NMO Problem 2
A circumference was divided in
n
equal parts. On each of these parts one number from
1
to
n
was placed such that the distance between consecutive numbers is always the same. Numbers
1
1
,
4
and
1
7
were in consecutive positions. In how many parts was the circumference divided?This problem is from the NMO.This problem is part of
this set
.
The answer is 20.
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2 solutions
A nice simple solution.+1). Yours is much better solution. I simply gave mine how I did it.
W e s o l v e f o r p ∗ q ( m o d n ) = X w h e r e p i s t h e d i s t a n c e b e t w e e n t h e n u m b e r s , n t h e d i v i s i o n . W e c h o o s e a ′ n ′ a n d t r y q = s a y 3 , 4 , 5 , 6 , e x c l u d i n g w h e n n , q h a v e a c o m m o n f a c t o r . W h e n v a l u e s o f X f o r q = 1 1 , 4 , 1 7 a r e i n c o n s e c u t i v e p o s i t i o n s w e g e t o u r s o l u t i o n . I t i s c l e a r t h a t n > 1 7 . A t n = 2 0 a n d p = 3 , w e g e t t h e s o l u t i o n .
This solution is not mine:
We start placing the numbers from 1, and move a certain number of sectors to place the subsequent numbers. After n moves, we will be back to the first sector, hence we can consider the numbers in modulo n . So the number of moves it takes from 11 to 4 equals that of from 4 to 17, hence 4 − 1 1 ≡ 1 7 − 4 ( m o d n ) . So n ∣ 2 0 , and from n ≥ 1 7 , we get n = 2 0 .