$${\mathtt{6}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\frac{{\mathtt{360}}}{{\mathtt{6}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{12}} = {\mathtt{62.353\: \!829\: \!072\: \!448}}$$

Any number (except 0) raised to the power of 0 is 1

Thanks Heureka, I want to look at this too :)

(A) Convince yourself that 50 is the inverse of 5 modulo 83,

If 50 is the inverse of 5 modulo 83 then 5*50 (mod83) must equal 1

5*50=250 = 83*3+1 = 1 (mod 83)        ok I'm convinced 

and that 12 is the inverse of 7 modulo 83.

12*7=84 = 83+1 = 1  (mod 83)             Yep - I am happy.

Find the inverse of 35 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

Well since 5 is  the inverse of 50 and 7 is the inverse of 12 THEN  5*7 is the inverse of 50*12 (mod83)

Check

35*(50*12)= 35*600 = 21000 = 253*83 + 1 = 1 mod83

So            The inverse of 35 mod 83 = 600

However

It seems to me that there must be an infinite number of inverses if 35 lets see if I can find some more.

600= 7*83+ 19

So any number of the form 83N+19 where N is an integer will be the inverse of 35 (mod83)

---------------------------------------------------

I'm still having problems with part B   

The answer is 1!!

do this on the calculator!

100^1

That's pretty cool, Radix. I probably wouldn't do it that way myself, but it is handy to know that there are other methods...!!!!

Numbers are infinite, if you can keep counting, then you keep going. Therefore, there is no such thing as a 'Highest Infinite Number'

(20%) of (500)  =  (0.20) x (500)  =  100

I really like this solution Alan, thank you.

For the last bit you just have to show that 9m/19 is not an integer for  0<m<19

can't you just say that 9 and 19 are relatively prime so 9m/19 will only be an integer if m is a multiple of 19.  Since m cannot be a multiple of 19, 9m/19 is not an integer.

That is, 9m is not divisable by 19.

More added 31/7/15

Vectors

Here the other multiplication, beginning with the last  digit :

Thanks Radix,

I have never seen it starterd with the highest power first.  I learn something every day :)

Hi Melody,

in Germany you may begin with 3  or  with the last 2  ! It is different from teacher to teacher. Both is allowed.

$${\frac{{\mathtt{1}}}{{\mathtt{26}}}} = {\mathtt{0.038\: \!461\: \!538\: \!461\: \!538\: \!5}}$$    near    $${\frac{{\mathtt{0.038\: \!461\: \!538}}}{{\mathtt{1}}}} = {\frac{{\mathtt{3.846\: \!153\: \!8}}}{{\mathtt{100}}}}$$  =  3.8461538 %

Hi Radix,    

That is back to front to how I would do it.  Is that how it is taught in Germany?   

(The answer is the same)

11. You deposit $200 each month into an account earning 3%(per annum) interest compounded monthly.

I am assuming that the money goes in at the beginning of the month and the interest is paid at the end of the month.

a. How much will you have in the account in 30 years?

b. How much total money will you put into the account?

c. How much total interest will you earn?

a)    You can also do this with the future value of an ordinary annuity formula 

C=200     n=30*12=360      i=0.03/12 = 0.0025

Amount after 30 years = $116547.38

$${\mathtt{200}}{\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{1.002\: \!5}}}^{{\mathtt{360}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.002\: \!5}}}}\right) = {\mathtt{116\,547.376\: \!919\: \!658\: \!203\: \!63}}$$

b)  $${\mathtt{200}}{\mathtt{\,\times\,}}{\mathtt{360}} = {\mathtt{72\,000}}$$

You  have put $72,000 into the account.

c)   Interest = $116547.38 - $72,000 = $44547.38

$${\mathtt{116\,547.38}}{\mathtt{\,-\,}}{\mathtt{72\,000}} = {\mathtt{44\,547.38}}$$

.

There are no bracket so this is your question

u-v/8v - 6u-3v/8v

$$\\u-(v/8)*v - 6u-(3v/8)*v\\\\
=u-\frac{v^2}{8}-6u-\frac{3v^2}{8}\\\\
=u-6u-\frac{v^2}{8}-\frac{3v^2}{8}\\\\
=-5u-\frac{4v^2}{8}\\\\
=-5u-\frac{v^2}{2}\\\\$$

Where should the closing bracket go?

Biotomic121, we still need a pic.  Your explaination made no sense at all.    

try looking here

a)200*1.03^30 = $485.45

b)

c)  200*1.03^30 minus $200

Try looking through this :)

Thanks Heureka :)   

I'm confused but hopefully a little concentrated effort (of my own) will sort me out .  

$$\small{\text{
Given $m = 2n + 1$, what integer between 0 and $m$ is the inverse of
$ 2 \pmod{ m}$
}}\\
\small{\text{
Answer in terms of $n$.
}}$$

$$\small{\text{
$2x \equiv1 \pmod{m}$
}}\\
\small{\text{
$2x \equiv1 \pmod{2n+1}$
}}\\
\small{\text{
$2x-1 = 2n+1$
}}\\
\small{\text{
$2x = 2n+2$
}}\\
\small{\text{
$\mathbf{x = n+1}$
}}\\$$

$$\small{\text{
The inverse of $a$ modulo 39 is $b$.
}}\\
\small{\text{
What is the inverse of $4a$ modulo 39 in terms of $b$? }}\\
\small{\text{
Give your answer as an expression in terms of $b$.
}}$$

$$\small{\text{
$a\cdot b \equiv 1 \pmod{39}$ and $4a\cdot x \equiv 1 \pmod{39}$ so $ab=4ax$ and $x=\frac{1}{4}b$
}}\\
\small{\text{
The inverse of $4a \pmod{39}$ is $\frac{1}{4}b$
}}$$