Three consecutive positive prime numbers have a sum that is a multiple of 7. What is the least possible sum?
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Three consecutive positive prime numbers have a sum that is a multiple of 7. What is the least possible sum?
5+7+11 = 18+3 = 23
7+11+13 = 18+13 = 31
11+13+17 = 30+1+3+7 = 41
13+17+19 = 30 + 3 + 7 + 9 = 49
i think that the true solution is:
2+3+5=14
14 is a multiple of seven
so the three numbers are 2:3:5:thumb:
Ummm,
2+3+5 <> 14
:confused:
How in the world did you get 14 from 2, 3 and 5?
2*5+3.. nope.. 3*5-2.. nope.. 5^2/3 .. nope
2^3 + 5... nope... 3^2 + 5.. hey there we go.
2 + 3 + 5 = 10
10 = 6 + 4
6 + 4 = 1*(6^1) + 4 * (6^0)
So,
2 + 3 + 5 = 14 base {modulo?} 6
however, thats a 14 that still isn't divisible by 7.
:wave:
:confused: i'll try to think it over again and then reply:confused: