Question

If a gym class is divided into 4 equal teams, 2 students have to sit out; 5 teams, 1 student has to sit out; 6 teams, 4 students have to sit out. If there are fewer than 70 people in the class, what is the minimum number of students that need to be added so that the class can be divided into 4, 5 and 6 equal teams with no students sitting out?
(A) 4 (B) 10 (C) 12 (D) 14 (E) 26

Answer 1

I would like to have an explanation and answer please. Thank you all for attempting and helping me answering this question

Answer 2

The answer to this problem is that you need common factors of 4, 5, and 6
since 4 factors into (2*2), and 6 factors into (2*3), and 5 is prime, the smallest number that is divisible by 4, 5 and 6 will be (2*2*3*5) = 60

Now the next part is to figure out what the number of students in the class actually is.

If we think about the factors, then 2, 3, and 5 are the 3 smallest primes.
Since we have remainder 2 on division by 4, and half way to 4 is 2, then we are divisible by 2.

we also know that we aren't divisible by 5, and since we have remainder 4 on division by 6, and 4/6 = 2/3 then we aren't divisible by 3 either.

so we know we have a factor of 2, but any other factor can't have 3 or 5, meaning that the only possible numbers are ones with prime factorization of 2*(prime > 5).

So we need to find all the numbers with prime factorization of (2*p, p>5) < 60

Possible numbers are:
2, 14, 22, 26, 34, 38, 46, 58

note:

2, 2*7, 2*11, 2*13, 2*17, 2*19, 2*23, 2*29

then find out ones that give us remainder 2/3 when dividing by 6:

22, 34, 46, 58

then find which gives us remainder 1/5 when dividing by 5:

46!

which is 14 less then 60, so you're answer is D.

Answer 3

Thank you for answering this question the explanation makes perfect sense. Kudos to you because others explained it in a confusing way

Answer 4

no problem.