - 4t + 3  =  0

    factoring:                             (t - 3)(t - 1)  =  0

    solving:                      either  t = 3  or  t = 1

It will take 1 second for the rocket to reach an altitude of 180 feet.

At 3 seconds, it will again be at 180 feet, only this is when it is coming down.

MONDAY

p = number of words that do not start with M

-- there are 5 choices for the first letter, then 5 choices for the second letter, 4 choices for the third letter,

       3 choices for the fourth letter, 2 choices for the fifth letter, and 1 choice for the last letter.

-- p = 5 x 5 x 4 x 3 x 2 x 1 = 600 arrangements

q = number of words that start with M but do not end with Y:

-- there is only one choice for the first letter and there are 4 choices for the last letter, and 4 choices for

       the second letter, 3 choices for the third letter, 2 choices for the fourth letter, and 1 choice for the

       fifth letter

-- q = 1 x 4 x 3 x 2 x 1 x 4 =  96 arrangements