The Saga of "54" and the Truth - Post #120

[baseballart]

Very simply, 3 was subtracted, the "." dropped, and "54" remained

[frankbmd]

Originally Posted by baseballart

Very simply, 3 was subtracted, the "." dropped, and "54" remained

Intellectually appealing, but flawed in the sixth line of the formula, which belies the result obtained.

[DavidG1966]

The Curse of 1940, also called Dutton's Curse, was a superstitious explanation for why the New York Rangers of the National Hockey League did not win the league's championship trophy, the Stanley Cup, from 1940–94.

Net = Hockey 54 Years = Duration of the Curse

1940 was also Pre WWII (for the US)

http://en.wikipedia.org/wiki/Curse_of_1940

Was anyone involved in the naming a Hockey or Rangers Fan?

[drmondobueno]

Originally Posted by baseballart

Very simply, 3 was subtracted, the "." dropped, and "54" remained

Silence. Head hurts.

[peterb69]

Emc2 = 54

[WWGjohn]

I have it on good authority that the name came from a meeting of "Monster set" collectors, 54 to be exact, who decided to create this forum where only posts pertaining to their beloved set would be allowed. Not sure how nonMonster set posts ever got started.

John

[Leon]

I just emailed Frank quite a long explanation and he can post it whenever he wants to. Nice thread and nice stories guys....

[novakjr]

Originally Posted by peterb69

Emc2 = 54

Interesting. In that equation. There appears to be multiple possiblities for the values of e m and c. They can all equal 3, although I'd lean toward that being unlikely, as it would've been more simple to assign a cubed singular varible. Although it is possible assuming that e, c and m could all be the same number, if they were representing different theoretical specifics. The more likely scenario being any combination that has one of them equaling 1, another equaling 3, and the third equaling 9..

In the context of that equation e definitely does not equal mc², in term of a fully whole number valued equation.

Unless, your large 2 represented "square". Then we're left with a different set of possibilities all-together. C could equal 1, while e and m could be interchangeably represent the numbers 2 and 27, 1 and 54, 3 and 18, or 6 and 9. It's possible that c could equal 3, while e and m represent 1 and 6, or 2 and 3. C could be represented by either positive or negative numbers. As it pertains to e and m. They could only both be negative or both be positive, but not 1 of each..

Again, we are not left with a solely whole number valued equation that can make e=mc².

Assuming that both emc2=54 AND e=mc², or emc²=54 and e=mc², fractions and decimals are the only possibilities.

[tschock]

I think it was from a 12 year old named Adam who was missing #54 from his set. He was definitely into older police partner TV shows like "DragNet" and "Car(d) 54 Where are you?".

[frankbmd]

Originally Posted by novakjr

Interesting. In that equation. There appears to be multiple possiblities for the values of e m and c. They can all equal 3, although I'd lean toward that being unlikely, as it would've been more simple to assign a cubed singular varible. Although it is possible assuming that e, c and m could all be the same number, if they were representing different theoretical specifics. The more likely scenario being any combination that has one of them equaling 1, another equaling 3, and the third equaling 9..

In the context of that equation e definitely does not equal mc², in term of a fully whole number valued equation.

Unless, your large 2 represented "square". Then we're left with a different set of possibilities all-together. C could equal 1, while e and m could be interchangeably represent the numbers 2 and 27, 1 and 54, 3 and 18, or 6 and 9. It's possible that c could equal 3, while m and c represent 1 and 6, or 2 and 3. C could be represented by either positive or negative numbers. As it pertains to e and m. They could only both be negative or both be positive, but not 1 of each..

Again, we are not left with a solely whole number valued equation that can make e=mc².

Assuming that both emc2=54 AND e=mc², or emc²=54 and e=mc², fractions and decimals are the only possibilities.

David, I used the numbers on my telephone keypad, squared the c and multiplied and came up with 72, so if Emc2 is correct, welcome all to

Net72

[novakjr]

Originally Posted by frankbmd

David, I used the numbers on my telephone keypad, squared the c and multiplied and came up with 72, so if Emc2 is correct, welcome all to

Net72

The first portion of that was assuming that the larger 2 was just a 2... The second portion addressed as if it represented a "square".

[frankbmd]

Originally Posted by novakjr

I could be wrong. But I'm pretty sure that the way it is written, you would square the value of c before doing anything else with it. Otherwise it would need to be written as (emc)²

3 * 6 * (2*2) = 72

Don't screw around with a math major.