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Tricky math problem is you are bored?
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I am not trying to get my homework done. This problem is provided as it is, and I do have the answer. So it is a challenge for anyone. Best answer will be awarded in approximatively 8 hours. Have fun gals and guys!
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In this problem, two prime numbers πβ<πβ are considered consecutive if no other prime π can be found inbetween them (there exists no prime π such as πβ<π<πβ).
Problem:
The sum of nine consecutive primes is of the form 10βΏ. Solve ELEGANTLY for all valid values of π.
Regards,
Dragon.Jade :-)
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Congratulations to Puzzling and Polyhymnio for their correct answers.
Jeff Aaron does have a point.
The BA goes to Puzzling for their swift and correct answer.
Dragon.Jade :-)
3 Answers
- PuzzlingLv 75 years agoFavorite Answer
Most prime numbers are odd. The sum of 9 *odd* numbers will always result in an odd number, so it is impossible to get an even number like 10βΏ using just odd numbers.
Hence, the only possible answer is one that includes the only even prime number (2).
Answer:
2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 10Β²
- PolyhymnioLv 75 years ago
Since the sum of 9 odd numbers is odd, the only sequence of 9 primes adding up to 10^n, an even number, must start with 2,, the only even prime.
2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100
- Jeff AaronLv 75 years ago
If n must be an integer, then Puzzling is correct: n = 2
However, if n can be any real number, then there are infinity solutions, for example:
The sum of the 2nd through 10th primes is approximately 10^2.1038037209559568642469874218273
The sum of the 3rd through 11th primes is approximately 10^2.1903316981702914844529652053939
The sum of the 4th through 12th primes is approximately 10^2.2718416065364989692903698655714
etc.
And if n can be non-real, there are even more solutions, for example n could be approximately 2.1038037209559568642469874218273 + 6.283185307179586476925286766559i, etc.