How can you simplify the quadratic formula using completing the square?

[Anamitra]

Answers:
1. $${10}{!}{-}{9}{!}$$
=3265920
2. Sum of digits=0+1+2+...+9
=45
45 is divisible by 9. Therefore all numbers according to the given prescription should be divisible by 9
3. Theorem to be used: The highest power of a prime,p, in n! is given by:
{ [n/p]+[n/(p^2)]+[n/(p^3)]-----------}
You may include as many terms[possible] as you can in the above series.
[x] denotes the greatest integer less than or equal to x

The highest power of 5 in 1000! is 249
The highest power of 2 is even greater.
Therefore we have 249 zeros.

 

[agentredlum]

Answers:
1. $${10}{!}{-}{9}{!}$$
=3265920
2. Sum of digits=0+1+2+...+9
=45
45 is divisible by 9. Therefore all numbers according to the given prescription should be divisible by 9
3. Theorem to be used: The highest power of a prime in n! is given by:
{ [n/p]+[n/(p^2)]+[n/(p^3)-----------}
You may include as many terms[possible] as you can in the above series.
[x] denotes the greatest integer less than or equal to x

The highest power of 5 in 1000! is 249
The highest power of 2 is even greater.
Therefore we have 249 zeros.

YES! VERY NICE SIR!

I'snt it amazing that ALL numbers constructed in this way are divisible by 9?

Just a humble observation...you don't need ! on n in your formula #3

by the way, the highest power of 2 in the prime factorization of 1000! is 994 but you only use 249 of them to produce factors of ten cause there are 249 5's

Do you know any problems like this?

 

[Anamitra]

Just a humble observation...you don't need ! on n in your formula #3:

Thanks for pointing it out. It was just due to inadvertence!
[I have edited the thing: It was looking extremely odd]

 

[agentredlum]

Thanks for pointing it out. It was just due to inadvertence!
[I have edited the thing: It was looking extremely odd]

LOL

I would pay good money to know the exact value of 1000!/5

If you make cubes with side Planck Length and you packed the entire known universe wth these cubes, i don't think you would come even remotely close to 1000! cubes.

I edited it out of the quote by you in my post

 

[agentredlum]

I have a question about ancestors that seems to be related to number theory. Let's consider my total ancestors. 1 generation before me there is my father and mother so that's 2 ancestors. My father had parents and so did my mother, so 2 generations before me is total 6 ancestors for me. My grandparents also had parents so 3 generations before me is total 14 ancestors for me. You get the idea, 4 generations before me is total 30 ancestors, 5 generations, 62 ancestors, n generations, 2^(n + 1) - 2 ancestors.

Assuming 30 years for every generation and assuming 100020 years of humanity on this planet, then by the above argument I am supposed to have about 2^3335 - 2 ancestors.

This has to be wrong but I can't figure out where the mistake in the logic is. The only way I know to reduce this humungous number is to assume that at many places along the line, many parents were brother and sister, or father and daughter, or mother and son, or grandfather and granddaughter, or grandmother and grandson.

This is very disturbing to me...HELP!

 

[Guffel]

Google "ancestor paradox" and I hope you'll get happier.

 

[agentredlum]

Google "ancestor paradox" and I hope you'll get happier.

WOW! I can't believe all these years it never occurred to me to google it. Thanx Guffel, or should i call you cousin?

 

[agentredlum]

IMHO I accept this explanation

http://www.genetic-genealogy.co.uk/supp/ancestor_paradox.html

The othrs seem to be by the same person and they are against the posted link in some respects. However i have only looked at 10 of 35000 hits.

Still not happy.

 

[Guffel]

Still not happy.

Read http://en.wikipedia.org/wiki/Charles_II_of_Spain" and I think you'll cheer up. With heroes like him, and his horizontal family tree, the ancestor paradox is easily explained.

 

[agentredlum]

Is it me or does google like to exaggerate their own importance? The first page claimed 35700 hits. By the third page it went down to 332. I only get 10 hits per page...

 

[agentredlum]

Read http://en.wikipedia.org/wiki/Charles_II_of_Spain" and I think you'll cheer up. With heroes like him, and his horizontal family tree, the ancestor paradox is easily explained.

LOL this is so funny i have to quote some of it.

'He is noted for his extensive physical, intellectual, and emotional disabilities along with his consequent ineffectual rule as well as his role in the developments preceding the War of the Spanish Succession'

'Empress Maria Anna was simultaneously his aunt and grandmother and Margarita of Austria was both his grandmother and great-grandmother.'

