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STEP I Question 5, have I completely dropped the ball?

A:Un=1+5(n1)A: U_{n}=1+5(n-1)
B:Un=2+5(n1)B: U_{n}=2+5(n-1)
C:Un=3+5(n1)C: U_{n}=3+5(n-1)
D:Un=4+5(n1)D: U_{n}=4+5(n-1)
E:Un=5+5(n1)E: U_{n}=5+5(n-1)

to prove that any term in B added on to a term in C is a term in E, I did:

2+5(n1)+3+5(t1)=5+10((nt1)1).2+5(n-1) + 3+5(t-1) = 5 + 10((n-t-1)-1).

This term clearly lies in E as 10 is a multiple of 5.

to prove that a term in B^2 is a term in D, I did:

(2+5(n1))2=(4+5(p1))(2+5(n-1))^2 = (4+5(p-1)) where pnp \neq n

4+20(n1)+25(n1)2=(4+5(p1))4 + 20(n-1) + 25(n-1)^{2} = (4+5(p-1))

this simplifies to:

4+5((5n26n+2)1)4 + 5((5n^{2}-6n+2) - 1)

4+5(p1)4 + 5(p-1)

therefore B2B^2 lies in DD.

----------------

(i) x2+5y=243723x^2 + 5y = 243723

x^2 can be any number, so let x2=r.x^2 = r.

y can be any number, so let y=(n1)y = (n-1)

so:

r+5(n1)=243723r + 5(n-1) = 243723

because the last digit in 243723 is 3, assume that this number belongs to the set of numbers, CC.

(n1)=243723r5(n-1) = \frac{243723-r}{5}

if n=1, then the RHS must be equal to zero.

However, this cannot be. Because r must be equal to 3 if it is to belong to the C progression.

243723350\frac{243723-3}{5} \neq 0

Hence x and y must belong to the negatives/non integers as it does not obey the normal rules for the 5 positive progressions.

am I wrong? any tips? I don't know what to do here!

this is from a hard paper, 72/120 for 1, 55/120 for 2.

this is STEP I 2004

EDIT: I don't want to attempt (ii) if my solution to (i) is completely off of the ball.

thanks
(edited 12 years ago)
Reply 1
bump

any people who are good at STEP online atm?
(edited 12 years ago)
Reply 2
ok, I'll bump later when all the mathmos are back
The answer is 4.
Reply 4
I only skim-read it, but I think the important fact is to note that x^2 must be in progression D. So adding 5y, you're still in progression D, contradiction.

Edit-as it is, I think your proof basically works, but you might need to tighten up a few bits.
(edited 12 years ago)
Reply 5
There is a separate STEP discussion thread where the papers from each year going back to the early 1990's are discussed in depth. In most case model answers are included.

This would be a good place to start looking for info on questions such as this.

One point to note is that when adding the term in B to the term C you can not assume that the n is the same.

So (2 + 5q) + (3 + 5r) = 5 + 5q + 5r = 5 (1 + q + r) a multiple of 5.
Reply 6
Original post by House Dagoth
ok, I'll bump later when all the mathmos are back
You should NOT bump a thread less than an hour after you posted it, let alone three times.
Reply 7
Original post by msmith2512
One point to note is that when adding the term in B to the term C you can not assume that the n is the same.

So (2 + 5q) + (3 + 5r) = 5 + 5q + 5r = 5 (1 + q + r) a multiple of 5.


thanks for the correction. I see what you mean.
(edited 12 years ago)
Reply 8
your solution looks pretty solid to me.
Have you found out which categories the square numbers live in? (note that you've been invited to consider one example earlier)

Numbers in some categories cannot therefore be squares.
Reply 10
Original post by ian.slater
Have you found out which categories the square numbers live in? (note that you've been invited to consider one example earlier)

Numbers in some categories cannot therefore be squares.


because there is an x^2 the term must lie in A, D, or E. This is because x^2 takes the value of 1,4,9,16,25,36 etc, these all lie in A, D or E... and adding on any multiple of five will not change that (as you can clearly see).

However, the last digit of the number of 243723 is 3, hence this number lies in the progression C.

Contradiction.

Hence there are no positive integer values of x and y as it does not obey the rules for the five positive progressions.

what do you think? right? wrong?

thanks,
Dagoth Ur.
(edited 12 years ago)
Reply 11
I think you should read the questions a bit more carefully.

Why you are asserting "contradiction" all the time, show a convincing result first.

You are correct, but you must show clearly the result.

x2+5y=243723x^2 + 5y = 243723

Therefore, as you have identified it to be an element of the sequence CC:

x2+5y=3+5(n1)x^2 + 5y = 3 + 5(n - 1)

as you say, xx and yy can be arbitrarily chosen, then:

x2+5(n1)=3+5(n1)x^2 + 5(n - 1) = 3 + 5(n - 1)
x2=3x^2 = 3

***

But is this a proof of your statement?

1. Have you shown that this number can't be represented by any of the other sequences?
2. Is this the only case that should be considered?
3. What about assuming that xx and yy are themselves part of the sequences? (rather than the result...)

You are not reading carefully, you were given all the positive numbers, and yet you are not using them!
There is no proof of two lines; you are not solving a question, but proving the solution.
Original post by House Dagoth
because there is an x^2 the term must lie in A, D, or E. This is because x^2 takes the value of 1,4,9,16,25,36 etc, these all lie in A, D or E... and adding on any multiple of five will not change that (as you can clearly see).The key thing here is showing that x^2 must always lie in A, D or E, and you haven't done that.
In this case I think you're safest proving the result for each category A to E.

E.g. Suppose x in A. Then x = 1 + 5n for some n>=0. Then x^2 = 1 + 10n+25n^2 = 1 + 5m (where m = 2n+5n^2). So x^2 is in A.
Similarly if x in B, x=2+5n, x^2 = 4 + 20n + 25n^2 = 4 + 5m. So x^2 is in D.
...

The rest of your argument is OK, but you'd lose most of the marks for not having shown the x^2 in A,D or E result.

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