A simple one

[knight247]

A teacher was trying to arrange the students of her class in rows with an equal number of students in each row. First, she tried to arrange 4 students in a row. Later, she tried to arrange 6 students in a two. In either case, she was left with one extra student. However, when she tried to arrange 7 students in a row, she was not left with any extra student. What is the minimum number of students in the class?

I don't have answer options or an OA. The answer is 49 as per my calculations. Detailed explanations would be appreciated. Thanks

[gmatboost]

If there was 1 student left over when she tried both 4 and 6 per row, that means that the number of students is both
A) 1 more than a multiple of 4
A) 1 more than a multiple of 6

What do we call a number that is both a multiple of 4 and a multiple of 6?
Well, 4 = 2*2 and 6 = 2*3
To be a multiple of both, a number would need to have both 2*2 and 2*3 in its prime factorization, which means it would need to have 2*2*3 as part of its prime factorization. Which means it would need to be a multiple of 2*2*3 = 12.

Another way of looking at that step is that we need the LCM of 4 and 6, which is 12.

So, the number of students is 1 more than multiple of 12.
Also, it is a multiple of 7.

Start checking numbers that are 1 more than a multiple of 12 to see if each is or is not a multiple of 7.

12 -> 13 -> NO
[spoiler]24 -> 25 -> NO
36 -> 37 -> NO
48 -> 49 -> YES[/spoiler]

[knight247]

Thanks Greg. I didn't wanna rack my brain too much so I did it by a simple formula substitution method.

Let N be the number of students
N-1=4j (Coz 1 subtracted from the Dividend would make it perfectly divisible by 4)
N-1=6k (Coz 1 subtracted from the Dividend would make it perfectly divisible by 6)

4j=6k
2^2j=3*2k

So, N-1 has both 2^2 and 3 among its prime factors
N-1=(2^2)(3)x
N-1=12x
N=12x+1

Only x=48 fits the requirement. So, N=49. I'm guessing this method was pretty accurate