Rom

I don't think you got this one quite right Melody.

Consider the 1 digit numbers.  The digit can't be 0, 4, 5.  That leaves 4 possible

2 digit numbers have 4C1 x (5C1) = 20 possibilities

3 digit numbers have 4C1 x (5C1)² = 100 possibilities

There is 1 4 digit number, 1000, and it is counted

This gives 4 + 20 + 100 + 1 = 125 numbers

Oh, duh.  Ok, These are the numbers that don't match what you want.

We have to subtract this from the 343 and we get

343 - 125 = 218

Just like Melody got.  Sorry for the confusion.

If the digits are all distinct and there are b digits for a 3 digit number we have (we don't start numbers with the digit 0)

(b-1)(b-1)(b-2) = 100

This has 1 real solution, b=6

For number 2 recall Fermat's Little Theorem says \(a^{p-1}\equiv 1 \pmod{p} \text{ for p prime, a a positive integer.}\)

so \(5^6 \equiv 1 \pmod{7}\)

\(5^{30}=\left(5^6\right)^5 \equiv 1^5 \pmod{7} = 1^5 = 1\)

I don't know what the committee forming argument is.

However proving this should be easy enough using the definition

\(k \dbinom{n}{k} = k \dfrac{n!}{k!(n-k)!} = \\ \dfrac{n!}{(k-1)!(n-k)!} = \\ n \dfrac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\ n\dbinom{n-1}{k-1}\)

certainly looks to be

\(s_n = (-1)^{n-1} n^2,~n = 1, ~2,~\dots\)

can you predict \(s_5\) ?

This was pretty clever thinking actually.  If you had multiplied instead of added you would have gotten it right.  Well done!

Suppose Jacky gets 0 apples.

Then we have 3 apples to distribute among 3 friends, and 3 oranges to distribute among 3 friends.

These are independently distributed so the total number of distributions will be the product of the

numbers of distributions of apples and oranges.

3 apples distributed among 3 friends can be done \(\dbinom{3+3-1}{3-1}=\dbinom{5}{2} = 10\) ways

3 oranges is the same so the total number of valid distributions when Jacky gets 0 apples is \(10\cdot 10 = 100\)

Now suppose Jacky gets 1 apple.  We have 2 apples to distribute among 3 friends, and 3 oranges.  This is

\(\dbinom{2+3-1}{3-1}\dbinom{5}{2} = \dbinom{4}{2}\dbinom{5}{2} = 6\cdot 10=60\)

Continuing in the same way there are \(\dbinom{3}{2}\dbinom{5}{2}=30\) ways to distribute the fruit if Jacky gets 2 apples, and

\(\dbinom{2}{2}\dbinom{5}{2}=10\) ways to distribute the fruit if Jacky gets 3 apples

Summing all these up we get \(100+60+30+10=200 \) different ways to distribute the fruit with Jacky getting no oranges

if this works it's a cheesy question

let's try

52*289 / 12 = 1252.33

choice A

it's dripping with cheese

I guess the first thing I would do is check for rational roots.

\(a_0 = -44,~a_4=1 \\ 44 = 2^2 \cdot 11 \\ \text{so the only possibly rational roots are} \\ \pm 2,~\pm 4,~\pm 11\)

checking these we see that none of them are roots and so you are left with going the full bore find the quartic roots method.

Let's say Stacey starts at point (0,0)

She walks 11 meters west to (-11,0)

She walks 30 meters north to (-11,30)

She walks 4 meters west to (-15, 30)

Finally she walks 22 meters south to (-15, 8)

she is

 \(d = \sqrt{(-15)^2 + (8)^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \text{ meters from her starting point}\)

ok it's wrong, not surprising as it was basically a wag based on a couple points.