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**Posts by NIH**

Here is a solution to this one.

a(a + 1) = b(b + 2) <=> a² + a + 1 = (b + 1)².

For positive a, a² < a² + a + 1 < (a + 1)², implying there is a perfect square between consecutive squares; contradiction.

For negative a, a(a + 1) = p(p - 1), where p = -a, so there are no solutions for p - 1 > 0; i.e., a < -1.

It is then easy to check for solutions with a = -1, 0.

These are: (a,b) = (-1,-2), (-1,0), (0,-2), (0,0).

mathsyperson wrote:

What an amazing coincidence! It must be coincidence because it stops working when you use any power higher than 5.

It's a consequence of the following even bigger coincidence!

a^n + (a + 4b + c)^n + (a + b + 2c)^n + (a + 9b + 4c)^n + (a + 6b + 5c)^n + (a + 10b + 6c)^n = (a + b)^n + (a + c)^n + (a + 6b + 2c)^n + (a + 4b + 4c)^n + (a + 10b + 5c)^n + (a + 9b + 6c)^n,

where a, b, c are any positive integers and n can be 1, 2, 3, 4, or 5.

See also http://nrich.maths.org/askedNRICH/edited/412.html and http://www.primepuzzles.net/puzzles/puzz_065.htm

ganesh wrote:

Problem # n+8

Can you find two numbers composed of only ones which give the same result by addition and multiplication?

How about

and . Works in any number base!wcy wrote:

how to differentiate log x ?

Out of interest, you can derive it from the derivative of the exponential function:

y = log x <=> x = e^y.

So dx/dy = e^y = x.

Hence dy/dx = 1 / dx/xy = 1/x.

I found this in a book...

1 + 6 + 7 + 17 + 18 + 23 = 2 + 3 + 11 + 13 + 21 + 22;

1^2 + 6^2 + 7^2 + 17^2 + 18^2 + 23^2 = 2^2 + 3^2 + 11^2 + 13^2 + 21^2 + 22^2;

1^3 + 6^3 + 7^3 + 17^3 + 18^3 + 23^3 = 2^3 + 3^3 + 11^3 + 13^3 + 21^3 + 22^3;

1^4 + 6^4 + 7^4 + 17^4 + 18^4 + 23^4 = 2^4 + 3^4 + 11^4 + 13^4 + 21^4 + 22^4;

1^5 + 6^5 + 7^5 + 17^5 + 18^5 + 23^5 = 2^5 + 3^5 + 11^5 + 13^5 + 21^5 + 22^5.

That works!

Here's another approach.

Consider f(n) = 11 x 14^n + 1, modulo 15.

Then f(n) = 11 x (-1)^n + 1 = 12 or 5 (mod 15).

Hence f(n) = 0 (mod 3) or 0 (mod 5), is greater than 5 for all n, and is always composite.

(f(n) = 12 (mod 15) <=> f(n) = 15t + 12, for some integer t <=> f(n) = 3(5t + 4) <=> f(n) = 0 (mod 3), etc.)

Or you could use induction.

f(0) = 12 is divisible by 3,

f(1) = 155 is divisible by 5.

Then f(n+2) - f(n) = 11 × 14^n(14^2 - 1) = 11 × 3 × 5 × 13 × 14^n is divisible by both 3 and 5, and the result follows by induction.

(f(n) divisible by 3 implies f(n+2) divisible by 3; f(n) divisible by 5 implies f(n+2) divisible by 5.)

wcy wrote:

a can only be 0

Or -1.

If you can show why those are the only two values, the problem is solved!

wcy wrote:

but what abt LHS ?

Then the LHS equals a² + a + 1.

For what values of a can this be a perfect square?

mathsyperson wrote:

There are no squares that end in 2 or 8, so if 4 + 16(4n²+7n+3) always ends in 2 or 8, then we have a proof.

But if n = 3 or 4 (mod 5), then 4 + 16(4n²+7n+3) ends in 4.

Here's a hint: add 1 to both sides of the original equation.

**NIH**- Replies: 3

Let n be a non-negative integer. Is 14^n + 11 ever a prime number?

Try graphing y = sin(x) + sin(x°). (y = sin(x radians) + sin(x degrees).)

Is the function periodic?

**NIH**- Replies: 13

Find all integer solutions of a(a + 1) = b(b + 2).

A simple proof that no *integer* of the form 4n-1 can be written as the sum of two squares.

If n is even, say n = 2r, (2r)² = 4r² = 0 (mod 4). (Leaves remainder 0 when divided by 4.)

If n is odd, say n = 2r + 1, (2r + 1)² = 4(r² + r) + 1 = 1 (mod 4.)

So the sum of two squares, mod 4, is always equal to 0 = 0 + 0, 1 = 0 + 1 or 1 + 0, or 2 = 1 + 1; never 3. (Of course 4n - 1 = 3 (mod 4.))

