Problem 141

[euler]

We are always keen to address wording concerns where the meaning of the problem is ambiguous or lacking clarity, but your concern is stylistic rather than grammatical.

Indeed there are some people who prefer to use "that" exclusively with restrictive clauses (where the meaning would be affected by its absence) and "which" with non-restrictive clauses (where the clause contains non-essential information), but the only grammatical rule I am aware of is that "that" should never be used with a non-restrictive clause. In other words, in sentences which contain restrictive clauses, like the ones you make reference to and this one, it is perfectly acceptable to use "which" or "that".

You may find the link below interesting as you will discover that you have more grammatical freedom than you realised in such contexts.
http://www.oxforddictionaries.com/words/that-or-which

[dharasty]

I think I've found 29 distinct progressive perfect squares below one hundred thousand... and I haven't even searched all combos yet. My sum (of the ones I've found so far) is way over the one stated in the problem.

My answers are legit: I can show the d,q,r values, the geometric ratio, and the integer square root (sr). d*q+r = sr * sr = n < 100000.

For example, I've found THREE distinct progressive perfect squares over 90000 and under 100000.... so those three alone sum to over 124657.

Obviously... some of my findings are NOT legit! Can I post a few of my "findings", and someone can tell me why these d, q, r values are wrong?

[TripleM]

You can PM me.

[Oliver1978]

My thoughts concerning the sample numbers from the description: 9 is in the list because with d=2 we get q=4, r=1. For this [1,2,4] there's the factor 2 forming this geometric sequence.

Am I right with this? Or am I completely going wrong?

[RobertStanforth]

My thoughts concerning the sample numbers from the description: 9 is in the list because with d=2 we get q=4, r=1. For this [1,2,4] there's the factor 2 forming this geometric sequence.

Am I right with this? Or am I completely going wrong?

Yes, you are correct.

[Treferwynd]

I keep getting the same numbers as Erf_Erf, I've tried a few different approaches and keep getting 23261047907 for 10^10.

Can I show my numbers to someone to see what I'm missing?

[jaap]

I keep getting the same numbers as Erf_Erf, I've tried a few different approaches and keep getting 23261047907 for 10^10.

Can I show my numbers to someone to see what I'm missing?

That number is too high. The difference to what it should be is a 10-digit square so probably you have one errant entry that shouldn't be included in the sum.

[Treferwynd]

That number is too high. The difference to what it should be is a 10-digit square so probably you have one errant entry that shouldn't be included in the sum.

Thank you very much, I've solved it! I had missed an incredibly simple but extremely important fact...

For anyone else getting the same numbers: don't be too clever when checking your results!

[PierrotLeFou]

This may be a stupid question, but can 1 (one) be a valid ratio?
Starting at 1: 1*1 = 1, 1*1 = 1^2
or 9*1 = 9 and 9*1 = 9 = 3^2
It doesn't correspond to a "progressive" sequence.
My understanding of the problem:
Square numbers are a product of prime numbers raised to an even exponent.
Is I have s = af^2 where f is any combination of factors.
I choose f as the numerator of the ratio and any integer d less than f as the denominator.
So, af^2 / (f/d) -> afd -> ad^2.
So the second backward term is also a square.
There are f-1 possible sequence for each choice of f.

[mdean]

I think you're misinterpreting something. d,q, and r can never be all equal. If you divide a number by 3, the largest possible remainder is 2. So the ratio can never be 1.

[PierrotLeFou]

Maybe I misunderstand something but if n=1, d=1, I will have q=1 and r=0
OK by reading again my post, I understand ... r must be at least 1 to have a geometric progression.
ex. 9 / 4 = 2, r = 1. 9 is an accident. I'm considering 4, 2, 1. Note that 4 and 1 are squares.

[PierrotLeFou]

Well, I'm rather frustrated ...
Can anyone tells me [ appeal for hints removed by moderator ]
Recently, I had little time to devote to this problem.
We can always consider the ratio between D, Q and R as a rational number but the denominator can be 1.
So it becomes an integer ratio.
Also, the numerator and the denominator must be relatively primes. Otherwise we will get duplicate solutions.
An example with 36. We have 36 = 2^2 x 3^2. So the factors/numerators are 2, 3 and 6
With 2 / 1: 36 -> 18 -> 9
With 3 / 1: 36 -> 12 -> 4
With 3 / 2: 36 -> 24 -> 16
With 6 / 1: 36 -> 6 -> 1
With 6 / 2: 36 -> 12 -> 4 the same as 3 / 1
With 6 / 3: 36 -> 18 -> 9 the same as 2 / 1
With 6 / 4: 36 -> 24 -> 16 the same as 3 / 2
With 6 / 5: 36 -> 30 -> 25