What is the smallest integer with a remainder of 1 when divided by 6, 8, and 10?

[PrudensOptimus]

What is the smallest integer greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10?

the answer is 121... how do you do this problem?

 

[ahrkron]

You are looking for a number n such that n = 6a + 1 for some integer a. Similarly, you get two more equations, all equal to n. You can eliminate n from all of them, and you get two equations with three unknowns. The remaining freedom comes allows you to choose non-integers and also higher and higher integers, but you only take the smallest integer solution for n.

 

[PrudensOptimus]

Can you explain how you got the equation?

 

[NateTG]

You're looking for the least common multiple of 6,8, and 10, and then you'll want to add one.

There are many methods for finding the least common multiple.

 

[HallsofIvy]

Saying that the number, x, has remainder 1 when divided by 6, 8, and 10 means x= 6i+ 1, x= 8j+ 1, and x= 10k+ 1 for some i, j, k. Putting those together, 6i+ 1= 8j+ 1= 10k+ 1. Subtract 1 from each part: 6i= 8j= 10k. Divide each part by 2: 3i= 4j= 5k. Since 3, 4, and 5 are mutually prime, the smallest values that will work are i= 4*5, j= 3*5 and k= 3*4: in other words, 6i= 8j= 10k= 6*20= 4*15= 5*12= 120 and so x= 6i+ 1= 8j+ 1= 10k+1= 121.

 

[PrudensOptimus]

Originally posted by HallsofIvy
Saying that the number, x, has remainder 1 when divided by 6, 8, and 10 means x= 6i+ 1, x= 8j+ 1, and x= 10k+ 1 for some i, j, k. Putting those together, 6i+ 1= 8j+ 1= 10k+ 1. Subtract 1 from each part: 6i= 8j= 10k. Divide each part by 2: 3i= 4j= 5k. Since 3, 4, and 5 are mutually prime, the smallest values that will work are i= 4*5, j= 3*5 and k= 3*4: in other words, 6i= 8j= 10k= 6*20= 4*15= 5*12= 120 and so x= 6i+ 1= 8j+ 1= 10k+1= 121.

Since when did they teach this in middleschool? This seems soo hard, seems like Number theory or some crap...

Where can I go for more help?

 

[NateTG]

I know that lcm and gcf are covered in middle school, and that will get you the answer. (As I pointed out above.)

The horde of equations that Hurkyl used is more aptly described as basic algebra. (Although the exercise could certainly be described as a number theory problem.)

 

[PrudensOptimus]

Originally posted by NateTG
I know that lcm and gcf are covered in middle school, and that will get you the answer. (As I pointed out above.)

The horde of equations that Hurkyl used is more aptly described as basic algebra. (Although the exercise could certainly be described as a number theory problem.)

I am not very good with number theories, because I don't deal much with those excercise:\

 

[selfAdjoint]

The problem is not actually high schol algebra because one of the constraints is that the answer contain only whole numbers, and there is no way to express that in high school algebra.

This is actually a question to pick out top performers. Some percentage of the top 1% of math brains will be able to figure out a procedure (something like what Hurkyl gave)to get an answer without knowing a general rule.

 

[Hurkyl]

Blarg, this is the second time (that I've noticed) that HallsofIvy has been accused of being me!

 

[NateTG]

Originally posted by Hurkyl
Blarg, this is the second time (that I've noticed) that HallsofIvy has been accused of being me!

Mea Culpa.

I get confused b/c you two Newmans look alike.

 

[HallsofIvy]

One of us is being insulted!

(I'm just not sure which of us.)



Why the remark about "middle school"? Who said this was a middle school problem?