In fact, for the square of an odd integer, we can write (2r + 1)² = 4r(r + 1) + 1.

Since one of r and r + 1 must be even, 4r(r + 1) is divisible by 8.

So the square of an odd integer leaves remainder 1 when divided by 8, which is sometimes useful.

mathsyperson wrote:

Random fact for the day: All primes of the form 4n+1 can be written as a²+b².

None of the primes of the form 4n-1 can be written in this way. (a, b, n ∈Z)

If you can prove/disprove this, you will be very rich.

Fermat proved this at the beginning of the 17th century! In fact, he proved that all primes of the form 4n+1 can be written as the sum of two squares in exactly one way. The proof is not trivial. However, there is a very simple proof that no primes of the form 4n-1 can be written as the sum of two squares.

wcy wrote:

haha this is meant to be a trick question...

Then you should have specified a circular orbit!

For an elliptical orbit, both the speed and the velocity change. But for a circular orbit, although the speed is constant, the velocity changes, thus greatly increasing your chances of pulling off the trick!

It's not possible to express y explicitly in terms of x using any of the standard elementary functions. However, there is a formal solution using something called the Lambert W function. This function can be evaluated using mathematical packages such as Maple and Mathematica, but no calculator currently has a button for it. See the references below for details.

http://mathworld.wolfram.com/LambertW-Function.html

http://www.americanscientist.org/template/AssetDetail/assetid/40804;_voi8-8bIm

http://www.orcca.on.ca/LambertW/

Another approach would be to use the Newton Raphson method. If a is an approximation to a root of

f(x) = 10^x - x^10 = 0, then a - f(a)/f'(a) will be a better approximation.

In this case, we have f'(x) = ln(10) * 10^x - 10x^9.

For example, if a = 1.4 is an approximate solution, then

1.4 - (10^1.4 - 1.4^10)/(ln(10) * 10^1.4 - 10*1.4^9) ~= 1.3744 is a better approximation.

This converges quite rapidly. The next two convergents are, to 10 decimal places, 1.3713296532 and

1.3712885814.

http://www.sosmath.com/calculus/diff/der07/der07.html

Finally, here's a page which will solve, for example, the equation 10^x = x^10, giving several answers in terms of the Lambert W function (aka the ProductLog function), along with the numeric values.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic

Here's a hint as to why the total is always 25.

In the examples above, color or shade (in the first two columns) the numbers 1, 2, 3, 4, and 5. Do you see a pattern?

mathsyperson wrote:

I think NIH might have had a typo when telling us the second one...

Thanks -- it's fixed now!

Of course, this comes from the well known result, first proved by Euler, that pi^2/6 = 1/1^2 + 1/2^2 + 1/3^2 + ...

MathsIsFun wrote:

Who could possibly think that comparing the distance your feet are apart to the length your foot travels would be a number that has digits going on forever ...

Indeed. And who could imagine that that ratio is equal to 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + ... ? Or imagine that it is equal to sqrt(6/1^2 + 6/2^2 + 6/3^2 + ... ) ?

ckvijayakumar wrote:

... log x=0 log0...

The flaw in your proof is that the domain of the log function does not include 0. In other words, log0 is undefined.

Zach wrote:

I'm not sure if I got that correct, based on the point of the idea. Perhaps I got it wrong.

The first column should be in *ascending* order.

mathsyperson wrote:

I'm annoyed at this because at a glance it seems easy to prove how it works, but on closer inspection it's not at all!

I agree; it *looks* trivial, but it's not!

**NIH**- Replies: 25

Take the first 10 natural numbers -- that's 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Pick any five of the numbers, and write them in a column, in increasing order. Then take the five remaining numbers -- the ones you didn't pick -- and write them in a second column, in decreasing order.

In each row, subtract the smaller number from the larger one, giving you a third column. (See the example below.) Now add the numbers in the third column.

1 9 8

4 7 3

5 6 1

8 3 5

10 2 8

Total: 25

What is your total? Everybody post their total...

You're almost there, mathsyperson!

It's better to leave the expression as A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16. If a, b, c are all odd then the numerator is the product of four odd numbers, and hence is itself odd. So A^2 is not an integer, and neither is A. The same argument applies if two of a, b, c are equal to 2 and the remaining side length is odd.

Milos has already shown that if a = b = c = 2, then A = sqrt(3). If a = 2 and b, c are odd, then, to form a triangle, we must have b = c, with b > 1. (Try drawing a triangle with sides 2, 9, 11!) So we have an isosceles triangle with base 2 and height sqrt(b^2 - 1). b^2 - 1 is not a perfect square for b > 1, so the height is irrational, and hence also the area.

So we've proved a stronger result, a triangle with all sides lengths equal to either 2 or an odd number does not have integral area.

See What is 0 to the 0 power? for a good discussion of this topic.

MathsIsFun wrote:

Good one, NIH! That sum *should* be equal to the height of the triangle, am I right?

Indeed it should!

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