Solutions to the 64th William Lowell Putnam Mathematical Competition Saturday, December 6, 2003

نویسندگان

  • Manjul Bhargava
  • Kiran Kedlaya
چکیده

expand both sides, and compare the terms on both sides in which k of the terms are among the ai. On the left, one has the product of each kelement subset of {1, . . . , n}; on the right, one has ( n k ) (a1 · · · an) · · · (b1 . . . bn), which is precisely ( n k ) times the geometric mean of the terms on the left. Thus AM-GM shows that the terms under consideration on the left exceed those on the right; adding these inequalities over all k yields the desired result.

برای دانلود رایگان متن کامل این مقاله و بیش از 32 میلیون مقاله دیگر ابتدا ثبت نام کنید

ثبت نام

اگر عضو سایت هستید لطفا وارد حساب کاربری خود شوید

منابع مشابه

The 63 rd William Lowell Putnam Mathematical Competition Saturday

B–4 An integer n, unknown to you, has been randomly chosen in the interval [1, 2002] with uniform probability. Your objective is to select n in an odd number of guesses. After each incorrect guess, you are informed whether n is higher or lower, and you must guess an integer on your next turn among the numbers that are still feasibly correct. Show that you have a strategy so that the chance of w...

متن کامل

Solutions to the 78th William Lowell Putnam Mathematical Competition Saturday, December 2, 2017

Remark: A similar argument shows that any secondorder linear recurrent sequence also satisfies a quadratic second-order recurrence relation. A familiar example is the identity Fn−1Fn+1 − F2 n = (−1)n for Fn the nth Fibonacci number. More examples come from various classes of orthogonal polynomials, including the Chebyshev polynomials mentioned below. Second solution. We establish directly that ...

متن کامل

Solutions to the 71 st William Lowell Putnam Mathematical Competition Saturday

A3 If a = b = 0, then the desired result holds trivially, so we assume that at least one of a, b is nonzero. Pick any point (a0, b0) ∈ R, and let L be the line given by the parametric equation L(t) = (a0, b0) + (a, b)t for t ∈ R. By the chain rule and the given equation, we have d dt (h◦L) = h◦L. If we write f = h◦L : R → R, then f (t) = f(t) for all t. It follows that f(t) = Ce for some consta...

متن کامل

Solutions to the 71st William Lowell Putnam Mathematical Competition Saturday, December 4, 2010

A3 If a = b = 0, then the desired result holds trivially, so we assume that at least one of a, b is nonzero. Pick any point (a0, b0) ∈ R, and let L be the line given by the parametric equation L(t) = (a0, b0) + (a, b)t for t ∈ R. By the chain rule and the given equation, we have d dt (h◦L) = h◦L. If we write f = h◦L : R→ R, then f ′(t) = f(t) for all t. It follows that f(t) = Ce for some consta...

متن کامل

ذخیره در منابع من


  با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید

عنوان ژورنال:

دوره   شماره 

صفحات  -

تاریخ انتشار 2003