Project Euler Forum

No, there isn't.

Thanks for the clarification.

A friend of mine spoiled the answer to this one for me, so I know my current wrong answer is close (my answer is 12 too small), but I refuse to type in the correct answer until I get it myself.

Can anyone confirm or deny the following partial results:

Let d be the common difference of the arithmetic progression in question. If you restrict yourself to arithmetic progressions that have d<=400, 550, 10^3, or 10^4, then the last three digits of the answers would be respectively.

It's killing me trying to find out where the missing 12 solutions are.

Kevin_L wrote:A friend of mine spoiled the answer to this one for me, so I know my current wrong answer is close (my answer is 12 too small), but I refuse to type in the correct answer until I get it myself.

Can anyone confirm or deny the following partial results:

Let d be the common difference of the arithmetic progression in question. If you restrict yourself to arithmetic progressions that have d<=400, 550, 10^3, or 10^4, then the last three digits of the answers would be

Expand

419, 770, 462, 970

respectively.

It's killing me trying to find out where the missing 12 solutions are.

you're letting z equal 0

z != 0

I am enjoying this problem very much.
I would like to verify my algorithm for a values of n other than 1155.
The next higher value of n I get with ten solutions is n=1755.
I suspect this is incorrect.
Am I incorrect in my calculation that n=1755 has ten solutions?
Does it really have more?
Thanks in advance.

Neilius wrote:..
The next higher value of n I get with ten solutions is n=1755.
I suspect this is incorrect.
...

It is you suspicion that is incorrect.

Tom

dawghaus4 wrote:
It is you suspicion that is incorrect.

Tom

Many thanks.
I finally worked it out.
Program runs in about 6 seconds.
A very enjoyable problem

n = a*b, a > b
Let n = 15 = 15*1 = 5*3
1. a = 15, b = 1 => d = 4, 19^2 - 15^2 - 11^2 = 15
2. a = 5, b = 3 => d = 2, 7^2 - 5^2 - 3^2 = 15
n = 15 has exactly 2 solutions and 15 < 27 (by condition). Where am I wrong?

DanielJackson wrote: Wed Feb 19, 2020 11:28 pm n = a*b, a > b
Let n = 15 = 15*1 = 5*3
1. a = 15, b = 1 => d = 4, 19^2 - 15^2 - 11^2 = 15
2. a = 5, b = 3 => d = 2, 7^2 - 5^2 - 3^2 = 15
n = 15 has exactly 2 solutions and 15 < 27 (by condition). Where am I wrong?

$5^2-3^2-1^2=15$. 15 has at least 3 solutions.

mdean wrote: Thu Feb 20, 2020 12:40 am $5^2-3^2-1^2=15$. 15 has at least 3 solutions.

I realized I hadn't considered it. Thanks