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42 =2 x 3 x 7 =3 "Prime Factors"
Total divisors =The exponent of each "Prime Factor" + 1, all multiplied together: (1+1) x (1+1)x(1+1) =8 divisors:
1 | 2 | 3 | 6 | 7 | 14 | 21 | 42 (8 divisors), or factors.
42 - 1 =41 Prime number
42+1 =43 Prime number.

CoolStuffYT is correct but here's a more rigorous analysis

I would go with that !

so 48 is the answer?

Cool!   You DO use the second '5' ...but it is idistiguishable from the other....so the numbers would look the same....so you must divide by two....

mean = (19x9 + 20x6 + 21x7 + 22x3 + 23x2+ 24x1 + 28x2 + 36x1 + 68x1 + 18x4)/(9+6+7+3+2+1+2+1+1+4)

mean = 403/18 ~22.39

Here is the graph:

hmm, I have one shot and I'm trying to figure this one out.  Rom, you say to divide by 2! but why do you divide by 2! when you don't use the second 5 at all?

4 choices for the 1st digit

4 choices for the 2nd digit

3 choices for the 3rd digit

2 choices for the 4th digit

However because there are 2 5's the number will be halved because of division by 2!

4x4x3x2/2 = 4x4x3 = 48 different 5 digit numbers

Regular cost = y = 40x

 Special cost =  y = 25x + 150

Graph to follow.....

That is the answer I find....maybe Rom will find different!

\(\dfrac{\dfrac{4}{2x-1}+\dfrac 1 4}{x+1}=\)

\(\dfrac{\dfrac{4\cdot 4 + 2x-1}{4(2x-1)}}{x+1} =\)

\(\dfrac{15+2x}{4(2x-1)(x+1)}\)

Way phat! 

\(-1=\sqrt{3x-2}-\sqrt{4x+1}\\ \sqrt{4x+1}-1 = \sqrt{3x-2}\\ (4x+1)-2\sqrt{4x+1}+1 = 3x-2\\ 4x+2 - 2\sqrt{4x+1} = 3x-2\\ x+4=2\sqrt{4x+1}\\ x^2+8x+16 = 4(4x+1) = 16x+4\\ x^2 -8x + 12 = 0\\ (x-6)(x-2) = 0\\ x=6,~2\)

Let's check these solutions

\(\sqrt{18-2} - \sqrt{24+5} = 4 - 5 = -1\\ \sqrt{6-2}-\sqrt{8+1} = 2-3 = -1\)

Both solutions are valid.

x^3-10x^2+37x-52

36 total marbles

a) P[green] = 8/36 = 2/9

b) P[red] = 10/36 = 5/18

c) P[!yellow] = 1 - P[yellow] = 1 - 12/36 = 1-1/3 = 2/3

deleted

3/6 are odd for each die

3/6 x 3/6 = 9/36 = 1/4 chance of both being odd   

\(\text{There are 20 total bills. Assume we choose any of them with equal probability.}\\ P[20] = \dfrac{3}{20}\\ P[10]= \dfrac{6}{20}=\dfrac{3}{10}\\ P[5] = \dfrac{7}{20}\\ P[1] = \dfrac{4}{20} = \dfrac{1}{5} \\ \text{Let }X \text{ be the value of the bill chosen.}\\ E[X] = 20\cdot \dfrac{3}{20} + 10\cdot \dfrac{3}{10} + 5 \cdot \dfrac{7}{20} + 1\cdot \dfrac{1}{5}\\ E[X] = \dfrac{159}{20} =$ 7.95\)

Twenty bills

3/20 ($20)  + 6/20 (10) + 7/20(5) + 4/20(1) = 7.95 expected value

I see now, thanks!