49th Putnam 1988

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Problem B3

αn is the smallest element of the set { |a - b√3| : a, b non-negative integers with sum n}. Find sup αn.

 

Solution

Answer: (√3 + 1)/2.

Evidently the elements of { |a - b√3| : a, b non-negative integers with sum n} are spaced a distance (√3 + 1) apart, so either smallest positive or the largest negative lies in the interval ( -(√3 + 1)/2, (√3 + 1)/2). Thus (√3 + 1)/2 is certainly an upper bound.

Let an, bn be the pair with |an - bn √3| = αn. All the values (an - bn √3) lie in the interval ( - (√3 + 1)/2, (√3 + 1)/2) ), so for any ε > 0, we can find two distinct values, corresponding to say m and n, which differ by less than ε. In other words αm-n < ε. But now consider the sequence of values αm-n, α2(m-n), α3(m-n), ... For kαm-n < (√3 + 1)/2 we have αk(m-n) = k αm-n. So we get a sequence of increasing values with spacing less than ε. Thus the sequence gives a value within ε of (√3 + 1)/2, which is therefore the least upper bound.

 


 

49th Putnam 1988

© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001