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I'm fine just hurt my knee

Sorry, I reposted it. Check that and PLEASE RESPOND

3 - The school bought 106 boxes of candies, because:

[106 * 100] mod 14 ==2

1 - Change 970,405 to 970,425 and you have a multiple of 225.

instead of , which caused

Missing numbers. Please revise the question.

I hope the following helps:

To be clear, when I wrote "There are an infinite number of possible values..."  I meant "There are an infinite number of possible values for r ..."

These are ALL the possible sums of 10 when rolling 3 colored dice:

136  145  154  163  226  235  244  253  262  316  325  334  343  352  361  415  424  433  442  451  514  523  532  541  613  622  631  Total =  27 different ways

Using Chinese Remainder Theorem + Modular Multiplicative Inverse:

i=0;j=0;m=0;t=0;a=(9, 7, 5);r= (5, 3, 1);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

N = 315 m +  311, where m =0, 1, 2, 3.......etc. And since they want "4-digit positive integer", then:

N =[315 * 3] + 311 =945 + 311 =1,256, which is the answer.

(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99)

1 - Add up all the numbers from 1 to 49 inclusive and that is the value of your "x"

2 - Add up all the numbers from 51 to 99 inclusive and that is the value of your "y"

3 - Subtract "x"  from "y" and that is the answer to your problem. Good luck.

Well that wasn't so bad. Thanks Alan.

Here is Binet's Formuls for finding the nth Fibonacci number:

F(90) =  ((1/2 (1 + sqrt(5)))^90 - (1/2 (1 - sqrt(5)))^90)/sqrt(5) 

= 2.880067194370816120 = 2.88 x 10^18

Just one careless mistake Melody, it should be \(-\frac{325}{e}+120\)

5x^4 * e^-x)dx

\(\displaystyle\int_0^1 (5x^4 * e^{-x})dx\\ \)

I did this with several applications of Integration by Parts.

I ended up with

\(\frac{-325}{e}-120\)         but I most likely made several careless mistakes along the way. 

nice diagram

Let w = a + bi and z = c + di.  Then

\(\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\)

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\(\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\)

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (w + z)/(1 + wz) is real.

corrections:

a = 13       b = 9      c = 7

13² = 9² + 7² - 2 * 9 * 7 * cos(A)

Sorry, had to eat dinner.

... And how did you know I was refering to AoPS. Whatever, irrelevent. 

Here is why I did that.

So, if we took the probability that the red is first, then it is 4/15. Because there is no replacement, we have to remove one value from both numerator and denominator: (4-1)/(15-1). One mpre time for the third: (4-2)/(15-2). Then we multiply them all together. Careful: if you subtract 1/1 instead of 1 from the numerator and 1 from the denominator, then you will get a wrong number. 

This in mind, tell me how I did the other 2 parts: white and blue. Just reply to this thread.

I don't take AoPS classes this is just a problem. I listen to video though.

Also tyty, ill try to get the reasoning?

Lol, I never gave the answer. 

Here is the answer: 

\((\frac {4}{15} \cdot \frac {3}{14} \cdot \frac {2}{13})+(\frac {5}{15}\cdot\frac {4}{14}\cdot\frac {3}{13}) + (\frac {6}{15}\cdot \frac {5}{14}\cdot \frac {4}{13})\)

Compute it using any calculator. Don't expect us to do that as well. (Next time, use the class message board.)

Lol, true. (Nauseated be like: YA THINK?)

I did but I still needed help. And btw your answer is wrong