## Problem 135

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TripleM
Posts: 382
Joined: Fri Sep 12, 2008 2:31 am

### Re: Problem 135

No, there isn't.

JMW1994
Posts: 43
Joined: Sat Apr 09, 2011 10:35 pm

### Re: Problem 135

Thanks for the clarification.

Kevin_L
Posts: 5
Joined: Sun Apr 01, 2012 12:18 pm

### Re: Problem 135

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
respectively.

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

alice0meta
Posts: 1
Joined: Mon Aug 05, 2013 7:31 pm

### Re: Problem 135

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
respectively.

It's killing me trying to find out where the missing 12 solutions are.
you're letting z equal 0

z != 0

Neilius
Posts: 8
Joined: Mon Aug 24, 2015 4:27 am
Contact:

### Re: Problem 135

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?

dawghaus4
Posts: 54
Joined: Fri Nov 29, 2013 2:22 am

### Re: Problem 135

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

Neilius
Posts: 8
Joined: Mon Aug 24, 2015 4:27 am
Contact:

### Re: Problem 135

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

DanielJackson
Posts: 2
Joined: Wed Feb 19, 2020 11:03 pm

### Re: Problem 135

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?

mdean
Posts: 156
Joined: Tue Aug 02, 2011 1:05 am

### Re: Problem 135

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.

DanielJackson
Posts: 2
Joined: Wed Feb 19, 2020 11:03 pm

### Re: Problem 135

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