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liubo4ka [24]
3 years ago
12

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.
Mathematics
1 answer:
REY [17]3 years ago
8 0

Answer: Hello!

first, we have m classrooms and n students, where 3<m<13<n

a) 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?

If this statement is true, then it should be true for all the m between 3 and 13, and all the n greater than 13.

If we want that every classroom has the same amount of students. then n/m should be an integer number.  

Let's find a counterexample!

if n is 17, then n is a prime number. this means that only can be divided by itself and 1. Then m doesn't divide n, and we cant divide 17 students in any amount of classrooms. This is only possible if n is a multiple of m.

b) if now we have 3n students, again we need to find a counterexample where 3n is not divisible by m. this is only possible if 3n is a multiple of n.

suppose that n = 20, then 3n= 60, and we need to find a number m that cant divide 60 to prove that this is false.

then if m = 7, 60/7 = 8.5 is not an integer, then, in this case, we cant divide 3n students in m classrooms.

the only situation where you can divide 3n students into m classrooms is if 3n is a multiple of m

c) now we have 13n students and m classrooms again.

if n = 13, then 13n = 13*13.

now, 13 is also a prime number, then 13*13 can only be divided by 1, 13 and itself.

then there is no m that divides this number.

in this case, because there is no m that divides 13 (again, because 13 is prime) the only situation where you can divide the 13n students into m classrooms is where n is a multiple of m.

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