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If

\(\large{ f(x)=\dfrac{a}{x+2}}\),

solve for the value of \(\mathbf{a}\) so that

\(\large{f(0)=f^{-1}(3a)}\).

\(\begin{array}{|rcll|} \hline f\Big( f^{-1}(x) \Big) &=& x \quad | \quad x = 3a \\ f\Big( f^{-1}(3a) \Big) &=& 3a \quad | \quad f^{-1}(3a)=f(0) \\ f\Big( f(0) \Big) &=& 3a \quad | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\ f\left( \dfrac{a}{2} \right) &=& 3a \quad | \quad f\left(\dfrac{a}{2} \right) = \dfrac{a}{\dfrac{a}{2}+2} = \dfrac{2a}{4+a} \\ \dfrac{2a}{4+a} &=& 3a \quad | \quad :a \\\\ \dfrac{2 }{4+a} &=& 3 \\\\ 4+a &=& \dfrac{2}{3} \\\\ a &=& \dfrac{2}{3} -4 \\\\ a &=& \dfrac{2}{3} -\dfrac{12}{3} \\\\ \mathbf{ a } &\mathbf{=}&\mathbf{ -\dfrac{10}{3}} \\ \hline \end{array}\)

Prove algebraically that the square of any odd number is always 1 more than a multiple of 8.

Let n stand for any integer in your working.

Let k be a non-negative integer.

\(\begin{array}{|rcll|} \hline n^2 &=& (2k-1)^2 \quad | \quad 2k-1 \text{ is any odd number } k > 0,\ k\in \mathbb{N} \\ &=& 4k^2-4k+1 \\ &=& 4k(k-1)+1 \quad | \quad \text{The product of two consecutive integers must be even so let k(k-1)=2m } \\ &=& 4\cdot 2m +1 \\ &=& 8m +1 \\ \mathbf{n^2} & \mathbf{=} & \mathbf{8m +1 \equiv 1 \pmod{8}} \\ \hline \end{array}\)

btw there's a small error in your answer.. I'll let you find it!

Nah I'd never beat you.  Taunt you relentlessly perhaps but never beat you!

If  \(f(0)=f^{-1}(3a)\)  ,  solve for the value of a so that    \(f(0)=f^{-1}(3a)\)

This looks kinda interesting   -  I will assume that a is a constant.

\(let\;\\ y=\dfrac{a}{x+2} \quad x \ne-2\\ x+2=\dfrac{a}{y}\\ x=\dfrac{a}{y}-2\\ f^{-1}(x)=\dfrac{a}{x}-2\\~\\ f^{-1}(3a)=\dfrac{a}{3a}-2=\frac{1}{3}-2=\frac{-5}{3}\\ \)

\(f(0)=\frac{a}{2}\)

if

\(f(0)=f^{-1}(3a)\\ \frac{a}{2}=\frac{-5}{3}\\ a=\frac{-5*2}{3}\\ a=\frac{-10}{3}\\ a=-3\frac{1}{3} \)

\(\text{Well the brute force way is }\\ y=\dfrac{a}{x+2}\\ \dfrac 1 y = \dfrac{x+2}{a}\\ x = \dfrac a y - 2\\ f^{-1}(x) = \dfrac{a}{x} - 2\)

\(f(0) = f^{-1}(3a)\\ \dfrac{a}{2} = \dfrac{a}{3a}-2\\ \dfrac a 2 = \dfrac 1 3 - 2 = -\dfrac 5 3\\ a = - \dfrac{10}{3}\)

ok thanks I get it.

You had to use a calc for that of course.   :)

10^100 mod 72 = 64   I just used the web2.0 calc for that.

So the biggest number possible will be  10^100 - 64

the smallest 2 digits will be 36 and all the rest will be 9.  

9*98+9 = 891.    

Ok I can see what you have done and I can see it has worked this time but

I question if it would work everytime.

--------

I am just thinking out loud now.

You have found the biggest number, but I am nost so sure that the biggest number would always have the greatest sum of the digits for a question like this.

I mean I can see that it has to work for this particular example because 

the 2 smallest have to add up to 9,   or

the 3 smallest have to add to 18   or

the 4 smallest have to add to 27 etc

And of course those small numbers can be placed anyway, they do not have to have the smallest place value.

but for other examples.......not so sure.

eg

997>989   however   9+9+7<9+8+9

I do like your logic though - thanks for sharing it :)

GOOD WORK.  I made a fundamental mistake.  I assumed that the three numbers had to be different.  Nowhere did the problem stipulate that.  Three 37's is a perfectly valid solution.

