SOLUTION: hi. i was assigned to show that there are 43411 consequtive composite numbers. ok, i know that i can do 43411! and if i`ll divide the whole thing by, say n!+3 I`d be left with t

Algebra ->  Divisibility and Prime Numbers  -> Lessons -> SOLUTION: hi. i was assigned to show that there are 43411 consequtive composite numbers. ok, i know that i can do 43411! and if i`ll divide the whole thing by, say n!+3 I`d be left with t      Log On


   



Question 29117: hi. i was assigned to show that there are 43411 consequtive composite numbers. ok, i know that i can do 43411! and if i`ll divide the whole thing by,
say n!+3 I`d be left with the composites. or should i do n! +1 but to start dividing from 43410! instead? thanks.

Answer by venugopalramana(3286) About Me  (Show Source):
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hi. i was assigned to show that there are 43411 consequtive composite numbers. ok, i know that i can do 43411!
NO YOU NEED TO TAKE (43411+1)!=43412! AND THEN THE 43411 CONSECUTIVE NUMBERS ARE
43412!+2,43412!+3,43412!+4,43412!+5,43412!+6,43412!+7,43412!+8,43412!+9........
.......43412!+43409,43412!+43410,43412!+43411,43412!+43412.


and if i`ll divide the whole thing by,
NO... NO DIVISION NEEDED ..JUST LIST THE NUMBERS AS GIVEN ABOVE.
say n!+3
NO... NO DIVISION NEEDED ..JUST LIST THE NUMBERS AS GIVEN ABOVE
I`d be left with the composites. or should i do n! +1 but to start dividing
NO... NO DIVISION NEEDED ..JUST LIST THE NUMBERS AS GIVEN ABOVE
from 43410! instead?
NO... NO DIVISION NEEDED ..JUST LIST THE NUMBERS AS GIVEN ABOVE