'The indolence of the young Charles was indulged to such an extent that at times he was not expected to be clean. When his half-brother Don John of Austria, a natural son of Philip IV, obtained power by exiling the queen mother from court, he covered his nose and insisted that the king at least brush his hair.[3]The only vigorous activity in which Charles is known to have participated was shooting. He occasionally indulged in the sport in the preserves of the Escorial.'

Well, you did it Guffel, you made me happy for a while.

I've heard of Don Juan, but Don John?!?

The history of humanity is one of trial and error, unfortunately, mostly error.

 

[math4math]

Cool tricks to do math quick

Thanks

 

[Anamitra]

Four Simple Problems:
1. You have a set of twelve natural numbers[arbitrary natural numbers].Show that a pair can be always located from this set so that their difference is divisible by 5

2. Prove that the following:
$${[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}$$
where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]

3. What is the last digit[unit's place] in the sum:

1!+2!+3!+....99!

4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12

 

[agentredlum]

Four Simple Problems:
1. You have a set of twelve natural numbers[arbitrary natural numbers].Show that a pair can be always located from this set so that their difference is divisible by 5

2. Prove that the following:
$${[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}$$
where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]

3. What is the last digit[unit's place] in the sum:

1!+2!+3!+4!+5!+6!+....99!

4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12

I got #3 pretty quick the answer is 3. The only factorials that contribute to the last digit are 1, 2, 3, 4 since 5! ends in zero all other factorials con tribute zero to the units position.

It is not too hard to find the next to last digit.

 

[agentredlum]

Posted by Anamitra

'4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12'

The last 2 digits of this sum are 13 so when divided by 4 the remainder is 1.
This gaurantees that the above expression must be of the form 12n + 1 or 12n + 5 or 12n + 9
When these are divided by 3 they leave remainder 1, 2, 0
So the best i can do for now is say the sum leaves remainder 1, or 5, or 9
Still working on #4

 

[agentredlum]

Anamitra posted

'2. Prove that the following:$${[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}$$where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]'

The expression in parenthesis must be of the form 5w, or 5w + 1, or 5w + 2, or 5w + 3, or 5w + 4

The exponent 35*36*37 is even

Carefull consideration of all 5 cases will reveal allowable forms.

(5w)^even is clearly of the form 5n

(5w + 1)^even is clearly of the form 5n + 1 since all terms in the binomial expansion will be multiples of 5 except the last term which will be 1^even or simply 1

(5w + 2)^even is clearly of the form 5n - 1 or 5n + 1 since 2^even ends in 4 or 6 and anything that ends in 4 is of the form 5n - 1, anything that ends in 6 is of the form 5n + 1. Again all terms in the binomial expansion will be multiples of 5 except the last term.

(5w + 3)^even By now it should be obvious that we only need to consider 3^even. This always ends in 9 or 1 and again this is of the form 5n -1 or 5n + 1

(5w + 4)^even Consider 4^even always ends in 6. This is of the form 5n + 1

This completes the proof

 

[pwsnafu]

Posted by Anamitra

'4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12'

The last 2 digits of this sum are 13 so when divided by 4 the remainder is 1.
This gaurantees that the above expression must be of the form 12n + 1 or 12n + 5 or 12n + 9
When these are divided by 3 they leave remainder 1, 2, 0
So the best i can do for now is say the sum leaves remainder 1, or 5, or 9
Still working on #4

All you need to know is that 4! = 24 and any larger factorial is a multiple of 4! (e.g. 6! = 6 times 5 times 4!). So only the first three terms are relevant.


While we are on the topic of divisibility: take any three digit number, repeat it to get a six digit number. This number is divisible by 7. Tends to surprise people who haven't seen it before.

 

[agentredlum]

All you need to know is that 4! = 24 and any larger factorial is a multiple of 4! (e.g. 6! = 6 times 5 times 4!). So only the first three terms are relevant. While we are on the topic of divisibility: take any three digit number, repeat it to get a six digit number. This number is divisible by 7. Tends to surprise people who haven't seen it before.

Oh WOW! why didn't i think of that? So the answer is 9, Thanx pwsnafu!

A*10^5 + B*10^4 + C*10^3 + A*10^2 + B*10 +C=A*10^2(10^3 + 1) + B*10(10^3 + 1) + C(10^3 +1)

(10^3 + 1)(100A +10 B + C) = 1001(100A +10 B + C) = 7*143(100A + 10B + C)

7*11*13(100A +10B + C) So this proves that your example is divisible by 7, 11, 13, 77, 91, and 143 in one proof.