Your contribution got me to thinking.  I noted that in my answer 17, 27, and 67, the sum of the 10's digit is 9.  Likewise in your answer 37, 37, 37, the sum of the 10's digit is 9.  Does this suggest the same thing to you that it does to me?

Hi Melody: He/she wants a 100-digit number which is divisible by 72.

I simply tried: 1E100 mod 72 =64 as the remainder. So, if I subtracted it from 1E100, the number would consist of all 9s and 36 and the end and be divisible by 72 with no remainder, which is the case. At the same time, it gives him/her the largest sum of the 100 digits, 98 of them being 9s and the last two being 36.

Yes this helped a lot I think I understand how this problem works now Thank you soo much!

Thank you so much these are a possible solution. I was also messing around with this problem and found that the number 37 also fits for all digits. Thanks for the help.

Yes for addition or subtraction.

BUT only if there are just numbers on the bottom - you have to be more careful if there are pronumerals (letters)

Multiply both sides of the equation but the smallest common denominator. 

2/3x +4=2/5

\(\frac{2}{3}x+4=\frac{2}{5}\\ \frac{2x}{3}+4=\frac{2}{5}\\ \text{The lowest common denominator is 3*5=15}\\ 15[\frac{2x}{3}+4]=15*\frac{2}{5}\\ [\frac{15*2x}{3}+15*4]=\frac{15*2}{5}\\ \frac{5*2x}{1}+60=\frac{3*2}{1}\\ 10x+60=6 \)

_{it is 2/3 x + 4=2/5}

 the x is not with the 3

\(\frac{2}{3x}+4=\frac{2}{5} \)

Multiply by 15 on both sides

\(\frac{10}{x}+60=6\)

Put 60 on the other side

\(\frac{10}{x}=-54\)

x=10/-54, or -5/27

I assume this is the correct positioning of x.

Guest, I know what mod is but I do not follow your logic. You lost me right at the beginning.

Where did 64 come from?  

I did get the same asnwer as you, so I guess your logic is sound but I cannot follow it.

Here are my thoughts.

A 100-digit positive integer is divisible by 72. What is the greatest possible value of the sum of the number’s digits?

72=9*8    So it must be divisable by both 9 and by 8

if a number is divisable by 9 the digits have to add to a multiple of 9

1000 is divisable by 8, any number ending in 000 will be divisable by 8.

so the first   97 digits could be 9

now I have to think about the units, 10s and 100 places.

The sum of the 3 digits must be add to 9,18 or 27.    (in order for the number to be disable by 9)

27 is no good because the number would end in 9 and so not be divisable by 8

How about 18     

864  that works    

so I am going to say that the greatest possible value for the sum of the digits is

9*97+18 = 891

We can say

C99+d=A6A

A6A+d=BC7

BC7+d=B91
First equation means d = A6A-C99

Second gives us d=BC7-A6A

BC7-A6A=A6A-C99

2(A6A)=BC7+C99

From this we see that the ones digit on the right side must be 6 (9+7)

So digit A must be 8 or 3.

From the third equation we can see that d must have a ones digit of 4

Therefore A must be 3.

Knowing this, C99+d=363

We can guess that C is one or two by saying d is positive(Big assumption-_-)

if c is two then d=64

363+64=B27

B must be four then if the assumption is correct

Then 427+64=491

WE ARE CORRECT!!!!!!

A=3

B=4

C=2

32+42+22=96

Hope this helps

Thank youi!!

is it the same for addition? I need this one too

2/3x +4=2/5

\(5 - \dfrac 3 4 x = \dfrac 3 8\\ \text{multiply both sides by }8\\ 40 - 6x = 3\)

Good work Babada,

I did it using GP formula and I got just 1 different from you.

this is ok because it us up to our guest asker to look at our working and decide who is correct :)

\(2^1+2^2+ ....+2^8 = \frac{a(r^n-1)}{r-1}=\frac{2(2^8-1)}{2-1}=2^9-2\)

\(2^{10}-(2^9-2)\\ =2^{10}-2^9+2\\ =2^9*(2-1)+2\\ =2^9+2\\ =512+2\\ =514 \)