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Semi-prime age, version 2

My age is a product of (exactly) two prime numbers. Also, the same was true before my last birthday and will be true after my upcoming birthday.

Assume that I am an elf, with unlimited lifespan.

What are my possible ages?

(See https://brilliant.org/discussions/thread/semi-prime-age-version-1/ for a simpler version of this problem.)

Note by Johan Falk
2 years ago

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Ok, so I just wrote a short script to check this problem out numerically. There are 16 valid ages to find below 1000 years, and 80 valid ages below 10 000 years. This probably covers the life span of the elves in Middle Earth, but it doesn't solve the problem completely.

EDIT: …and there are 2415 valid ages below 1 000 000 years. I am starting to think that the pattern is really difficult to find. Johan Falk · 2 years ago

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Note: I do not yet know if this problem has more than one solution. I have come a bit on analyzing the problem. My steps so far:

Call the age N. Every second number is divisible by two. Every third number is divisible by three. Thus, there is at least one of (N–1), N or (N+1) that is divisible by two , and exactly one is divisible by three.

Every second even number is divisible by 4, so if two numbers in the sequence is divisible by two, one of them is divisible by 2*2. This would then be the only two prime factors in that number. But this cannot be the case, since 4 doesn't fit into a sequence of three semi-prime numbers.

Thus, only one number is divisible by two. Thus the middle number (N) is even.

So we are looking for a number N such that… \(N = 2P_1\) either \((N–1)\) or \((N+1)\) is \(3P_2\) and either \((N+1)\) or \((N–1)\) is \(P_3 \cdot P_4\)

where \(P_1\), \(P_2\) and \(P_3\) are unique prime numbers, but \(P_4\) may be equal to \(P_3\). Johan Falk · 2 years ago

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