Which completes the proof of your VERY COOL observation.

 

[Anamitra]

What pwsnafu has said is correct.In fact agentredlum has been shown diligent efforts with this one[the fourth sum] .All factorials onwards from 4! are divisible by 12. Agentredlum did the third one quite well and I was expecting him and many others, to do the fourth one by the far simpler method which pwsnafu has mentioned just now.Incidentally the second sum[in the set I have given] has an interesting alternative[solution technique].

 

[Anamitra]

If you take a number N=abcabc ,where a,b c are digits,you may write

N=abc*1000+abc
=abc[1001]
1001 is divisible by 7.

 

[agentredlum]

If you take a number N=abcabc ,where a,b c are digits,you may write

N=abc*1000+abc
=abc[1001]
1001 is divisible by 7.

Oh WOW! Your solution is more elegant.

 

[Anamitra]

Regarding sum 2 of post 153:

The square of any number[integral number] can be expressed as 5n or5n+1 or 5n-1

If you square a number the possible last digits[units place] are 0,1,4,6,5 and 9
If the last digit is one or nine the squared number is at a unit's distance from a number ending with zero.
If the squared number ends with 6, it is again at a unit's distance from a number ending with 5.
If the squared number ends with zero or five it is of the form 5n

This result,alternatively, may be established by using Fermat's Theorem
$${N}^{p-1}{-}{1}$$ is divisible by p[a prime] if N does not contain p as a factor.

Therefore,
$${N}^{4}{-}{1}$$ is a multiple of five[if N does not contain 5]
But
$${N}^{4}{-}{1}{=}{(}{N}^{2}{+}{1}{)}{(}{N}^{2}{-}{1}{)}$$
Hence the result.

 

[Anamitra]

Problem 1 of Post #153:Solution

The last digit[unit's digit] of any number[natural number] can occur in 10 ways from 0 through 9.We don't have eleven choices for it.
So, if we have eleven or more numbers[say 12 numbers] of we can always locate a pair which have the same value for the last digit.
Take their difference--you have a zero in the unit's place. So it must be divisible by five.

 

[agentredlum]

Problem 1 of Post #153:Solution

The last digit[unit's digit] of any number[natural number] can occur in 10 ways from 0 through 9.We don't have eleven choices for it.
So, if we have eleven or more numbers[say 12 numbers] of we can always locate a pair which have the same value for the last digit.
Take their difference--you have a zero in the unit's place. So it must be divisible by five.

Very nice sir, PLEASE POST MORE!

 

[Anamitra]

Here is simple tip for framing your own problems.

You raise a natural number to the fifth power. The digit in the unit's place does not change.
n and n to the power 5 have the same digit in the unit's place.
Therefore expressions like,
$${n}^{5}{-}{n}$$ are multiples of 5[and 2 of course]

You can always substitute n by some frightening formula that represents a natural number.

 

[micromass]

Question: find me 10 values for $t\in [0,2\pi[$ such that both cos(t) as sin(t) are rational?
Further question, find me 10 rational values for t such that both cos(t) as sin(t) are rationals.

If you did that question, then this should become not so hard anymore:
Find 10 values for t such that $\sqrt{x^3-2}$ is rational.

 

[agentredlum]

Question: find me 10 values for $t\in [0,2\pi[$ such that both cos(t) as sin(t) are rational?
Further question, find me 10 rational values for t such that both cos(t) as sin(t) are rationals.

If you did that question, then this should become not so hard anymore:
Find 10 values for t such that $\sqrt{x^3-2}$ is rational.

All pythagorean triples will solve The first half of the first question.

For question 3 do you mean 10 values of x such that sqrt(x^3-2) is rational?

Do you want x rational?

Great questions!

 

[cubzar]

post #166
1. cos(t) and [itex]\sqrt{}1-cos²(t)[/itex] are rational if cos(t)=a/c, with c=largest member of a pythagorean triple and a=one of the other members.
t=0, $\pi$, cos^-1(3/5), cos^-1(4/5), cos^-1(5/13), cos^-1(12/13), cos^-1(8/17), cos^-1(15/17), cos^-1(12/20), cos^-1(7/25), cos^-1(24/25)

 

[SteveL27]

All pythagorean triples will solve The first half of the first question.

