A school administrator will assign each student in a group of n students to one of m classrooms. If 3<M<13<N, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?
(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.
(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
Solution:
(1) m=3n so 3n/m is an integer. Given that n can be any number bigger that 13, n can be 14, 15, 16, 17, 18, 19... . m can be 4, 5, 6, 7,8, 9, 10, 11, 12. If m = 10 and n=15, 3*15/10=4,5 which is not an integer, BUT if I have m=10 and n= 30, 3n/m is an integer. NOT SUFFICIENT
(2) 13n = m so 13n/m is an integer. If n =15 and m =5, 13n/m is an integer, so it s ok. BUt if n= 15 and m = 4 the ratio is not an integere. Not Sufficient.
Check together 3n = m and 13n = m.
so 3n/m = 13n/m which can only happen if n=0, but it is not possible.
I end up wioth E.
(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.
(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
Solution:
(1) m=3n so 3n/m is an integer. Given that n can be any number bigger that 13, n can be 14, 15, 16, 17, 18, 19... . m can be 4, 5, 6, 7,8, 9, 10, 11, 12. If m = 10 and n=15, 3*15/10=4,5 which is not an integer, BUT if I have m=10 and n= 30, 3n/m is an integer. NOT SUFFICIENT
(2) 13n = m so 13n/m is an integer. If n =15 and m =5, 13n/m is an integer, so it s ok. BUt if n= 15 and m = 4 the ratio is not an integere. Not Sufficient.
Check together 3n = m and 13n = m.
so 3n/m = 13n/m which can only happen if n=0, but it is not possible.
I end up wioth E.
riba made
