The prime factors of N are both 2 mod 3, where N = 117968734185852909111122793179458044429748833609671492956099236663454458021167483523103665043186849896340566668709082216630299186083868726474403014230450576527498336056399517500373699648384503014218007972984281676376443077883955580519873153351211473492074726331214866950482499437571095445498914388647218496094203506927341398236714690311062488254644318997750013131645597009546941216705983462580669754031513
15
146Ṁ7582resolved May 27
Resolved
NO1H
6H
1D
1W
1M
ALL
N is a number that I generated by multiplying together two large primes. N = 1 mod 3, so therefore the prime factors of N must either:
- Both be 1 mod 3 (question resolves to no)
- Both be 2 mod 3 (question resolves to yes)
This question is managed and resolved by Manifold.
Get
1,000 to start trading!
🏅 Top traders
# | Name | Total profit |
---|---|---|
1 | Ṁ316 | |
2 | Ṁ271 | |
3 | Ṁ86 | |
4 | Ṁ2 |