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.

Please read (and follow!) the guidelines. Your subject should be the first 5 to 8 words of the question.

4x+1=n
or
5y+3=n
where x and y represent the number of groups (and are, therefore, integers). I can't just solve algebraically here because the point is that there are multiple possibilities for n. (And, also, I have three variables and only two equations.)

I'd personally prefer to try numbers first because the numbers in the answer choices are not that large (for me, anyway) and because they ask for the two SMALLEST possibilities.

For 4x+1, some possibilities are:
5, 9, 13, 17, 21, 25, 29, 33, 37, ... (note the pattern is just to add four each time)

For 5y+3, some possibilities are:
8, 13, 18, 23, 28, 33, 38, ... (note that the pattern is just to add five each time)

Any matches? 13 is on both lists. 33 is on both lists. I need the two smallest matches, so I'm done. (I would've added more to my lists if I hadn't had two matches.) 13+33 = 46

Or I could realize that I'm only looking for those circumstances in which the two equal each other (that is, n is the same), so
4x+1 = 5y+3
4x = 5y + 2
I'm still looking at trying some numbers now but, again, I've been asked for the SMALLEST two possibilities, so I should know the numbers aren't going to get too big. And I'm going to start at 1 for the integers I'm trying because the point is that I'm trying the smallest possible divisors to get the smallest possible dividend... and that means starting at 1.

If I plug in the integer 1 for y, will I get an integer for x? No.
If I plug in 2 for y, will I get an integer for x? Yes! x=3. So the circumstance when x=3 and y=2 should yield the same n if I plug those two numbers back into my original equations. Try it - 4(3)+1=13 and 5(2)+3=13. There's one. Keep going. (I strongly prefer the first method by the way - I'd rather just start trying numbers right away!)

so heres my question. I took the same approach, but was off on both numbers. I used:

4k=5m+2

and got that 12 worked and 32 worked ... i did that by just listing out numbers for both sides of the equation, and see what overlapped. adding 12 and 32 gives me 44 though ..

This is a good start, but you haven't finished off the problem yet. The fact that you got 12 means that k=3 and m=2. But n = 4k+1 = 13. Alternatively, you could say n = 5k + 3 = 13. Either way, the smallest such value of n is 13.

Repeat this for 32, and you get k=8 (and m=6). So n = 4k+1 = 33 (or n = 5m+3 = 33).

13 + 33 = 46

Rey

Cheers, Rey ! I forgot that we were looking for 'n' ... aaagh !

These questions are so annoying sometimes, aren't they? :)