For question 3 do you mean 10 values of x such that sqrt(x^3-2) is rational?

Do you want x rational?

Great questions!

Micromass must want x rational. Otherwise it's too easy: Any x such that x^3 is two more than a square works. For example x = cube root of {2, 3, 6, 11, 18, ...}.

 

[dimension10]

Four Simple Problems:
1. You have a set of twelve natural numbers[arbitrary natural numbers].Show that a pair can be always located from this set so that their difference is divisible by 5

2. Prove that the following:
$${[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}$$
where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]

3. What is the last digit[unit's place] in the sum:

1!+2!+3!+....99!

4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12

1.

If the numbers are a, b, c d, e, f, g, h, i, j, k and l, then

$$a \equiv m (mod 5)$$
$$b \equiv n (mod 5)$$
$$c \equiv o (mod 5)$$
...
$$l \equiv x (mod 5)$$

Then the question just requires that the difference between a certain 2 of the set {m, n, o,...,x} is divisible by 5. And we know that the {m, n, o,...,x} are either 0, 1, 2, 3 or 4. To fulfil the premises, two elements of the set {m, n, o,...,x} should be equal. This has to be as we have 12 numbers which is greater than 5.

3. 1!=1 (mod 10)
2!=2 (mod 10)
3!=6 (mod 10)
4!=4 (mod 10)
5!=0 (mod 10)
6!=0 (mod 10)
(I know I should put the congruence sign but I can't find that one on the keyboard :) )

Then all the factorials would have a remainder of 0. 1+2+6+4=13. The last digit of 13 is 3.

4.

1!=1 (mod 12)
2!=2 (mod 12)
3!=6 (mod 12)
4!=0 (mod 12)
5!=0 (mod 12)
...
1+2+6=9

The remainder is 9.

 

[agentredlum]

What is a rational angle? Please post any expression that represents a rational angle

 

[micromass]

All pythagorean triples will solve The first half of the first question.

For question 3 do you mean 10 values of x such that sqrt(x^3-2) is rational?

Do you want x rational?

Great questions!

Yes, I want x rational
You solved the first problem by using pythagorean triples, but that wasn't what I had in mind. What I had in mind is to find a formula that generates the pythagorean triples. The method of finding such a formula is useful in finding the rational values x such that $\sqrt{x^3-2}$ is rational.

 

[pwsnafu]

Further question, find me 10 rational values for t such that both cos(t) as sin(t) are rationals.

Not many choices! What an understatement.

Even if you replace "both" with "either" you still won't have many choices!

 

[Anamitra]

If $${\theta}{,}{cos}{(}{\theta}{)}{,}{sin}{(}{\theta}{)}$$ are rational simultaneously,
$${n}\theta{,}{cos}{(}{n}\theta{)}{,}{,}{sin}{(}{n}{\theta}{)}$$
are also rationals provided n is integral.This may not be true for the sub-multiple angles[I have assumed the angle to be in radians]

We can use Pythagorean triplets to generate the rationals sin[theta] and cos[theta] but the problem is to get a rational theta[not equal to zero] corresponding to the ratios of sin and cos.If we get one such set we can get a huge number of valid sets.
If $${tan}{(}\frac{\alpha}{2}{)}{=}\frac{m}{n}$$
Where m/n is a rational[m, n are integers, n not equal to zero]
$${sin}{(}\alpha{)}{=}\frac{2nm}{{m}^{2}{+}{n}^{2}}$$
$${Cos}{(}\alpha{)}{=}\frac{{n}^{2}{-}{m}^{2}}{{m}^{2}{+}{n}^{2}}$$

Here Sin and cos of the angle are rational--but the angle itself may be rational or irrational.
[This method fails to locate a rational theta for rational values of sin and cos]
A relevant link:
http://www.yaroslavvb.com/papers/olmsted-rational.pdf

The link discusses the situation in relation to degrees. May be things are more favorable in terms of radians.

 

[Anamitra]

From the last post:
If $${\theta}{,}{cos}{(}{\theta}{)}{,}{sin}{(}{\theta}{)}$$ are rational simultaneously,
$${n}\theta{,}{cos}{(}{n}\theta{)}{,}{,}{sin}{(}{n}{\theta}{)}$$
are also rationals provided n is integral.This may not be true for the sub-multiple angles[I have assumed the angle to be in radians]Now, if theta=a/b where a and b are integers and n=b we have an integral value a for which
Sin(a) and Cos(a) are rational