Difference between revisions of "2004 Pan African MO Problems/Problem 3"
Rockmanex3 (talk | contribs) (Solution to Problem 3 -- numbers in a circle) |
Rockmanex3 (talk | contribs) m (Adjusted difficulty to Intermediate NT.) |
||
Line 29: | Line 29: | ||
{{Pan African MO box|year=2004|num-b=2|num-a=4}} | {{Pan African MO box|year=2004|num-b=2|num-a=4}} | ||
− | [[Category: | + | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 10:46, 27 March 2020
Problem
One writes 268 numbers around a circle, such that the sum of 20 consecutive numbers is always equal to 75. The number 3, 4 and 9 are written in positions 17, 83 and 144 respectively. Find the number in position 210.
Solution
Let be the 268 numbers. First, we will prove that a number twenty positions after said number would be the same number.
Lemma: for integers
We know that , where . By subtracting and adding from both sides, we must have . Therefore, , so .
Thus, we know that , where is an integer. If , then . If , then . Note that is a factor of , so we can know that if , then .
With this in mind,
Thus, since , we must have .
See Also
2004 Pan African MO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All Pan African MO Problems and Solutions |