Melody

I get 27 too

P(divisable by 8) = \(\frac{27}{6^3}= \frac{3^3}{3^3*2^3}=\frac{1}{8}\)

So I agree with Fiora

NOTE

27 is not the number that are multiples of 8 and 6^3 is not the total number of numbers.

The fact is that only the last 3 rolls have any relevance to this problem.

So Fiora and I just chose to completely ignore the first 5 rolls.

If 7n+1 is a multiple of 4, then what is n mod 4 

7n+1 = 0 (mod 4)

4n+3n+1 = 0 (mod 4)

3n+1 = 0 (mod 4)

3n = -1 (mod 4)

3n = 3 (mod 4)

n = 1 (mod 4)

a)

\(\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\ \text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\\ Where \;\;\;n\ge3\;\;\;and \;\;\;n\in Z\)

I am going to prove this using Induction.

STEP 1

Prove true for n=3

\(LHS=n^3 \\ LHS=3^3 \\ LHS=27\\ RHS= n + 3n(n - 1) + 6 \binom{n}{3}\\ RHS= 3 + 3*3(3 - 1) + 6 \binom{3}{3}\\ RHS= 3 + 18 + 6 \\ RHS=27\\ RHS=LHS\\ \text{So the statement is true for n=3}\)

STEP 2

If true for n=k prove true for n=k+1

so we can assume

\(k^3=k+3k(k-1)+6\binom{k}{3}\\ k^3=k+3k^2-3k+6\binom{k}{3}\\ k^3=3k^2-2k+6\binom{k}{3}\\\)

Now I need to prove it will be true for n=k+1

i.e.  Prove

\((k+1)^3=(k+1)+3(k+1)(k+1-1)+6\binom{k+1}{3}\\ (k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\ \qquad \qquad \qquad \text{An aside:}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)!}{3!(k+1-3)!}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{k!(k+1)}{3!(k-3)!(k-2)}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\frac{(k-2)+3}{(k-2)}*\binom{k}{3}\\ \qquad \qquad \qquad \binom{k+1}{3}=\binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\\\)

\(LHS=(k+1)^3\\ LHS=k^3+3k^2+3k+1\)

\(RHS=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\ RHS=(k+1)+3(k+1)(k)+6\left[ \binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\right]\\ RHS=k+1+3k^2+3k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=1+3k^2+4k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=3k^2-2k+1+6 \binom{k}{3}+6k+\frac{6*3}{k-2}*\binom{k}{3}\\ substituting\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\binom{k}{3}\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{k!}{3!(k-3)!}\\ RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{(k-3)!(k-2)(k-1)k}{3!(k-3)!}\\ RHS=k^3+6k+1+\frac{6*3}{1}*\frac{(k-1)k}{3!}\\ RHS=k^3+6k+1+3k^2-3k\\ RHS=k^3+3k^2+3k+1\\ RHS=LHS\\~\\ \text{So if it is true for n=k then it will be true also for n=k+1} \)

Step 3

Since it is true for n=3 it must be true for n=4, n=5, n=6,  ....

Hence it must be true for all integer n where n greater or equal to 3.

LaTex:

\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\
 \text{by counting the number of ordered triples (a, b) of positive integers}, 
\\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\\
Where \;\;\;n\ge3\;\;\;and \;\;\;n\in Z

LHS=n^3        \\                              
LHS=3^3 \\
LHS=27\\
RHS= n + 3n(n - 1) + 6 \binom{n}{3}\\
RHS= 3 + 3*3(3 - 1) + 6 \binom{3}{3}\\
RHS= 3 + 18 + 6 \\
RHS=27\\
RHS=LHS\\
\text{So the statement is true for n=3}

k^3=k+3k(k-1)+6\binom{k}{3}\\
k^3=k+3k^2-3k+6\binom{k}{3}\\
k^3=3k^2-2k+6\binom{k}{3}\\

