Problem 135
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See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
Re: Problem 135
Thanks for the clarification.
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 respectively.
It's killing me trying to find out where the missing 12 solutions are.
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
It's killing me trying to find out where the missing 12 solutions are.

 Posts: 1
 Joined: Mon Aug 05, 2013 7:31 pm
Re: Problem 135
you're letting z equal 0Kevin_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 berespectively.Expand
It's killing me trying to find out where the missing 12 solutions are.
z != 0
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?
Thanks in advance.
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.
Re: Problem 135
It is you suspicion that is incorrect.Neilius wrote:..
The next higher value of n I get with ten solutions is n=1755.
I suspect this is incorrect.
...
Tom
Re: Problem 135
Many thanks.dawghaus4 wrote:
It is you suspicion that is incorrect.
Tom
I finally worked it out.
Program runs in about 6 seconds.
A very enjoyable problem

 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?
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?
Re: Problem 135
$5^23^21^2=15$. 15 has at least 3 solutions.DanielJackson wrote: ↑Wed Feb 19, 2020 11:28 pmn = 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?

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