All Answers 358095 Answers

1 - If each twin gets "0" candies, then there are: [10 + 3 -1] C [3 - 1] ==66 ways

2 - ........................   1  ........................................: [8 + 3 - 1] C [3 - 1] ==45 ways

3 - ........................  2  .........................................: [6 + 3 - 1 ] C [3 - 1] ==28 ways

4 - ........................  3  .........................................: [4 + 3 - 1] C [3 - 1] == 15 ways

5 - ........................  4...........................................: [2 + 3 - 1] C [3 - 1] == 6 ways

6 - ........................  5  ..........................................: [0 + 3 - 1] C [3 - 1] == 1 way

7 - Total number of ways ==[66 + 45 + 28 + 15 + 6 + 1] ==161 ways

Using a calculator, I got  .... 9.9998575128 are the ten digits after the decimal point.

Thankyou for the feedback GA, ..... and I shall endeavour to find out what hypergeometric distribution counting is.

Tiggsy

No problem, Guess I'll help...

We let one of our integers be x

and the other integer be y

xy + x + y = 39

We want to maximize the distance between the product of the sum and the product...

(Didn't really understand this part, so my answer could be wrong)
 

So basically, we want our sum to be big, and our product to be big, to get the largest product????

Anyway, to get the largest product, there is a trick...

Try this out yourself:

93 * 95

94 * 94

When evaluating both of them, you will notice that the second expression is greater. This is because the closer the numbers are, the greater the product.

So, when using this theory in our problem, we want the sum and the product of the two integers to be big.

In doing so, we have to think, and for this, I'm pretty sure to just use trial and error.

We have 5 and 4, product 20, sum 9, so total product 180. Pretty good... Except, that they don't sum to 39

We have 19 and 1, product 19, sum 20, so total product = 380. Pretty sure this answer is correct, from what I understand, because the numbers:

19 and 20

Are basically the closest you can go...

So, my answer is 380 as the total product.

Yup, the answer is. I'm so disappointed that the guest can't even do THIS PROBLEM themselves. @Guest, I don't want to be rude or anything, but you really are ruining your future by asking this question here (If you can't do this one ofc)... 

Oh, this looks like a pythag problem, to better understand pythag go to:

Isn't the answer just 2/12 or simplified 1/6? 

She had 12 then gave 2 away and she had 10/12 left, meaning she gave away 2/12 or 1/6

Sorry guest(s), I believe those are both the wrong answers. Thanks for your help, but can someone (preferably a user) please help still?

Try this:

[19 x 1] + 19 + 1 ==39

The answer is actually 28/5, but thanks for your help.

That was wrong, sorry.

159  167  235  333  Total =  4

Then the sum of all possible values of z is 9 + 7 + 5 + 3 = 24.

a=listfor(n, 1, 11, (11+n) nCr 4);print a, "=", sum a

(495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315) = 32857 - total number of solutions.

=   \(\sqrt[3]{(-57)^4}\)   =  -57 \(\sqrt[3]{-57}\)         Not sure what form you are needing

If by "AA" you mean the digit is repeated such as "44",  "66",  "99"......etc., then there is no such number that is divisible by 19. Here are ALL the multiples of 19 between 3,000 and 4,000:

(3002, 3021, 3040, 3059, 3078, 3097, 3116, 3135, 3154, 3173, 3192, 3211, 3230, 3249, 3268, 3287, 3306, 3325, 3344, 3363, 3382, 3401, 3420, 3439, 3458, 3477, 3496, 3515, 3534, 3553, 3572, 3591, 3610, 3629, 3648, 3667, 3686, 3705, 3724, 3743, 3762, 3781, 3800, 3819, 3838, 3857, 3876, 3895, 3914, 3933, 3952, 3971, 3990) Total = 53 such multiples of 19.

Note: if your number is: 3AA0, then the last multiple of 3,990 will satisfy the condition. Check your numbers carefully.

a=listfor(n, 0/2, 18/2, (19 -2*n) nCr 2);print a, "=", sum a

(171, 136, 105, 78, 55, 36, 21, 10, 3 ) = 615 ways of distributing 17 candies to 3 kids + 2 twins.

171 ways of distributing the candies to the 3 remaining kids when the twins get "0" candies each

136 ............................................................................................................................ 1   ...................

105............................................................................................................................. 2   ..................

78 .............................................................................................................................. 3   ..................

55 .............................................................................................................................. 4   ..................

36 .............................................................................................................................. 5   ..................

21 ............................................................................................................................... 6   ..................

10 ............................................................................................................................... 7   ..................

3 ................................................................................................................................. 8   ..................

7,  17,  37,  47,  67, 97 ==6 such primes

 7^n = sqrt(7)/2^(2/3), solve  for n

Sqrt(7) / 2^(2/3) ==1.666718885

7^n ==1.666718885   [Take the log of both sides]

n ==Log(1.666718885) / 1og(7)

n ==0.262528542