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Hi ,

Never mind for my answer I found it 

AG=AG

BG=GJ+JB

CG=GJ+JC=0 

SO we have AG+AG+AB+AG+AC

which give:AG=1/3(AB+AC) 

Since J is the middle of BC we can write that AB+AC=AJ

So AG=1/3(AJ)=2/3AJ

SO AG=2/3AJ

 So A,G,J are aligned  

Thank you very much for your help appreciate it ❤😀

Nice job, EP....!!!!

A slight correction for asdf's answer......the second point is  (4, 0)

And the slope of the line is  [ -5 - 0 ] / [ 0 - 4] = -5 / -4 =   5 /4    ....just as asdf found !!!!

Also  (-4, 0)  can't be an alternative solution because this would mean that the line has a negative slope....

Very well done, PM!!!!

Here's another way

217 * 45 =  9765

9765 = 1627 (6) R 3

3 is the units digit

Here's another way

In the form   ax^2 + bx + c, the equation of the x coordinate of the vertex is given by   -b/ 2a

So.....   2 =  -b / [2a]  ⇒   4a = - b  ⇒  b = - 4a

And we know that

1 = a(2)^2 - 4a(2) + c           and    -3  =  a(-4)^2 - 4a(-4)  + c 

1  = 4a - 8a + c                              -3 = 16a + 16a + c

1 = -4a + c                                      -3 = 32a + c    ⇒    3 = -32a - c

Add the last two equations and we have that

4 = -36a

-4/36 = a

-1/9 = a

And   b = -4a =  -4(-1/9) = 4/9 

To find c, we have

1 = (-1/9)(2)^2 + (4/9)(2) + c

1 =  (-4/9) + 8/9 + c

1 = 4/9 + c

5/9 = c   

So.....the function is

y = (-1/9)x^2 + (4/9)x + 5/9

To find tthe zeroes, set this = 0

(-1/9)x^2 + (4/9)x + 5/9  = 0           multiply through by  - 9

x^2 - 4x -  5   = 0            factor

(x - 5) ( x + 1)  = 0

Setting each factor to  0 and solving for x and we get  that   x = - 1  or x = 5

So   m = 5  and n = -1

So

m - n =   5 -  (-1)   =   6

I suspect you may have wanted a Venn Diagram for this:

The modular multiplicative inverse of an integer "a" modulo "m" is an integer "b" such that:
ab ≡ 1 mod m
If n = 1, a = 13^(-1)
If n = 2, a = 85^(-1).........and so on.
As a consequence of the above, no matter what the value of n, the multiplicative inverse of "a" will always be a 7. And as a result, 7a ≡ 1 mod 9.

Here is my answer.......

place 6 on each side of the see-saw....    Pick the lighter side 6

Now place 3 on each side     Pick the lighter side 3

place one on each side.....    Like guest said, if they balance , then the islander NOT on the see saw is the lightweight......if they do NOT balance, the lightweight is up in the air.      Done.

1, 2, 3, 4, 5, 7,  9, 11, 13, 17, 19, 23......only prime numbers can be written from here on to 1009, which is a prime number.
So, you only have: 4, 9 the only two composite numbers that you can write between 1 and 1009.

Mmm - beats me how to do it manually though. 

We know \(50-8=42\) students played at least one sport. Now, we see that \(33+24=57\), but there couldn't be 57 players with only 42 students! Therefore, some play both. In fact, \(57-42=15\) students played both.

You are very welcome!

:P

15 play both

1.) 30,35,40

5n+5

2.) s+2.7

For k I said:

AJ=AB+BJ

⇔AJ=AB+1/2BC

AJ=AB+1/2BA+1/2AC

AJ=1/2AB+1/2AC

so 1/2 × 2/3 =1/3 

We have 2/3 × AJ = AG 

which prove that AG=2/3AJ

So I said :

if AB and AC are collinear ABC are aligned

show that AG and AJ are collinear

goal: we want to get AG =KAJ 

mean to express AG and AC. .with the help of AB(AI=AB) and AC(AK=AC)

SO: AJ=IB+BJ

AJ=IA+AB+1/2BC (since BJ=1/2 of BC)

⇔1/3AB+AB+1/2(BA+1/2AC) since BA+AC=BC

⇔1/3AB+AB+1/2AB+1/2AC(because we have multiplied by 1/2 for AB and AC)

⇔5/6AB+1/2AC( I did 1/3+1-1/2)

now for AG I did 

AG=IA+AK 

⇔(1/3AB)+(1/2AC)

 Now I am blocked. 

Is it correct for AJ?

Yes …well, I’m not a moderator, but I am a Genetically Enhanced Chimp, and …

(continued on your post…)  

So, maybe I could test this theory anyway…

But without proper credentials, it’s unethical…and probably illegal. Worst of all, the real moderators might be really, really pissed at me. I would have to stay in the high treetops to avoid the incoming missiles of banana peels, mangos, and small coconuts they’d throw at me.

Living in the treetops may seem like the highlife, but it would become tiresome, quickly.  It is a nice place to visit, though.    

The brilliant, bear-feeding-troll-post link now works.

GA

It would have been better to have started with  AG = AB + BG and AG = AC + CG etc., 

but your final equation is correct and you can make use of it by multiplying throughout by a negative sign.

 - GA = - (- 1/3 AB - 1/3 AC) , so   AG = (1/3)( AB + AC ).

Now you should see that AJ and AG are both scalar multiples of the same vector AB  + AC , 

in which case A G and J must be aligned.

Finally, work out the value of k in order that  AG = k.AJ .

Hmm.
I have a very similar question as a matter of fact.
The only difference is that I am trying to compute the remainder when a_2017 is divided by 45.
Any thoughts?
I couldn't find where the 44 was implemented in guest's answers, otherwise I would have attempted to solve this problem by plugging in 2017 in to his/her's method. 
It is clearly not viable to type all the first 2017 numbers

You could sum up all the digits from 0 to 9 as follows:
(0, 1 608, 1 236, 1 806, 2 408, 3 010, 3 606, 4 207, 4 808, 5 409)
Total = 28 098. If you sum up this number, it =27 which is divisible by 9. And since it ends in 5(2015), it is also divisible by 5.
So now, you only have 2 numbers to worry about:
20162017 mod 45 =37

Here's the equation of the curve:

Can you take it from here?

Ginger, if you were a moderator then you could set a few bear traps and see how this policy might work.

BTW the link to the brilliant bear-feeding troll post doesn’t work. 

Genetically Enhanced Chimps Rule while Gorillas drool!!  

Nice answer