PROBLEMS FOR MAY
Please send your solution to
Edward J. Barbeau
Department of Mathematics
University of Toronto
Toronto, ON M5S 3G3
no later than June 30, 2003. It is important that your complete mailing address and your email address appear on the front page.
Notes: The notation [a, b] refers to the closed interval { x : a £ x £ b }. A function f defined on a closed interval is strictly increasing iff f(u) < f(v) whenever u < v. Such a function has an inverse function g defined on the image of f that satisfies y = g(x) if and only if x = f(y).
- 227.
-
Let n be an integer exceeding 2 and let
a0, a1, a2, ¼, an, an+1 be positive real numbers for which
a0 = an, a1 = an+1 and
for some positive integers ki, where 1 £ i £ n.ai-1 + ai+1 = ki ai
-
Prove that
2n £ k1 + k2 + ¼+ kn £ 3n .
- 228.
-
Prove that, if 1 < a < b < c, then
loga (loga b) + logb (logb c) + logc (logc a) > 0 .
- 229.
- Suppose that n is a positive integer and that 0 < i < j < n. Prove that the greatest common divisor of (n || i) and (n || j) exceeds 1.
- 230.
-
Let f be a strictly increasing function on the
closed interval [0, 1] for which f(0) = 0 and f(1) = 1.
Let g be its inverse. Prove that
9
å
k=1æ
èf æ
èk 10ö
ø+g æ
èk 10ö
øö
ø£ 9.9 .
- 231.
- For n ³ 10, let g(n) be defined as follows: n is mapped by g to the sum of the number formed by taking all but the last three digits of its square and adding it to the number formed by the last three digits of its square. For example, g(54) = 918 since 542 = 2916 and 2 + 916 = 918. Is it possible to start with 527 and, through repeated applications of g, arrive at 605?
- 232.
-
(a) Prove that, for positive integers n and
positive values of x,
(1 + xn+1)n £ (1 + xn)n+1 £ 2(1 + xn+1)n .
-
(b) Let h(x) be the function
defined by
Determine a value N for whichh(x) = ì
ï
í
ï
î1, if 0 £ x £ 1; x, if x > 1.
whenever 0 £ x £ 10 and n ³ N.|h(x) - (1 + xn)1/n | < 10-6
- 233.
- Let p(x) be a polynomial of degree 4 with rational coefficients for which the equation p(x) = 0 has exactly one real solution. Prove that this solution is rational.