Tricky math problem is you are bored?

Question

Hello,

DISCLAIMER:
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!

NOTE:
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In this problem, two prime numbers 𝑝₁<𝑝₂ are considered consecutive if no other prime 𝑃 can be found inbetween them (there exists no prime 𝑃 such as 𝑝₁<𝑃<𝑝₂).

Problem:
The sum of nine consecutive primes is of the form 10ⁿ. Solve ELEGANTLY for all valid values of 𝑛.

Regards,
Dragon.Jade :-)

Puzzling · Accepted 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 10ⁿ 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 = 10²

? · Answer

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 Aaron · Answer

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.

Tricky math problem is you are bored?

3 réponses