All Answers 358187 Answers

So then the possible combinations are:

1-0-0  1

1-1-1  1

1-2-2  3

1-3-3  3

1-4-4  3

1-5-5  3

1-6-6  3

1-7-7  3

1-8-8  3

1-9-9  3

2-0-0  1

2-2-4  3

2-3-6  6

2-4-8  6

3-0-0  1

3-3-9  3

4-0-0  1

5-0-0  1

6-0-0  1

7-0-0  1

8-0-0  1

9-0-0  1

Total: 52

Probability is 52/900=26/450=13/225.

I see my mistake. Thanks Rom!!!

thx i got it by meself 

Thanks for your great answer!

How did you get this?

Thankyou Rom

Let x be the number:
x =(n - 1) * (1 + 110%)
x =(n - 1) * (1 + 110/100)
x =(n - 1) * (1 + 1.10)
x =(n - 1) * (2.10)
x =2.1n - 2.1

I'd like to try it with other similar questions.

It looks like an easy way that would not normally require a lot of trial and error, just a bit I think.

Let the numbers be

a, b,c,d, and  X

But it is better to replace c+d with Y

So our positive integer numbers are    a, b, Y, X       where   a < b < Y/2 < X

A+b+Y/2+X = 320

Y/2+X=283

so

a+b = 37

283/3 = 94.3...

93, 94, 96 are the closest integers that add to 283

93+94 = 187

so

a+b = 37

Let

  \(a=18-\alpha,\quad b=19+\alpha,\quad Y=187-\delta, \quad X=96+\delta\)

Now

\(a+X=119\\ 18-\alpha+96+\delta=119\\ \delta=5+\alpha\)  

\(a=18-\alpha,\quad b=19+\alpha,\quad Y=187-5-\alpha, \quad X=96+5+\alpha\\ a=18-\alpha,\quad b=19+\alpha,\quad Y=182-\alpha, \quad X=101+\alpha\\\)

the smallest value of alpha is 0 and the largest is 17

So  the difference between the greatest and least values of X is 17

A fair six-sided number cube with the digits 1-6 on its faces is rolled three times.
The positive difference between the values on the first two rolls is equal to the value of the third roll.
What is the probability that at least one 3 was rolled?
Express your answer as a common fraction.

\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 1} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(0) & \text{_ _ _ _ _ _} \\ & 2~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & {\color{red}3}~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 4~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 5~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & 6~(5) & \text{_ _ _ _ $\color{blue}x$ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{2} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 2~(0) & \text{_ _ _ _ _ _} \\ & {\color{red}3}~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 4~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & 5~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 6~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ {\color{red}3}} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 2~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & {\color{red}3}~(0) & \text{_ _ _ _ _ _} \\ & 4~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 5~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 6~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 4} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 2~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & {\color{red}3}~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 4~(0) & \text{_ _ _ _ _ _} \\ & 5~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 6~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 5} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & 2~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & {\color{red}3}~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 4~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 5~(0) & \text{_ _ _ _ _ _} \\ & 6~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 6} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(5) & \text{_ _ _ _ $\color{blue}x$ _} \\ & 2~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & {\color{red}3}~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 4~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & 5~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 6~(0) & \text{_ _ _ _ _ _} \\ \hline \end{array}\)

The probability that at least one 3 was rolled is \(\mathbf{ \dfrac{14\ {\color{red}x} }{30\ {\color{blue}x} } = \dfrac{7}{15} }\)

I think that the reason 35 is wrong is that your method assumes that the 4 stars all all distinct and also assumes that the bars are not.

In this question neither are distinguishable

You can work it out like i have by assume all are distinct or you can probably do it some other way where neither are distinct but you can not go half way like you have done.

I am going to work with the idea that all the cookies are somehow different from each other and all the chips are as well

There are 4 places to put the first chip

4 for  the second

4 for the third and 4 for the forth.

That  is  4^4 = 256 possible combinations.

Now how many of those have one cookie with three and one cookie with one and 2 cookies with none?

There are 4 ways to chose 3 chips frm the four, ill melt those together so they cannot seperate.

Now there are 4 posible cookies to put that melted mess into.

Now there are 3 possible cookies left for the other one.

So that is 4*4*3 = 48 possibilies

How many ways can all 4 chips can go into one cookie.  That is easy .. 4 ways.

