All Answers 358512 Answers

Hi Melody, yes guilty, that's one of my answers - Bertie.

You ask about modular arithmetic.

The basic rules are pretty simple and are easy to prove using some simple algebra.

Two integers x and y are said to be congruent modulo n, written

\(\qquad x\equiv y (\text{mod } n)\)

if they each give the same remainder ( less than n) when they are divided by n.

That is, if the remainder is r, (equal to or greater than zero, but less than n), then

\(x = k_{1}n + r\)

and

\(y = k_{2}n + r\)

for some integers \(k_{1} \text{ and }k_{2}. \)

Subtracting one from the other gets us \(x - y = (k_{1}-k_{2})n = kn,\) for some integer k.

So, often more convenient in practice, x - y is a multiple of n.

The basic rules are then, if   \(x\equiv y(\text{mod }n) \text{ and } u\equiv v(\text{mod }n),\)

then

\(x + u \equiv y + v(\text{mod }n)\)

(just one (mod n) at the end of the line is normal practice)

and

\(x\times u\equiv y\times v(\text{mod }n)\).

They are both easy to prove if you write them out in the equation form used earlier.

(  x - y = k(1)n, u - v = k(2)n, adding, (x + u) - (y + v) = (k(1) + k(2))n = kn, etc., multiply the two equations to prove the multiplcation result.)

As an example/confirmation of the two rules, given that  \(N = (22\times 31) + (11\times 17) + (13\times 19)\),

calculate the remainder when N is divided by 7.

Working to modulo 7,

\((22\times 31)+(11\times 17)+(13\times 19)\equiv (1\times3)+(4\times 3)+(6\times 5)\equiv45\equiv 3.\)

Check, N = 1116 = 159*7 + 3.

Division is a bit more awkward, since the division of one integer by another need not be an integer. Subtraction is ok though.

The most recent Nauseated one was a physics answer. Not intended to be funny and I doubt i would be able to find it.

Here is an old one that you will appreciate. :)             It certainly brought  smile to my face. :)

1/3 * 360   =  120 

its fine dont worry... your answer was helpful as well 

thank again for your help and yes b)  5^(2x+1)  ,your note is so important as well I have to be careful the next time 

I really appreciate your working and helping.

Sorry Sara,

I misread your question.   

y = 5^x

y^2 = (5^x)^2 = 5^(2x)

5y^2 = 5*5^(2x) = (5^1)*(5^2x) = 5^(2x+1)

As long as b) is meant to be 5^(2x+1) not 5^(2x) + 1  then b) is the answer.  Please use brackets to ensure an unambiguous interpretation.

I guess it is the right answer 

thank you for your help Alen!!!  

3^m+3^m+3^m → 3*3^m → (3^1)*(3^m) → 3^(m+1)

Hi...

it's nice to see you too 

thank you for your information but what i actually asked to add them not multiplication 

3^m+3^m+3^m=

3 ✖️ n ✖️ n ✖️ n ✖️ d ➗ w =

\(\frac{3n^3d}{w}\)

What is zero times infinity

It is undefined.

If we assume there are 8 pieces of paper with an 'a' on them, 8 pieces with a 'b', 8 pieces with a 'c' and 7 pieces with a 'd', 7 with an 'e', 7 with an 'f' and 7 with a 'g', then there are 3*8 + 4*7 = 52 pieces of paper in total, 8 of which have a 'c'.  The probability of drawing a piece with a 'c' on it is therefore 8/52 = 2/13 ≈ 0.154

Hi Sara,

It is nice to meet you :)

\(3^m*3^m*3^m=(3^m)^3=3^{3m}=(3^3)^m=27^m\)

The only item common to both terms is c so:  bcd + ac = c(bd + a)

My interpretation is as follows:

Another different answer here:

I saw this one get answered yesterday (or earlier today) by 2 different guests.  Didn't you like their answers?

