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.
2519
2519 is the first such number, what is the 100th such number?
@jackellyn how did u find that?
@elisaab 2520*100 -1?
I believe the 100th such number should be 251999.
Yeah. Is it correct?
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., \(n\pmod{x}\equiv x-1\)
The logic I think should work is that, if the number be N. Then N+1 would be divisible by all those numbers. Yeah.
Exactly what you said. :)
Therefore, \(2520n-1\) satisfies this property for all integers \(n\).
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)
Actually the first number is the LCM(1,23,4,5,6,7,8,9)-1= 2520-1= 2519
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