If 4 students were added to a dance class, would the teacher

First statement says: "if 12 students were added, the teacher could divide the students evenly into teams of 8", which means that x+12 is a multiple of 8. Now, x+12=(x+4)+8=(x+4)+{multiple of 8}. So, we have that the sum of x+4 and some multiple of 8 is a multiple of 8, which means that x+4 must also be a multiple of 8.

To elaborate more:
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it's clear.

Evenly divided into =/= divided into even numbers

So for the numbers you picked, bare minimum is 4. so is 4+4 divisible by 8? Yes. If 12, 12+4 divisible by 8? Yes. So on and so forth.

The red part is not correct.

Evenly divisible = divisible.

That's what I meant. I can't makes the does not equal sign, but I figured =/= was fine. From what vkm stated, he seemed to be under the impression that x+4/8 would be an even number like 2,4,8 as opposed to being simply divisible.

Thanks guys.

I am clear with the explanation.

Consider the no. of students present is X

Now the question is whether (x+4)/8 is integer or not. Y or N?

1. If 12 is added to X it is divided by 8; (x+12)/8 = integer.
so the values of X = 4,12,20,.... = 4m(m is a odd no.)
Now using this in basic equation , (4m+4)/8 => (m+1)/2 which is always an integer as m is odd.
So sufficient.

2. Considering X=1,2,3,4,5,6,7 we cannot find an exact answer as for diff X , (x+4)/8 satisfies and for some it doesn't

Hence A

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