OR is it just that you did not look for ages and notw you cannot find your question?  If you were a member it would have been automatically put in your watchlist and you would have been able to find it really easily!

Let Pedro's age today be Y

and  Petra's age today be   X

If twice the present age of pedro (2Y) is decreased from petra's age 5 years ago (X-5), the result is 4.

(X-5) - 2Y = 4

X-5 - 2Y = 4

X - 2Y = 9     times 4 because I am going to use it with (2) below.

4X - 8Y = 36    (1)

Nine times pedro's age 3 years ago [9(Y-3)]   exceeds 4 times Petra's present age [ 4X]  by 4.

9(Y-3) - 4X = 4

9Y-27 - 4X = 4

9Y - 4X = 31     

- 4X +9Y = 31  (2)

(1)+(2)

0+Y = 67

Pedro is 67 now

4X-8*67=36

4X=36+536

4X=572

X=143

Petra is 143 and Pedro is 67

check

If twice the present age of pedro (2Y) is decreased from petra's age  years ago  (X-5) , the result is 4.

138 - 2*67 = 138-134 = 4 good

Nine times pedro's age 3 years ago [9(Y-3)] 9*64 = 576  exceeds 4 times Petra's present age [ 4X=4*143=572]  by 4.

576-572=4 good

So that is correct I believe.

So last year Petra was 142 and Pedro was 66 years old.   Maybe they are sea turtles. Sea turtles live a long time.

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Here is the earlier answer It was only a few post back so maybe the answer did not suit you.  It is different from what I got.  Words such as these can be difficult to interprete.

Hi Melody

I have been a member on here for a long while. I came to learn some mathematics but I always look for the funny posts because they always made the day lighter like the comics do in a newspaper.   It is easy to miss posts now because there are so many and the search function is poor.

 With some effort I found some postings from both of them.  The ones I found are a year or so in age. I started laughing again while reading them. Nauseated is one of the names but I don’t remember and couldn’t find the name of the chimp because his early posts were made as a guest most of the time.  It’s driving me bananas trying to remember his name.

Posting a greeting to the chimp and Nauseated is a good idea. Do you think they will see it or be able to find the post?

Have you the link to the recent post made by the Nauseated dog?

1=.5x^(-1/2) 

\(1=.5x^{(-1/2) }\\ 1=\frac{0.5}{x^{(+1/2) }}\\ x^{(+1/2) }=0.5\\ (x^{(+1/2) })^2=0.5^2\\ x=0.25\)

How can a swan be a duck?

10(f-3)=2(f-20)-50

10f-30 = 2f-40-50

10f-30 = 2f-90

10f-30+30 = 2f-90+30

10f = 2f-60

10f-2f = 2f-60-2f

8f = -60

8f /8= -60/8

f = -7 and 4/8

f = -7 and 1/2

e^2 = 7.3890560989306502

There is an approximation - I just used the web2 calc.

Yes, the new search function is really poor and the trolls don't come often.

The dog was here last week but it didn't stay long long.  

Maybe he ate too much cat food and is very unwell :(

Maybe you should send him a get well and return quickly card?

I mean you could post it on the forum for him to see.  

A lot of members do not log on very often but they often watch without logging on. 

I miss the 'fun' comedy that used to be here too.  :((

Richard works for a cell phone retailer. He receives a monthly base salary of $1500 and he has a monthly sales quota of  $5000. He is paid a commission of 5% on the next $5000 in sales and 10% on any additional sales.  What were his gross earnings if he had sales amounting to $34,350? 

Need steps please that would be great

My understanding is that Richard has a quota of $5,000 WITHOUT any commission. Therefore

$34,350 - $5,000=$29,350 on which Richard will get commission as follows:

$5,000 X 5%=$250.00, and $29,350 - $5,000=$24,350 X 10%=$2,435.00, and we have:

$1,500 + $250 + $2,435=$4,185.00 Richard's gross earnings.