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I always want to remind any poster here that if they ever felt in despair you can always reach out to me privately..

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Nits74

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Oct 19, 2021
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Mr. Potter said:
send me an email address or contact and I will reach out directly.

I do this because I never want anyone to feel that the weight of the world is on their shoulders, I'll share the "burden" with you. You are loved.

Shalom
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Let's hope you don't get bombarded with calls tomorrow night. Seriously though, very nice gesture. I remember when you made this offer previously. You're a good man, Mr. Potter.
 
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PSUAXE70

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Oct 12, 2021
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LionJim said:
Sweet gesture, perfectly in character.

If anyone needs help with a math problem, I’m your man.
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My grandfather used to live on a street with houses on one side only and numbered consecutively. He said the sum of the house numbers of the houses on one side of his house equaled the sum of the house numbers on the other side. I guessed his house number was 6 out of 8 houses because 1 through 5 adds to 15 and 7 +8 add to 15. He said nice try but there were between 100 and 500 houses on his street. He never told me the exact number of houses on the street. What was his street number?
 
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Got GSPs

Got GSPs

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PSUAXE70 said:
My grandfather used to live on a street with houses on one side only and numbered consecutively. He said the sum of the house numbers of the houses on one side of his house equaled the sum of the house numbers on the other side. I guessed his house number was 6 out of 8 houses because 1 through 5 adds to 15 and 7 +8 add to 15. He said nice try but there were between 100 and 500 houses on his street. He never told me the exact number of houses on the street. What was his street number?
Click to expand...
I was told there would be no maths…
 
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Got GSPs

Got GSPs

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Grant Green said:
What is "excess bacon"?
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I don’t know personally. I hear tell that some people don’t eat bacon, so maybe there are people who only eat some bacon and not all bacon. It’s a longshot, but this really is my niche. I could excel in an environment where people with unlimited capacity for bacon eating was valued. I’m envisioning something like Newman from Seinfeld with the muffin bottoms.
 
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LionJim

LionJim

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PSUAXE70 said:
My grandfather used to live on a street with houses on one side only and numbered consecutively. He said the sum of the house numbers of the houses on one side of his house equaled the sum of the house numbers on the other side. I guessed his house number was 6 out of 8 houses because 1 through 5 adds to 15 and 7 +8 add to 15. He said nice try but there were between 100 and 500 houses on his street. He never told me the exact number of houses on the street. What was his street number?
Click to expand...
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
 
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PSUAXE70

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Oct 12, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...
Thank you. Very nicely done and well explained. I had read the answers before in a book about Ramanujan but had no idea how to start.
 
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LionJim

LionJim

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Oct 12, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...
I was able to remove the caveat. Since n/2 and n+1 are relatively prime (because -2(n/2)+(n+1)=1), if
(n/2)(n+1) is a perfect square, then so must both n/2 and n+1 be perfect squares. Also, you could assume that n was odd, not even , but then for this to work you’d have to have n to be larger than 500, which is larger than the given upper bound for the problem.

I really don’t have much of a mathematical imagination, never did, but my proofs were always watertight.
 
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Connorpozlee

Connorpozlee

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Oct 29, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...
Holy ****! I got to “here’s how it works” and advanced directly to this response. I’m going to go out on a limb and say well done!
 
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Midnighter

Midnighter

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Oct 7, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...

 
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PSU87

PSU87

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Oct 12, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...
Blah blah blah...

That's cute and all, but how 'bout some winning Powerball numbers for Friday?
 
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LionJim

LionJim

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Oct 12, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...
I hate to leave any loose threads, so here’s why the sum of the first n positive integers is equal to n(n+1)/2.

Set this sum equal to x:

x = 1 + 2 + 3 + … +n;
now do it backwards:
x = n + (n-1) + (n-2) + … + 2 + 1;
add these two column by column, and you get

2x = (n+1) + (n+1) + (n+1) + … + (n+1), n terms of (n+1), so you get

2x = n(n+1), dividing through by 2 gives

x = n(n+1)/2.
 
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ODShowtime

ODShowtime

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Nov 1, 2021
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LionJim said:
I hate to leave any loose threads, so here’s why the sum of the first n positive integers is equal to n(n+1)/2.

Set this sum equal to x:

x = 1 + 2 + 3 + … +n;
now do it backwards:
x = n + (n-1) + (n-2) + … + 2 + 1;
add these two column by column, and you get

2x = (n+1) + (n+1) + (n+1) + … + (n+1), n terms of (n+1), so you get

2x = n(n+1), dividing through by 2 gives

x = n(n+1)/2.
Click to expand...

Big deal, I know how to make my calculator say "hell".
 
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NittanyLionFan

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Oct 6, 2021
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LionJim said:
I hate to leave any loose threads, so here’s why the sum of the first n positive integers is equal to n(n+1)/2.

Set this sum equal to x:

x = 1 + 2 + 3 + … +n;
now do it backwards:
x = n + (n-1) + (n-2) + … + 2 + 1;
add these two column by column, and you get

2x = (n+1) + (n+1) + (n+1) + … + (n+1), n terms of (n+1), so you get

2x = n(n+1), dividing through by 2 gives

x = n(n+1)/2.
Click to expand...
I'm still lost with gazinta math ....
 
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PSU Mike

PSU Mike

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Or drop it into Excel and do it brute force in a few minutes.
 
F

Fac

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Oct 12, 2021
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LionJim said:
There were 288 houses on his street and he lived at 204.

Here's how it works. The sum of the first n consecutive positive integers is n(n+1)/2. Suppose that you were able to split the house numbers in such a way that the first (k-1) numbers and the last (n-k) numbers add up to the same sum. This implies that twice the sum of the first (k-1) numbers plus k is equal to the sum of the first n numbers, that is,

k(k-1)+k = n(n+1)/2. But the left hand side is equal to k squared. So for this to work, n(n+1)/2 has to be a perfect square, which will happen when both n/2 and n+1 are perfect squares. (Caveat: number theory is not my forte, and I haven't yet verified that this is the only way it can work. What you see here is not an acceptable mathematical proof.) In your example, you had n=8; both n/2 = 4 and n+1 = 9 are perfect squares. The next largest n where this works is n=288: n/2=144=12^2 and n+1=289=17^2. And 288*289/2=41616=204^2. Of course, 204=12*17.
Click to expand...
You lost me at "Here's how it works".
 
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Fac

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Mr Potter. Every night for the past 2 weeks I have been dreaming about wigwams and teepee's.

Can you explain why to me? I need to get better rested.

TIA
 
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SurgeOne

SurgeOne

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Oct 30, 2021
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Mr. Potter said:
send me an email address or contact and I will reach out directly.

I do this because I never want anyone to feel that the weight of the world is on their shoulders, I'll share the "burden" with you. You are loved.

Shalom
Click to expand...
I might after tonight’s game
 
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