Question

MEDAL FOR HELP!
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?

Answer 1

60 is the LCM...I suppose you could include multiples of 60

Answer 2

Choices are:
4 , 10, 12, 14, 26

Answer 3

lol i just answered this question earlier today

Answer 4

So, we come back to modular arithmetic:
From the problem we have
\[x \equiv 2 \text{mod(4)} \]\[x \equiv 1 \text{mod(5)} \]\[x \equiv 4 \text{mod(6)} \]

And we wish to add some number k to be
\[x +k\equiv 0 \text{mod(4)} \]\[x+k \equiv 0 \text{mod(5)} \]\[x +k\equiv 0 \text{mod(6)} \]

So our number can be 2 + multiple of 4, 4 + multiple of 5, and 2 + multiple of 6.

4 wont work since 4-2 = 2
10 wont work since 10-4 = 6
12 wont for the same reason
14 will satisfy 14-2 = 12 14-4 = 10 and 14-2 =12 which satisfies all our conditions

Answer 5

14...Because 46 = 60 - 14 = 5x9 +1 = 4x11 + 2 = 6x7 + 4

Answer 6

How did you get 9 from?

Answer 7

Groups of 5--1 left out....9*5 = 45 + 1 = 46

Answer 8

And why minus 14?

Answer 9

To put it a bit more formally:
\[ k -2 \equiv 0 \text{mod(4)}\]\[ k -4 \equiv 0 \text{mod(5)}\]\[ k -2 \equiv 0 \text{mod(6)}\]
Then you can test your values for k to show that 14 is the only solution that works.

Answer 10

If you hate modular arithmetic:

60 is the number you are looking for, this is because prime factorization of:
4 = 2*2
5 = 5
6 = 2*3

60 = 2*2*3*5
so it is divisible by all of the numbers

to figure out the number of students in the class, you can figure out a couple of things:
1 - the class divisible by 4 is remainder 2... note this means we are divisible by 2 because 4 - 2 = 2

2 - we aren't divisible by 3 either because we have remainder 4 on 6 which is 2/3

3 - we have remainder 1 when divisible by 5

now you can look at this 2 ways:
we know we have a factor 2: but we aren't divisible by 3 or 5, so all of our other prime factors are greater then 5.

We then are looking for all numbers 2*p where p>7 but (2*p) < 60

so we can start here:

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

which is

14, 22, 26, 34, 38, 46, 58

now, we should have remainder 1/5 on division by 5, so 46 is the only number that fits.

Answer 11

which is 14 less then 60

Answer 12

the other way to look at it is to start with 1 and add 5 until you're greater then 60

1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56

then get rid of odd ones:

6, 16, 26, 36, 46, 56

then get rid of ones divisible by 3:

16, 26, 46, 56

get rid of ones divisible by 4:
26, 46

now we want remainder 1 dividing by 5, so 46 is the number.