Question

how many students are there in a class if 2 students remain after 4 equal rows of seats are filled and 1 student remains after 3 equal rows of seats are filled?

Answer 1

There's multiple answers for this.

Answer 2

@oldrin.bataku

Answer 3

How?

Answer 4

\(10+12k\) for \(k\in \mathbb{Z}^+\) should work. If there are 10 students, then 2 rows of 4, and there are two left over. 3 rows of 3, and one is left over.

For 22, there are 5 rows of 4, or 7 rows of 3.

Answer 5

Every time you add 12, just add three more rows of 4, and four more rows of 3.

Answer 6

You're right. Totally got your point. Thanks.

Answer 7

You're welcome. In general, to solve more complicated problems like this, you can use the Chinese remainder theorem.
http://en.wikipedia.org/wiki/Chinese_remainder_theorem

Answer 8

If you put one more qualifier such as "there are 3 more rows of 3 than 4," then this problem would have a unique solution.

Answer 9

$$n\equiv2\mod 4\\n\equiv1\mod3$$just use the Chinese remainder theorem

Answer 10

@oldrin: What's this "mod" thing?

Answer 11

Are you familiar with congruence relations? modular arithmetic?

Answer 12

Nopes. :/

Answer 13

where did you come across this problem then? :/

Answer 14

It's a SAT-type question. I guess?

Answer 15

mod is just the remainder when you divide. So \(22 \pmod{4}\) is 2, since \[22=5\cdot 4+{\color{red} 2}\]

Answer 16

OHH^`

Answer 17

solving by inspection, you could just look at the first couple multiples of \(3\): \(0,3,6,9,\dots\) so then our candidates are \(1,4,7,10,\dots\). We want only even results so we inspect only every other multiple \(4,10\dots\) and we wish for them to not be multiples of \(4\) hence we're left with every other item again, \(10,\dots\)

Answer 18

@KingGeorge not really... \(\mod n\) is not really an operator in this context, \(
\equiv\mod n\) is though

Answer 19

mod is a little more general than that though. So 22 mod 4 is also -2, since \[22=6\cdot 4-\color{red}2\]It takes a little getting used to, but that's all mod is.

But to solve this problem, solving the way oldrin just described is probably the fastest.