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Dadeyemi

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Yeah, it doesn't actually affect your proof I think.
Nice question;

Let A be the real sequence (a_1, a_2, a_3 ..) and define del(A) to be sequence (a_2 - a_1, a_3 - a_2, ... ). Suppose that all the terms of the sequence del(del(A)) are 1 and that a_(19) = a_(94) = 0. Find a_1

Sauce; Heinz - no wait AIME 1994
Reply 1982
DeanK22
Nice question;

Let A be the real sequence (a_1, a_2, a_3 ..) and define del(A) to be sequence (a_2 - a_1, a_3 - a_2, ... ). Suppose that all the terms of the sequence del(del(A)) are 1 and that a_(19) = a_(94) = 0. Find a_1

Sauce; Heinz - no wait AIME 1994


I think

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My method:

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Reply 1983
I'm getting

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Reply 1984
SimonM
I'm getting

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So do I, after hastily checking my arithmetic :rolleyes:
Reply 1985
Yes, although not the method I'd apply in the AIME.

I would have gone. Sequences, differences, polynomial, roots, divide by 2: 12(119)(194)\frac{1}{2}(1-19)(1-94)
SimonM
Yes, although not the method I'd apply in the AIME.

I would have gone. Sequences, differences, polynomial, roots, divide by 2: 12(119)(194)\frac{1}{2}(1-19)(1-94)
I think some people might appreciate it if you unpacked that comment slightly!
Reply 1987
DFranklin
I think some people might appreciate it if you unpacked that comment slightly!


Probably, I was trying to show that I'd not treat the question 'rigorously' (and how you could get there quickly)

Essentially, (from GCSE coursework)

We know that if the nth row of finite difference tables is constant, we're looking at a polynomial of degree n. They give us the roots of this polynomial in the question, so all that remains is to determine the leading coefficient. For x2x^2 if the constant row is 1, then the coefficient will be 12\frac{1}{2}
Reply 1988
Show that if p2(x)xq2(x)=xr2(x)p^2(x)-xq^2(x)=xr^2(x) then p=q=r=0p=q=r=0 (not too hard, so don't do it if it's obvious for you)
SimonM
Show that if p2(x)xq2(x)=xr2(x)p^2(x)-xq^2(x)=xr^2(x) then p=q=r=0p=q=r=0 (not too hard, so don't do it if it's obvious for you)


That's not necessarily true

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Reply 1990
tommm
That's not necessarily true

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Sorry, p,q,rR[x]p,q,r \in \mathbb{R}[x] (polynomials)
Reply 1991
Couple of questions I came across earlier this week.

1) what is the sum of all 4 digit positive integer palindromes eg. 3223

2) In a school there are 1000 lockers and 1000 students. On the first day of school, the first student opens all the lockers. The second student then closes all the lockers with an even number. The third student will 'reverse', every third locker (if it is open, the student closes it. If it is closed, the student will open it). The fourth student reverses every fourth locker, and so on until all students have passed through. Which lockers will be open at the end?


The first is probably a fair bit harder than the second.
miml
Couple of questions I came across earlier this week.

1) what is the sum of all 4 digit positive integer palindromes eg. 3223

2) In a school there are 1000 lockers and 1000 students. On the first day of school, the first student opens all the lockers. The second student then closes all the lockers with an even number. The third student will 'reverse', every third locker (if it is open, the student closes it. If it is closed, the student will open it). The fourth student reverses every fourth locker, and so on until all students have passed through. Which lockers will be open at the end?


The first is probably a fair bit harder than the second.


Take any one four digit palindrome;

1000a + 100b + 10b + a. a can be from the set {1,2,3,..9} and b from the set {0,1,2,...,9}.

If a = 1 then you are clearly summing

1000 + 1, 1000 + 100 + 10 + 1, ... , 1000 + 900 + 90 + 1

Let us call this sum S.

If a=2 we have a similar situation - but note that we only have to add 10(1000) + 10 to S to get his sum.

(2) Can you consider a much smaller scenario, say 10 lockers. Do you notice anything going one here say anything to do witht he divisors?
Reply 1993
DeanK22
Take any one four digit palindrome;

1000a + 100b + 10b + a. a can be from the set {1,2,3,..9} and b from the set {0,1,2,...,9}.

If a = 1 then you are clearly summing

1000 + 1, 1000 + 100 + 10 + 1, ... , 1000 + 900 + 90 + 1

Let us call this sum S.

If a=2 we have a similar situation - but note that we only have to add 10(1000) + 10 to S to get his sum.

(2) Can you consider a much smaller scenario, say 10 lockers. Do you notice anything going one here say anything to do witht he divisors?

Ahh, I have done them. Just put them here for people to try.
In know the locker one was on the oxford 2008 entry paper but where di you get the first?
Reply 1995
The Muon
In know the locker one was on the oxford 2008 entry paper but where di you get the first?

I didn't know it was in the Oxford paper. It was also in MAO 1992.

The first one is from Mandelbrot 1.

Both are actually in the Art of Problem Solving Books.
Reply 1996
miml
I didn't know it was in the Oxford paper. It was also in MAO 1992.

The first one is from Mandelbrot 1.

Both are actually in the Art of Problem Solving Books.


Apparently it was also used as an interview question according to a random someone I sat next to at lunch in Keble :>

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ShortRef
Apparently it was also used as an interview question according to a random someone I sat next to at lunch in Keble :>

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It is to do with squares.
Reply 1998
DeanK22
It is to do with squares.


yes, I know :smile:
Reply 1999
For all math lovers, this is a MUST see: Finite Simple Group (of Order Two)

This is what happens when Maths and Art combines :biggrin:

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