help with writing & solving equations

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mathgrind

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Feb 22, 2016
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Here is the problem text:
"When the positive integer n is divided by 3, the remainder is 2 and when n is divided by 5 the remainder is 1. What is the least possible value of n?"

I got the answer by just subbing in values, 11 is the answer. But what I'm trying to do now is write an algebraic solution. So I tried to figure out how to represent each part of the problem with an equation and this is what I have:

n is divided by 3, the remainder is 2: (3n + 2) / 3
n is divided by 5, the remainder is 1: (5n + 1) / 5

When I subbed in some values, the equations seemed to kick out the correct remainder. But, how do I sove the two equations for n? If I set them equal to each other and try to solve for n, I don't get a solution.

1) (3n + 2) / 3 = (5n + 1) / 5
2) 15 * (3n + 2) / 3 = 15 * (5n + 1) / 5 (clear the denominator by multiplying by 15)
3) 15n + 10 = 15n + 3
4) 15n - 15n = 3 - 10
5) 0 = -7
This is where I'm stuck - obviously this doesn't return a solution. Are my equations wrong? Is trying to solve them for each other wrong?

Thanks for any assistance!
 
mathgrind said:
Here is the problem text:
"When the positive integer n is divided by 3, the remainder is 2 and when n is divided by 5 the remainder is 1. What is the least possible value of n?"

I got the answer by just subbing in values, 11 is the answer. But what I'm trying to do now is write an algebraic solution. So I tried to figure out how to represent each part of the problem with an equation and this is what I have:

n is divided by 3, the remainder is 2: (3n + 2) / 3
n is divided by 5, the remainder is 1: (5n + 1) / 5

When I subbed in some values, the equations seemed to kick out the correct remainder. But, how do I sove the two equations for n? If I set them equal to each other and try to solve for n, I don't get a solution.

1) (3n + 2) / 3 = (5n + 1) / 5
2) 15 * (3n + 2) / 3 = 15 * (5n + 1) / 5 (clear the denominator by multiplying by 15)
3) 15n + 10 = 15n + 3
4) 15n - 15n = 3 - 10
5) 0 = -7
This is where I'm stuck - obviously this doesn't return a solution. Are my equations wrong? Is trying to solve them for each other wrong?

Thanks for any assistance!
Click to expand...

the positive integer n is divided by 3, the remainder is 2 → n = 3u + 2

the positive integer n is divided by 5, the remainder is 1 → n = 5v + 1

where v and u are non-zero integers.

3u + 2 = 5v + 1 (= n)

5v - 3u = 1

Now what method/s you have been taught for solving u and v such that n is minimum.
 
mathgrind said:
Here is the problem text:
"When the positive integer n is divided by 3, the remainder is 2 and when n is divided by 5 the remainder is 1. What is the least possible value of n?"
any assistance!
Click to expand...
If I were doing this, I would start with \(\displaystyle n=6,~7,~8,\cdots\)
You will soon find it.
 
mathgrind said:
Here is the problem text:
"When the positive integer n is divided by 3, the remainder is 2 and when n is divided by 5 the remainder is 1. What is the least possible value of n?"

I got the answer by just subbing in values, 11 is the answer. But what I'm trying to do now is write an algebraic solution. So I tried to figure out how to represent each part of the problem with an equation and this is what I have:

n is divided by 3, the remainder is 2: (3n + 2) / 3
n is divided by 5, the remainder is 1: (5n + 1) / 5

When I subbed in some values, the equations seemed to kick out the correct remainder. But, how do I sove the two equations for n? If I set them equal to each other and try to solve for n, I don't get a solution.

1) (3n + 2) / 3 = (5n + 1) / 5
2) 15 * (3n + 2) / 3 = 15 * (5n + 1) / 5 (clear the denominator by multiplying by 15)
3) 15n + 10 = 15n + 3
4) 15n - 15n = 3 - 10
5) 0 = -7
This is where I'm stuck - obviously this doesn't return a solution. Are my equations wrong? Is trying to solve them for each other wrong?

Thanks for any assistance!
Click to expand...
About like Subhotosh Khan's solution but slightly different: Starting with 5v - 3u = 1, we have
u = \(\displaystyle \frac{5v-1}{3}\, =\, v\, +\, \frac{2v-1}{3}\)
So what is the smallest v so that 2v-1 is a multiple of 3. Then compute n=5v+1.
 
pka said:
If I were doing this, I would start with \(\displaystyle n=6,~7,~8,\cdots\)
You will soon find it.
Click to expand...

That's what I did to get the solution, except I started with numbers that divided by 3 left a remainder of 2. Thanks!
 
I do, thanks! I think I was headed in this direction but wasn't able to state it so succintly. This is definitely the fastest way to find n. Using this logic,
I listed:
5 +1 = 6
10 +1 = 11
15 +1 = 16
20 +1 = 21
25 +1 = 26

Do the same thing with 3:
3 + 2 = 5
6 + 2 = 8
9 + 2 = 11

Ding, ding, 11 is the answer. Since this was a question from a practice GRE test, I need to find solutions fast and this is the fast way. I was originally thinking there was a simple solution via simultaneous equations, but that doesn't seem to be the case. Thanks everyone for your assistance. It is much appreciated!!
 
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