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**Test for Divisibility by 7**

- Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, so will be the number

Example: 4991

Steps:

499 - 2(1) = 497

49 - 2(7) = 35

Since 7|35, the number is divisible by 7

**Test for Divisibility by 13**

- Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, so will the number

Example: 50661

Steps:

5066 + 4(1) = 5070

507 + 4(0) = 507

50 + 4(7) = 78

7 + 4(8) = 39

Since 13|39, the number is divisible by 13

**simplyjasper**- Replies: 11

First the slightly easier ones...

Definition:

2|10 means 2 divides 10 or 10 is divisible by 2 and leave no remainder

**Test for Divisibility by 5**

- Check if the last digit is 0 or 5

Example: 123512341325

Since the light digit is 5, the number is divisible by 5

**Test for Divisibility by 2**

- Check if the last digit is even (0, 2, 4, 6, 8)

Example: 12312348

Since the last digit is 8, the number is divisible by 2

**Test for Divisibility by 4**

- Check if the last 2 digits is divisible by 4

Example: 123124124

Since 4|24, the number is divisible by 4

**Test for Divisibility by 8**

- Check if the last 3 digits is divisible by 8

Example: 9798743648

Since 8|648, the number is divisible by 8

**Test for Divisibility by 16**

- Check if the last 4 digits is divisible by 16

Example: 7849758432

Since 16|8432, the number is divisible by 16

**Test for Divisibility by 11**

- By subtracting the sum of all the even position of the digit by the sum of all the odd position of the digit

Example: 19203182

Odd sum: (1 + 2 + 3 + 6) = 12

Even sum: (9 + 0 + 1 + 2) = 12

Difference is 0 and 0 is divisible by 11. Hence the digit is divisible by 11...

3rd line itself is wrong...

If y = x,

x + y = 2x or 2y not x^2

bobbym wrote:

Hi;

(2) is divergent. But I am having a little trouble proving that.

I was thinking it divergent too...

It's one of the four unseen questions in my paper which consists of more than 30% of our grades =/...

He kinda made the class adopt a phobia in Mathematics

Isn't it just by using the identity and taking away from both side

(a . b)^2

Or you need to prove the identity before you can use it?

By Lagrange Identity,

ineedhelpwithcalc wrote:

OK, I have loads of questions to do with Calc.

1) I don't completely understand definite integration. Do we only need to use the rectangle/trapezium rules if it specifies a number of portions to cut the area under the function into?

[/math]

Rectangle/Trapezium Rule and there's an additional Simpson's Rule... (Or that's what I have came across through my limited years of studies in Math)

They are usually used for the estimation of an integration and don't usually provide the exact value

How these all came out is due to the Riemann sum, you may wish to read up a little bit on them ...

ineedhelpwithcalc wrote:

2)

(x[sup]2[/sup]) + (1)And then I integrate both? What answer should I get?

If you do it your way, it should go

You can check where u made a mistake in your integration and don't do it during your exam

BeachRum1962 wrote:

Ok simplyjasper please post your method but I almost had it using soroban's method.

Posted by editing my previous post

bobbym wrote:

Hi simplyjasper;

That is what I would have done. That is definitely the way to sum that series, if he doesn't give you full credit than it isn't your fault.

Yeah I think my professor is a little bit psycho... He enjoys seeing us fail his paper ...

Any answers for (2) and (3)?

I did mine through partial fractions and manage to get the result ...

You can try mine if you think you are more comfortable with partial fractions...

Solving For Partial Fractions, u will get

Using B(t):

Taking Common Factor:

bobbym wrote:

Hi simplyjasper;

I am not making my point clear.

Shouldn't be no more difficult than differentiation. That is how you get the Maclaurin series for Arctan(x). So if you can form a maclaurin series and you know that Arctan(1) = pi/4 you just put x=1 an you have that series. You have summed it.

This is the series you were given to sum:

This is the maclaurin series for Arctan(x):

Now just put 1 into x and the 2 series are the equal.

Since you know that Arctan(1) = pi / 4 you are done.

Guess that's the only way... I couldn't find anything on the website that prove why is the series going towards to the value except for the fact that when u use 5 values of it u get it corrected to 2 decimal places and if u take 7 u get 3 decimals place and so on...

I'm totally not sure what he wants... I just wrote the series = tan-1(1) = pi/4...

Hopefully i will get some marks for that...

It was in my number theory or algebra module

I think there's something wrong with the diagram too...

Still with the information, it is sufficient to solve it..

Basically since ACD is an isosceles triangle, AC and CD is equidistant hence u should be able to get the angle for CAD and ADC

Then u should be able to find angle BAD

Finally, by using sine rule, you should be able to get BC

I think there is a need for restriction?

Things like

- u can't copy only a portion of the 'a's u type

- deleting is counted as 1 operation

Else it is pretty simple ...

What mathsyperson did was the best

or in another way would be type 4 'a's and do that 11 times

As typing 4 'a's and typing 2 'a's then copy and paste yield 4 operations

bobbym wrote:

Hi simplyjasper;

That is the MaClaurin series for Arctan(x) when you plug x=1 into it.

So it is easy to sum:

Arctan(1) = π / 4

So:

Hi Bobby,

Yes this is a well known series that is equals to

However, it's a 10 marks questions, I think he's looking into how did this formula came about...

**simplyjasper**- Replies: 9

Just came back from the Calculus exam and I obviously didn't fair well

Some questions that I had difficulties:

(1) Prove that

(2) Determine whether the following series is absolutely convergent, conditionally convergent or divergent

(3) The function

is defined by if andor

ifProve that the domain of

Is continuous on Justify your answer.

**simplyjasper**- Replies: 4

How do I prove by limits that this sequence is less than 2 for all values of n?

Appreciate your reply...

Did you use 'telescoping'? As after a great deal of time and random trial and error, I did manage to get this partial fractions...

But I doubt I have such a luxury of time during my exam...

I was just wondering is there any technique in helping you find a the relationship faster or even another way of solving the sum of this series

Since this is not Monty Hall....

I don't see why u need to have 1000 end branches...

Considering that every number of different suits yield the same points

First row should only include 1 - 9 pts with probability of 1/13 each and (10, J, Q, K) = 10pts with a probability of 4/13

Firstly approaching such question u should have some basic knowledge such as

There are many ways of solving these questions but I guess this formula would solve any angles

For example:

Another method:

Of course there's other ways such as drawing a right-angle isosceles triangle (that is 2 angles at 45 degrees and 1 angle at 90 degrees)

- the length and breadth would be at 1 unit and the hypotenuse is

A picture can be seen in the website below:

http://www.themathpage.com/aTrig/isosceles-right-triangle.htm

bobbym wrote:

Hi simplyjasper;

That is an indefinite summation. For those you must use the summation by parts formula.

To use that formula you must understand both the difference calculus and the summation calculus. Notice that it is very similar to the integration by parts formula.

If it were like this, a definite summation, you would have a few more options.

This one is much easier, is this what you intended?

Yes that is what i'm intending to find. Had no idea how to do that as i'm still new to posting these formula in this forum... How do you approach this question and manage to find it's 1?

**simplyjasper**- Replies: 5

How should I approach finding the sum of this infinite series

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