Easy math puzzles

Quote: teliot

Okay ... new puzzle. A sum of consecutive squares equaling a sum of the following successive consecutive squares.

What's the longest one you can find?

Quote: ThatDonGuy

Follow-up: as of now, it's 2,914,365 terms:
483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695
This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

Quote: teliot

of course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.

Quote: ThatDonGuy

Let there be A numbers on the left, and B on the right
(N - (A - 1))^2 + (N - (A - 2))^2 + ... + (N - 1)^2 + N^2 = (N + 1)^2 + (N + 2)^2 + ... + (N + B)^2

Quote: ThatDonGuy

N (N + 1) (2N + 1) - (N - A) (N - A + 1) (2 (N - A) + 1) = (N + B) (N + B + 1) (2 (N + B) + 1) - N (N + 1) (2N + 1)

Quote: teliot

of course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.

Quote: ThatDonGuy

Does the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.

Quote: ThatDonGuy

Does the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.

Quote: ThatDonGuy

Follow-up: as of now, it's 2,914,365 terms:
483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695
This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

Quote: teliot

Is there a solution that starts at N = 1?

Quote: teliot

Is there a solution that starts at N = 1?

Quote: ThatDonGuy

Pretty much every odd number from 3 through 99 has a sequence that has that many terms. I think the smallest even number size is 52.

Quote: GM

My infinite family gives a solution for every odd number at least 3. The smallest even number I have found was also 52.

Quote: ThatDonGuy

Any ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

If it is A^3 + (A+1)^3 + ... + N^3 = (N+1)^3 + ... + B^3, then this becomes 2 ((N-1) N / 2)^2 = ((A - 1) A / 2)^2 + (B (B+1) / 2)^2, which can be expressed as 2 Q^2 = P^2 + R^2, where P < Q < R and all three are triangular numbers.
The {3, 4, 5, 6} solution has A = 3, N = 5, B = 6; 2 • (5 • 6 / 2)^2 = 450 = 9 + 441 = (2 • 3 / 2)^2 + (6 • 7 / 2)^2; P = 3, Q = 15, and R = 21.

Quote: teliot

Is there a solution that starts at N = 1?

Quote: ThatDonGuy

Any ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?

Quote: unJon

I assume the winner wins the amount of money he has left. But does does the loser get $0 or some other amount?

Quote: GM

I am quite confident that there is not. If you assume that 1^2+2^2+...+N^2=(N+1)^2+...+(N+B)^2, then after using the summation formula for squares, you get the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0, which has degree 3 in both B and N. By using the the change of variables x=2(4B+2N+1)/(2B-2N-1), y=-6/(2B-2N-1), this equation can be transformed to the form y^2=x^3-12x+20. (This is not black magic, it is called the Weierstrass form and the curve defined by this equation is an elliptic curve.)

If the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0 has an integer solution for B and N, we get a rational (but not necessarily integer) solution of y^2=x^3-12x+20. The good news is that elliptic curves have been studied extensively and there is a description of the rational solutions of this equation, although it is not as simple as substituting numbers into a formula.

If we know x, y, then B and N are given by B=-(x+2)/(2y), N=(4-x-y)/(2y). y=-6/(2B-2N-1) means that y has to be a (possibly negative) integer dividing 6 or a fraction whose numerator in its lowest terms is a (possibly negative) integer dividing 6.

The integer solutions of y^2=x^3-12x+20 are (x,y)=(-2,6),(4,6),(2,2),(1,3),(10,30),(4,2),(22,102),(89,839) and the same x values with the sign of y changed. None of these give feasible values for B and N. The numerators and denominators of the rational solutions increase quickly, so numerator will never be a divisor 6, but a rigorous proof would require someone who knows more about this topic than I do.

Quote: ChumpChange

Super Old School episode:
JEOPARDY! 1974-75 Nighttime Syndicated Season - YouTube
https://www.youtube.com/watch?v=GtyWD9QpFME
If they all won in Final Jeopardy, there would have been a tie. Would they have given out 2 cars and had 2 returning champions?

Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?

Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?

Quote: GM

If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.

Quote: unJon

Show Spoiler

This is a simple game theory problem with no pure Nash equilibrium solution. That’s because if Amy goes High, Bob is best off going Low (winning 50% for Bob). But if Bob goes low, then Amy does best going low (winning 100% for Amy). But if Amy goes low, Bob does best going high (winning 50%). But if Bob goes high, Amy does best going high (winning 75% for Amy).

In other words, Amy wants to do the same thing as Bob: go high or low together. And Bob wants to do the opposite of Amy: go low when she goes high and vice versa.

The mixed Nash equilibrium is found by choosing the probability of going High and Low such that the opponent is indifferent between the choice.

For Amy that means: 25x + 50(1 - x) = 50x + 0(1 - x)
x = 2/3

For Bob that means: 75y + 50(1 - y) = 50y + 100(1 - y)
y = 2/3

Both Amy and Bob should randomly choose high 2/3 of the time and low 1/3 of the time. That leads to Amy winning the show 2/3 of the time and Bob winning the show 1/3 of the time. Importantly, this is the mixed Nash equilibrium because neither Amy nor Bob can improve their chance of winning by varying their strategy so long as their opponent follows the 2/3 high and 1/3 low random strategy.

Note: that these %s would be different if Amy and Bob were trying to maximize the dollars they win based on Jeopardy rules.

Quote: GM

If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.

