Melody

Hi Madyl,

I like the big picture, and the title.

I'm not good on ellipses though.......

Just use the sine rule.

Find the number of ways of distributing 20 chocolates, 20 gummy bears, and 20 lemon drops among two people, so that each person gets 30 candies. (The candies of each type are indistinguishable.)

If the computer program is properly written and that is the answer then I accept that.

However I am going to attempt if for myself.

Let one of the halves be

x chocolates, y gummy bears and 30-x-y  lemon drops
\(0\le x \le20\\ 0\le y\le20\\ 0\le 30-x-y \le20 \)

the number of dots in the whole square is 21*21 = 441

The number of dots in the triangles is 2(1+2+3+...10) = 2(55)=110

441-110 = 331 combinations ..... maybe

I think there is some double counting here.  

Lets see.

What I have done below is pinpointed all the dots that have been double counted.

The point in the middle has been counted four times.

The lines are 

x=y

x=30-x-y

and 

y=30-x-y

There are 11 points on each of the 3 lines which makes  33 overcounts

331-33= 298 combinations altogether.  This is the same answer that guest got with his/her program.  

What big blue box?

A general comment:

 Jenny it is worth putting in effort to understanding the relevance of the unit circle (in relation to trig)

I remember lacking all understanding of the relevance of it when I was at school but when I worked it out later on it was like a light turned on for me.

This comment is not specifically relevant to this question it's just a general comment.

Thanks Chris :)

Thanks Jenny.       

You are very welcome Jenny.

When 11^4 is written out in base 10, the sum of its digits is 16 = 2^4. What is the largest base b such that the base-b digits of 11^4 do not add up to 2^4? (Note: here, 11^4 in base b means that the base-b number 11 is raised to the fourth power.)

\(11_{base \;b}=b+1\\ (b+1)^4=b^4+4b^3+6b^2+4b+1\\ \text{This will be true, as written, for any base where the base is greater than 6}\\ \text{so for all bases (b) greater than 6 the sum of the digits will be 16 base 10 }=2^4_b \)

IF  b=6   then

\(b^4+4b^3+6b^2+4b+1\\ =b^4+4b^3+b*b^2+4b+1\\ =b^4+5b^3+0b^2+4b+1\\\)

so the sum of the digits will be  1+5+0+4+1=11 which is definitely not 2^4 

So the largest base b that does NOT work is 6

If it is odd then it has 180 degree rotational symmetry around the origin.

This means that   Cot(theta) = -Cot(-theta)

If this is true then it is odd.

If theta is acute then it is in the 1st quad and   -theta is in the 4th quad

The absolute value of cot(theta) and cot(-theta) will be the same.

Tan is pos  in the 1st quadrant  and negative in 4th quadrant

so Cot is pos in positive in 1st and negative in 4th quadrant

therefore the statement is true

Cot is an odd function.