Divisibility of Integers, remainders

[calvinshmoe]

The question is:

Integer n is divided by 3 the remainder is 2, when integer n is divided by 4 the remainder is 1, what is the remainder when n is divided by 12?

Help guys, and when it comes to this kind of question how to we approach first?

Thanks a million!

[stud.jatt]

You have to find the series of numbers for which both the statements hold true and then divide any one of them by 12 to see what the remainder would be.

First Statement
When integer n is divided by 3 the remainder is 2
let a be the quotient when n is divided by 3, then n is of the form n = 3a+2

plugging a = 0,1,2,3,4...
we have

n = 2,5,8,11,14,17,20....

Second Statement
When integer n is divided by 4 the remainder is 1
let b be the quotient when n is divided by 4, then n is of the form n = 4b+1

plugging b = 0,1,2,3,4,5...
we have

n = 1,5,9,13,17,21.....

Now note that the only numbers in the series that satisfy both these equations are 5,17,29..... and if you divide any of these by 12 the remainder will always be 5.

Actually you do not have to plot both the series to a lot of terms, I did that just to make the explanation easier and lucid. Just find the first common term in the two series (for e.g. 5 here) and that will give you the answer

Hope this helps

[jnelson0612]

Excellent explanation, thank you!

[djs53]

Can the algebraic solution for this be posted as well?

I ask because this charting solution works for when you can find a satisfying number for both remainder constraints, but for a question like "Question Bank-Dividing By 11 and 19" a satisfying number is not easily found.

Instead, for this problem, I believe you'd end up with:
n=3m+2 and n=4z+1, no?

and then
3m+2=4z+1

and then, I'm lost! Puhlease help!

[RonPurewal]

I can't think of any reasonable algebraic solution.

[RonPurewal]

The "Dividing by 11 and 19" problem is a different story, because both remainders are the same. If that happens, then it's not very hard to find a number -- just find a common multiple of 11 and 19, and then add the number.
For instance, if the remainder of x/11 is 5, and the remainder of x/19 is also 5, then x must be 5 more than a number that's a multiple of both 11 and 19. So, x = 11*19 + 5 would work just fine there.

On the other hand, if the remainder of y/19 is 5, and the remainder of y/11 is, say, 4, then you'll have to make a list.
This is still not as bad as you seem to think, and it can very easily be done within a moderate amount of time. Just list numbers satisfying y/19 with a remainder of 4, and check them until the other statement is also true:
24 --> nope (y/11 gives remainder of 2)
43 --> nope (y/11 gives remainder of 10)
62 --> nope (y/11 gives remainder of 7)
81 --> y/11 gives remainder of 4. Win.
There's no way to argue that this is somehow "excessively time-consuming".

--

Psychology observation of the day:
In 99.999% of cases when someone doesn't want to make lists because "it will take too long", that's actually not the issue in the person's mind at all.
The issue is that the person doesn't want to make the list -- for whatever reason -- and is making up random (and ultimately indefensible) justifications.

The reasons are varied, but one of the most common reasons -- as silly as it ultimately is -- is that people "feel dumb" if they can't think of a theory-based solution to something.
There's no sense in thinking this way. If something solves a problem, then it's just as good as anything else that solves the problem.