Quote: teliot

A friend wrote:

"I played quickly and see that Magma can convert the elliptic curve to a Weierstrass form, which is nice. The problem is that when one does such a conversion, the integer points from the original equation can be rational points on the new equation that are not integers. So this may not be as simple an exercise as I originally thought. Again, I have not done much with elliptic curves myself, so perhaps it is not that difficult in the end."

He suggested another professor, who I will write to and report back.

Quote: charliepatrick

Show Spoiler

Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.

Quote: charliepatrick

Show Spoiler

Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.

Quote: teliot

The other professor wrote back to me. He is an expert in Diophantine Equations. He wrote a rather long email to me, he definitely gave this some deep thought, and wrote in part,

"The problem therefore is to ensure these ratios are integral, for (x,y) a rational point on y^2=x^3-12x+20. This may or may not be easy: I don't have ideas off the top of my head. I checked that there are no solutions to your equation for |b|,|n|<10^5, so likely there are none.

But hope is not lost. Nikos Tzanakis has written a memoir devoted exclusively to finding integral points on curves of genus 1, particularly helpful when the curve of genus 1, as (*) above, is not in standard form."

Quote: charliepatrick

Show Spoiler

Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.

Quote: teliot

Show Spoiler

The unique prime divisors of 12 are 2 and 3, so at the very least it is clear that it balances with all multiples of 2 and 3 that are <= 12. And 5 = 2+3, 7 = 2+2+3 so you can do both of these since balanced + balanced = balanced. Of course, this relies on open spots being available to put the balanced configurations, which is why 11 fails.

Quote: charliepatrick

Further to other responses, I agree...

Show Spoiler

...that 5 would also work using 12h and 6h for 2, then adding (say) 11h 7h 3h. This also means 7 would work. So it looks as if only 1 and 11 don't work.

Comment

Yes, on a 12 tube centrifuge, only 1 and 11 tube spins need a dummy.



Note that the empty spots for a 5-tube spin become the full spots for a 7.

Joke

Quote: Gialmere

Despite the Wizard playing a lab tech on TV, both his and CP's answers are incorrect.

Quote: GM

I will try to get hold of Nikos Tzanakis' memoir. I have also noticed that he has a paper on sums of consecutive squares being a perfect square, which also involves elliptic curves.

In the meanwhile, I have come up with the following idea, which could reduce the problem to checking finitely many points on the elliptic curve y^2=x^3-12x+20 if someone were willing to do some analytic number theory.

As I wrote earlier, integer solutions for N and B would give a rational point on the elliptic curve such that the numerator of y-coordinate divides 6. Let's assume that we have such a rational point.

The co-ordinates of the rational points on the elliptic curve are of the form x=r/t^2, y=s/t^3, where r, s, t are integers, r and s are coprime to t, but not necessarily to each other.

Let u be the real root of x^3-12x+20 (it's about -4.11), and v, w the non-real roots. Then x^3-12x+20=(x-u)(x-v)(x-w), so ((r/t^2)-u)((r/t^2)-v)((r/t^2)-w)=s^2/t^6. r/t^2 is close to u, certainly between -4.11 and -4, which means that ((r/t^2)-v)((r/t^2)-w)>37 as v is about 2.05+0.8i and w is its conjugate, while on the right-hand side, s^2 is at most 36, so r/t^2-u<1/t^6. r/t^2-u>0, therefore |r/t^2-u|<1/t^6. This means that r/t^2 is a very good approximation to u.

Roth's theorem says that for any epsilon>0, there are only finitely many integers p, q (q>0) such that |u-p/q|<1/q^(2+epsilon). Unfortunately, this theorem is not effective, but there is an effective version due to Feldman, which says that there exists a kappa<3 (3 is degree of the polynomial which u is a root of) and a constant c>0, such that |u-p/q|>c/q^kappa for all integers p, q (q>0), and kappa and c can be computed in principle. This, combined with |r/t^2-u|<1/t^6 would give a bound on t.

The problem is that there are formulas which kappa and c which work for any algebraic number, but these are useless for practical purposes (the bound on t would be at least 10^1000000000000), however, much better kappa and c can be obtained for any concrete number if one is willing to work through many pages of analytic number theory. For example, |2^(1/3)-p/q|>(q^2.47)/4, and there are also good values of kappa and c for cube roots of other integers less than 100, but I don't know to what extent these calculations could be adapted to deal with the root of a more general cubic equation.

It is known that the logarithm of the maximum of the absolute values of the numerator and denominator of m times (-2,6) in the group structure on the elliptic curve grows like Cm^2, where C is a constant, and it is possible to give an effective lower bound for it. This means that given a bound on t, m can be bounded effectively and then it is a finite calculation to check all the possibilities. The bound on t would not have to be very good, something like 10^100000 would be OK, probably even 10^1000000, because it would still only involve checking a few thousand multiples of (-2,6).

Quote: GialmereA random number generator generates integers in the range 1...n, where n is a parameter passed into the generator. The output from the generator is repeatedly passed back in as the input.

If the initial input parameter is one googol (10^100), find, to the nearest integer, the expected value of the number of iterations by which the generator first outputs the number 1.[/b

That is, what is the expected value of x, after running the following pseudo-code?

Quote: Ace2

Show Spoiler

What ThatDonGuy said plus EM constant.

So 100 * ln(10) + .5772 + 1 =~ 231.84

Quote: ThatDonGuy

Quote: Ace2

Show Spoiler

What ThatDonGuy said plus EM constant.

So 100 * ln(10) + .5772 + 1 =~ 231.84

Okay, I'll ask: what's the EM constant?