Rom

assuming you're not counting 0....

4 is the smallest 5 digit base 2 palindrome, 10 is the 2nd smallest.

in base 3 10 is the length 3 palindrome 101

from equation 2 \(t=2s+3 \\ \\ \text{plugging this into the 1st equation we get} \\ \\ \dfrac{s}{4} - 2(2s+3) = 2 \\ \\ \dfrac{s}{4} - 4s - 6 = 2 \\ -\dfrac{15}{4}s = 8 \\ \\ s = -\dfrac{32}{15} \\ \\ t = 2s + 3 = -\dfrac{64}{15} + \dfrac{45}{15} = -\dfrac{19}{15} \\ \\ (s,t) = \left(-\dfrac{32}{15},~-\dfrac{19}{15}\right)\)

1 book is in the library, 1 book is checked out.

That leaves 4 books that can either be checked out or in the library.

\(\text{thus there are }2^4 = 16 \text{ ways for the books to be arranged as described.}\)

if direction

assume (x+9y) is divisible by 19.

x+9y = 19k

9x+5y = 9(19k-9y)+5y =

19(9k) - 76y =

19(9k - 4y)

and thus we see (9x+5y) is divisible by 19

you try the only if direction

I'm not going to answer this one as you need to start doing your own work but I will tell you how to look at it to solve it.

Put the 5 candies down on a table.

Use 2 straws to divide them up into 3 groups.

Now figure out how many different ways you can place the straws.

As they are independent

P[rain on Saturday AND rain on Sunday] = P[rain on Saturday] P[rain on Sunday] = (0.6)(0.25) = 0.15 = 15%

If you treat each couple as a single entity there are 3!=6 ways of seating the 3 couples.

There are 2 ways that the individuals can be ordered within the couple.

Thus there are 6*2 = 12 ways the couples can be seated optimally.

For this problem it's probably easiest to determine the number of 4 letter words with no vowels, and subtract that from the total number of possible words.

The number of words with no vowels is simply \(3^4\)

(do you see why?)

The total number of possible words is \(5^4\)

so the total number of words with at least 1 vowel is

\(n = 5^4 - 3^4 = 544\)

I suppose as a check we can do it the hard way.  The number will be the sum of all words with 1, 2, 3, and 4 vowels.  I.e.

\(2^1 \dbinom{4}{1}3^3+2^2 \dbinom{4}{2}3^2+2^3 \dbinom{4}{3} 3^1 + 2^4 = 544\)

-2(-a+4)=18

\(-2(-a+4)=18 \\ \\ -a + 4 = -9\\ \\ -a = -13\\ \\ a=13\)