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Mathematics 20 Online
OpenStudy (anonymous):

Here is a problem that my nephew asked me on Facebook: I just found a number with an interesting property : When I divide it by 2 , the remainder is 1 When I divide it by 3 , the remainder is 2 When I divide it by 4 , the remainder is 3 When I divide it by 5 , the remainder is 4 When I divide it by 6 , the remainder is 5 When I divide it by 7 , the remainder is 6 When I divide it by 8 , the remainder is 7 When I divide it by 9 , the remainder is 8 When I divide it by 10 , the remainder is 9 It's not a small number , but it's not really big either . Find the smallest number with such number.

OpenStudy (jackellyn):

2519

OpenStudy (anonymous):

2519 is the first such number, what is the 100th such number?

OpenStudy (btaylor):

@jackellyn how did u find that?

OpenStudy (anonymous):

@elisaab 2520*100 -1?

OpenStudy (kinggeorge):

I believe the 100th such number should be 251999.

OpenStudy (anonymous):

Yeah. Is it correct?

OpenStudy (kinggeorge):

It certainly has the correct remainders for 2-10. And jackellyn found the first one by finding the lcm of 2-10, and subtracting 1. This works because if you subtract 1, all the remainders will go to the highest possible remainder. I.e., \(n\pmod{x}\equiv x-1\)

OpenStudy (anonymous):

The logic I think should work is that, if the number be N. Then N+1 would be divisible by all those numbers. Yeah.

OpenStudy (anonymous):

Exactly what you said. :)

OpenStudy (kinggeorge):

Therefore, \(2520n-1\) satisfies this property for all integers \(n\).

OpenStudy (kinggeorge):

So really, \(-1\) might be a better answer than 2519 since it's smaller, and it's the number where \(|n|\) is minimal. (Unless we're only looking at positive numbers)

OpenStudy (anonymous):

Actually the first number is the LCM(1,23,4,5,6,7,8,9)-1= 2520-1= 2519

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