Remainders

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Remainders

by AleksandrM » Sat Jul 26, 2008 11:24 am
I guess this is a bit of an odd request. I made up the numbers for the following question, but I remember that I came across a question such as this one and did know how to proceed beyond a certain point.

When N is divided by 7 the remainder is 5, and when N is divided by 3 the remainder is 2. What is the remainder when N is divided by 15.

I got to the following:

I chose Z for quotient. Therefore:

7z + 5 = N and 3z + 2 = N Therefore:

7z + 5 = 3z + 2

4z = - 3

That is where I ran into a problem. I ended up with a negative. I suspect that the problem might be the Z I chose for the quotient.

Help is appreciated. However, I can certainly understand if it is difficult to explain without the original problem or the answer choices.
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by acecoolan » Sat Jul 26, 2008 12:10 pm
I am not sure if the question is complete

take for example the number 26 which satisfies both the conditions
remainder with 7 = 5 (7 * 3 + 5)
remainder with 3 = 2 (3 * 8 + 2)

26 divided by 15 gives a remainder of 12

At the same time take the number 47
remainder with 7 = 5 (7 * 6 + 5)
remainder with 3 = 2 (3 * 15 + 2)

But the remainder by 15 in this case is just 2

So theres something missing.

But the problem with the way you are approaching is that you are assuming that quotient with both 7 and 3 is Z ..which I guess is a wrong start.

Did I make sense?

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by AleksandrM » Sat Jul 26, 2008 1:21 pm
I am sure that your approach is absolutely correct. You probably cannot come up with a unique answer because I simply pulled the numbers out of the air to show how I ended up with a negative.

Otherwise, you do make sense. I suspected that my choice of z for both cases was a wrong turn, as you have confirmed.

However, how would you approach this question algebraically, without picking a number?

Thanks.

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by preetha_85 » Sat Jul 26, 2008 10:49 pm
Hi

N= 7Q + 5.
Also N = 3Q1 +2

Q1,Q-> Quotients

7Q + 5 = 3Q1 + 2.
7Q + 3 = 3Q1. (this shows that 7Q +3 must be divisible by 3)

So put values for Q from 1 etc to get the values in sych a way tat 7Q +3 is divisible by 3. (The first no. tat satisfies this is 3 )

So number is (7*3)+5 =26.

P.s : I dont think we can find a unique no. (6,9,12...etc will satisfy the equation)