3 - the formula looks like this: http://math.stackexchange.com/questions … 527#738527

Will consider here the special case when:

Then the solutions are of the form:

- integers asked us. It is clear that if you will satisfy the condition we can always write such a simple solution. It is easy enough to see how it turns out.]]>

Hi Agnishom,

Tulip is my dog.

This link, and the links there, and other posts on the "I have the cheese!" thread in the Members Only forum, will acquaint you more with Tulip.

Very Good, now I have confirmed that you really are phrontister.

]]>For all complex z, the upper limit point of sn= sum((-1)^k (k^(1/k) -z), k=1..n) is the the MRB constant.

]]>since x = x mod 4 + 4K

i^x = i^(x mod 4 + 4k)

= i^(x mod 4) * i^(4k)

= i^(x mod 4) * (i^4)^k

= i^(x mod 4) * 1^k

= i^(x mod 4)

Yes, this is always true ...

]]>{

x² - y² = (x+y)(x-y)

where x, y are positive integers

and (x-y) is positive

x² - y² is therefore a composit number (non prime) with factors (x-y) and (x+y)

}

?

But are you sure every non-prime can be generated with this? How about one that is compost of an odd factor and an even factor?

For example:

70 = 10 * 7 = (x+y)(x-y)

x = ½ ( 10+7 ) = 8.5

y = ½ ( 10-7 ) = 1.5

I think you have to work with decimals in order for this to produce 'any non-prime' ...

]]>I think you meant to say "up to a limit" (e.g. find all primes up to 29) instead of "in a given range" (e.g. find all primes in the range 5,208,079 and 6,206,197).

]]>2. Ratio of Division of Gains:

(i)When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.

Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at end of the year

(A's share of profit) : (B's share of profit) = x:y

(ii) When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time).Now, gain or loss is divided in the ratio of these capitals.

Suppose A invests Rs. x for p months and B invests Rs. y for q months, then

(A's share of profit) : (B's share of profit) = px : qy

3. Working and Sleeping Partners:

A partner who manages the business is known as a working partner and the one who simply invests the money is a sleeping partner.

I hope these concepts will be helpful for you to solve some of your math problems.

]]>Many more, alternate methods, and examples: Wikipedia: Divisibility rule

Thanks for adding other methods also..

]]>1 No. not factorable by 2 in (2)

There are 2 No.'s not factorable by 2 or 3 in (6)

There are 8 No.'s not factorable by 2 or 3 or 5 in (30)

There are 48 No.'s not factorable by 2 or 3 or 5 or 7 or in (210)

times 48 by (prime -1) to get the next number of no.'s. i.e. =480 no.'s in (2310) not factorable by 2,3,5,7, or 11.....and so on.]]>

because a no. divisible by 7 minus a no. that isn't, always = a no. NOT divisible by 7 this will work for any no. because the remainder remains when you minus from the no. which has a remainder of zero.

]]>BODMAS rule decides the order of operations for add, subtract, multiply, divide, etc as below shown

B-Brackets (do all operations contained in the brackets first)

O-Orders (powers and square roots etc)

D-Division

M-Multiplication

A-Addition

S- Subtraction

Let’s see an example and check how BODMASS Rule works

30-(2*6+15/3) +8*3/6

Step1: Brackets

2*6+15/3 = 12+5 = 17

Step2: Division

30-17+8*1/2

Step3: Multiplication

30-17+4

Step 4: Addition and Subtraction

30-17+4=17

20+30-5/4*2+ (1+6)

(5-6/4)+9*7

1-6*(2+9)/8

BODMAS Rule is very helpful in solving various algebra math problems.

]]>Example: 2,4,6,8,10…..

Arithmetic Series : The sum of the numbers in a finite arithmetic progression is called as Arithmetic series.

Example: 2+4+6+8+10…..

nth term in the finite arithmetic series

Suppose Arithmetic Series a1+a2+a3+…..an

Then nth term an=a1+(n-1)d

Where

a1- First number of the series

an- Nth Term of the series

n- Total number of terms in the series

d- Difference between two successive numbers

Sum of the total numbers of the arithmetic series

Sn=n/2*(2*a1+(n-1)*d)

Where

Sn – Sum of the total numbers of the series

a1- First number of the series

n- Total number of terms in the series

d- Difference between two successive numbers

Example:

Find n and sum of the numbers in the following series 3 + 6 + 9 + 12 + x?

Here a1=3, d=6-3=3, n=5

x= a1+(n-1)d = 3+(5-1)3 = 15

Sn=n/2*(2a1+(n-1)*d)

Sn=5/2*(2*3+(5-1)3)=5/2*18 = 45

I hope the above formulae are helpful to solve your math problems

]]>