Divisibility (Remainders): A group of n students can be divi

II wrote:A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A) 33
B) 46
C) 49
D) 53
E) 86

What would be your primary (or best) approach to answer this ?
What would be your secondary approach to answer this question ?

Thanks.

I'd probably brute force this one.

Start with multiples of 5, since 5 is bigger than 4 and therefore the jumps are larger.

So, the numbers that give us a remainder of 3 when divided by 5 are:

3, 8, 13, 18, 23, 28, 33, 38, 43, 48

(the biggest answer is 86, so that should be more than enough options).

Now we divide those numbers by 4 and see which ones give us a remainder of 1.

13 is the first winner.
33 is the second winner.

13 + 33 = 46: choose (b).

The answer is 46 . I wud explain the way I did , which is very mathematical and a sure answer.
Given:
N = 4k+1 ( as 4 with 1 student left over )
N = 5a+3 (as 5 with 3 student left over )
therefore ,
4k+1=5a+3
k=(5a+2)/4
the first two smaller values of a (i.e 2,6) that satisfies this equation gives the answer for N ( coz N = 5a+3 )

hence answers are 13 +33 = 46 .

Thanks
Senthil

If it were easy then everybody would do it and it wont be crowned as an ACHIEVEMENT

Hi Senthil,

Can you explain what you mean by "the first two smaller values of a (i.e 2,6) that satisfies this equation gives the answer for N ( coz N = 5a+3 )"?

It seems like you just chose 2 and 6 randomly and they seem to get you the answer. I would appreciate an explanation of how you decided to choose these numbers.

Thanks!

adt29 wrote:Hi Senthil,

Can you explain what you mean by "the first two smaller values of a (i.e 2,6) that satisfies this equation gives the answer for N ( coz N = 5a+3 )"?

It seems like you just chose 2 and 6 randomly and they seem to get you the answer. I would appreciate an explanation of how you decided to choose these numbers.

Thanks!

It doesn't look like Senthil comes around here much anymore, but I'm sure all he (or she) did was to plug in values of a, starting with 0 and then increasing by one, into k=(5a+2)/4 until he found two numbers that made k an integer. The first two are 2 and 6.

However, if you're going to just guess and check anyway, why bother messing around with the equations in the first place? Stuart's approach above, or some slight variation, is surely the best method for anything like this that comes up on the GMAT.

Also, note that once you find the first number that works, you can find the subsequent numbers by adding the least common multiple of 4 and 5, 20. So all of the numbers that satisfy both equations in this case would be of the form 13+20n where n is a non-negative integer.

Go by brute force method......
First # divisble by 4 and 5 and leaving remainders 1 and 3 resp. is 13 and second such number is 33.It takes hardly 1 minute to solve.