(k+1)^3=(k+1)+3(k+1)(k+1-1)+6\binom{k+1}{3}\\
(k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\
\qquad \qquad \qquad \text{An aside:}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)!}{3!(k+1-3)!}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{k!(k+1)}{3!(k-3)!(k-2)}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k+1)}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\frac{(k-2)+3}{(k-2)}*\binom{k}{3}\\
\qquad \qquad \qquad \binom{k+1}{3}=\binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\\

LHS=(k+1)^3\\
LHS=k^3+3k^2+3k+1

RHS=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\
RHS=(k+1)+3(k+1)(k)+6\left[ \binom{k}{3}+\frac{3}{k-2}*\binom{k}{3}\right]\\
RHS=k+1+3k^2+3k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=1+3k^2+4k+6 \binom{k}{3}+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=3k^2-2k+1+6 \binom{k}{3}+6k+\frac{6*3}{k-2}*\binom{k}{3}\\
substituting\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\binom{k}{3}\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{k!}{3!(k-3)!}\\
RHS=k^3+6k+1+\frac{6*3}{k-2}*\frac{(k-3)!(k-2)(k-1)k}{3!(k-3)!}\\
RHS=k^3+6k+1+\frac{6*3}{1}*\frac{(k-1)k}{3!}\\
RHS=k^3+6k+1+3k^2-3k\\
RHS=k^3+3k^2+3k+1\\
RHS=LHS\\~\\
\text{So if it is true for n=k then it will be true also for n=k+1}

I would prove this with induction.  Have you done induction yet?

a)

\(\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\ \text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}\)

So far I do not understand what this question is on about.

You are asked to prove something, where the only letter used, is n

then a, b and c are introduced to use in the solution.  Where did a, b and c come from?

\text{Prove that   }  n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\\
\text{by counting the number of ordered triples (a, b) of positive integers}, \\\text {where }   1 \le a,b,c \le   \text{in two different ways.}

I have not worked through it yet but I like the smiley face context.  :)

Happy, I may take a look for you but I keep telling people, one post = 1 question.

Normally I will not look at posts with more than one question.

Here is the pic

The similar triangles are shown.

So the answer is easy to find.

Hi Loki,

Most of your logic looks spot on.

Have a look at mine and see if you can work out what is different (I have not looked that hard)

Thanks Heureka, 

Here is my contribution. 

It is not much different from yours, but the explanation is maybe a little more different.

If a 7-digit number 13ab45c is divisible by 792, what is b?

\(792=8*9*11\)

For this number to be dividable by 8, the last 3 digits must be divisible by 8

456/8=57  no other digits will work so  c=6

\(13ab45c\\ becomes\\ 13ab456\)

Now a number is divisible by 11 if

| (sum of even digits) - (sum of odd digits) | = multiple of 11

Sum of odd digits = 1+a+4+6 = 11+a

Sum of even digits = 3+b+5 = 8+b

absolute value of difference = | 11+a - 8-b | = |3+a-b |  

so 3+a+b can be negative but it must be a multiple of 11

3+a-b = 11k    where k is an integer.

\(-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ a=8+b\;\;or \;\; a=-3+b\)

Now for a number to be divisible by 9, the sum of the digits must be a multiple of 9

so

\(1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad =27+2b\\ \quad b=0\;\;or\;\;9\\ b=9 \;\;doesn't \;\;work\\ so\;\; b=0,\;\;a=8 \; \text{is a possible solution}\\ ~\\ 1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad =16+2b\\ \quad b=1,\;\;a=-2 \;\text{which is invalid}\)

If b=9 then a=17 or 6     17 is not good so

If b=9  a=-3+9=6

If b=0  a=8+0=8

So the number could be

1380456   or

1369456

So the number is

\(13ab45c= 13ab456 =1380456\\ a=8\\ b=0\\ c=6\)

LaTex:

-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ 
a=8+b\;\;or \;\; a=-3+b

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad  =27+2b\\ \quad b=0\;\;or\;\;9\\
b=9 \;\;doesn't \;\;work\\
so\;\;   b=0,\;\;a=8 \; \text{is a possible solution}\\
~\\

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad  =16+2b\\
 \quad b=1,\;\;a=-2 \;\text{which is invalid}