So there are 48+4 = 52 ways out of  256 ways that 3 or more chips are in any one cookie.

So   There must be  256-52 = 204 ways to distribute the chips so that there is NOT more than 2 chips in any one cookie.

ie       204/256 = 51/64

This answer is the same as your given one so I guess it is right. 

NOTE: I had had not been trying to reproduce the given answer I most likely would have come up with some other solution.

To quote myself:

With probabillity it is easy to see why the correct answer is correct but often (it seems ) impossible to see why the wrong answer is wrong!  That is why it is so frustrating!

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\(\binom{3+4-1}{3}*\frac{3}{4}*(0.01+0.05+0.1+0.25)=6.15\)

Thanks Melody.  This approach still requires some trial and error of course.  In this case that worked out ok at the first guess.  It won't necessarily do so for every problem!

They are not hard to count.

|a-b|=c

a and b are interchangeable so there are really twice as many oucomes in the sample space

So that is 30 possible outcomes

Count those with a 3 and double it as well.    7*2=14 favourable outcomes

Prob is 14/30 = 7/15

How about

71,72,72,73,79,79,79    

Um I am sorry but this answer is incorrect

Modular arithmetic

Alan, I like your method.  It is much more simple than mine. 

I'd like to try and use it for other similar problems to make sure I fully understand.  

I think I do but I am not completely sure.

I think...

n-m = any positive even number     will give an integer solution for n and m, it is just a matter of chosing the correct solution given the other constraints of our question.

Yes I think i have it!   Thanks!

I'd still like to try doing more problems like this your way!

Find the remainder when \(24^{50} - 15^{50}\) is divided by \(13\).

\(\begin{array}{|rcll|} \hline &&\mathbf{24^{50} - 15^{50} \pmod{13}} \\\\ && \boxed{24\pmod{13}\equiv 24-13 \\ \equiv 11 \\ \equiv 11-13 \\ \equiv -2\pmod{13} \\ 15\pmod{13}\equiv 15-13 \\ \equiv 2 \pmod{13} } \\\\ &\equiv& (-2)^{50} - 2^{50} \pmod{13} \\ &\equiv & (-1)^{50}2^{50}- 2^{50} \pmod{13} \quad | \quad (-1)^{50}=1 \\ &\equiv & 1\cdot 2^{50}- 2^{50} \pmod{13} \\ &\equiv & 2^{50}- 2^{50} \pmod{13} \\ &\equiv & \mathbf{ 0 \pmod{13} } \\ \hline \end{array}\)

you left out all the multiples of 100

PSL4#18

\(\begin{array}{|c|c|c|c|c|} \hline & \text{penny} & \text{nickel} & \text{dime}& \text{quater} \\ & 1\ \text{cent} & 5\ \text{cent} & 10\ \text{cent}& 25\ \text{cent} \\ \hline &0&0&0&3 \\ &0&0&3&0 \\ &0&3&0&0 \\ &3&0&0&0 \\ \hline &1&2&0&0 \\ &1&0&2&0 \\ &1&0&0&2 \\ &0&1&2&0 \\ &0&1&0&2 \\ &0&0&1&2 \\ \hline &2&1&0&0 \\ &2&0&1&0 \\ &2&0&0&1 \\ &0&2&1&0 \\ &0&2&0&1 \\ &0&0&2&1 \\ \hline &1&1&1&0 \\ &1&1&0&1 \\ &1&0&1&1 \\ &0&1&1&1 \\ \hline sum &15&15&15&15 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{sum} &=& 15\times 1\ \text{cent}+ 15\times 5\ \text{cent} + 15\times 10\ \text{cent}+ 15\times 25\ \text{cent} \\ &=& 15 + 75 + 150 + 375 \\ &=& 615 \\ &=& \mathbf{$6.15} \\ \hline \end{array}\)

Two circles of radius 1 are centered at (4,0) and (-4,0).

How many circles are tangent to both of the given circles and also pass through the point (0,5)?

OSL4#29

\(\begin{array}{|c|c|c|c|} \hline \color{red}1 & 12 & 3 & 8 \\ \hline 6 & 9 & 14 & 11 \\ \hline 13 & \color{red}2 & 7 & 4 \\ \hline &5& 10 & \color{blue} 15 \\ \hline \end{array} \)

Thank you, Logic !