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Problem 1# Evaluate ∫ 1 2 ln x 2 − 2 x + x 2 d x \displaystyle \int_1^2 \frac{\ln x}{2 - 2x + x^2} dx ∫ 1 2 2 − 2 x + x 2 ln x d x .
Problem 2# Determine the real numbers k k k such that ∑ n = 1 ∞ ( ( 2 n ) ! 4 n n ! n ! ) k \displaystyle \sum_{n=1}^\infty
\left( \frac{(2n)!}{4^n n! n!} \right)^k n = 1 ∑ ∞ ( 4 n n ! n ! ( 2 n )! ) k is convergent.
Problem 3# Let n n n be a positive integer and let M n ( Z 2 ) M_n (\mathbb{Z}_2) M n ( Z 2 ) denote the n n n by
n n n matrices with entries from the integers mod 2. If n ≥ 2 n \ge 2 n ≥ 2 , prove that the
number of matrices A A A in M n ( Z 2 ) M_n (\mathbb{Z}_2) M n ( Z 2 ) satisfying A 2 = 0 A^2 = 0 A 2 = 0 (the matrix
with all entries zero) is an even positive integer.
Problem 4# For a positive integer a a a , let P ( a ) P(a) P ( a ) denote the largest prime divisor of
a 2 + 1 a^2 + 1 a 2 + 1 . Prove that there exist infinitely many triples ( a , b , c ) (a, b, c) ( a , b , c ) of
distinct positive integers such that P ( a ) = P ( b ) = P ( c ) P(a) = P(b) = P(c) P ( a ) = P ( b ) = P ( c ) .
Problem 5# Suppose that m , n , r m, n, r m , n , r are positive integers such that
1 + m + n 3 = ( 2 + 3 ) 2 r − 1 1 + m + n \sqrt{3} = (2 + \sqrt{3})^{2r-1} 1 + m + n 3 = ( 2 + 3 ) 2 r − 1
Prove that m m m is a perfect square.
Problem 6# Let A , B , P , Q , X , Y A, B, P, Q, X, Y A , B , P , Q , X , Y be square matrices of the same size. Suppose that
A + B + A B = X Y A X = X Q P + Q + P Q = Y X P Y = Y B .
\begin{align*}
A + B + AB & = XY & \hspace{1cm} AX & = XQ \newline
P + Q + PQ & = YX & \hspace{1cm} PY & = YB.
\end{align*}
A + B + A B P + Q + PQ = X Y = Y X A X P Y = XQ = Y B .
Prove that A B = B A AB = BA A B = B A .
Problem 7# Let q q q be a real number with ∣ q ∣ ≠ 1 |q| \ne 1 ∣ q ∣ = 1 and let k k k be a positive integer.
Define a Laurent polynomial f k ( X ) f_k (X) f k ( X ) in the variable X X X , depending on q q q and
k k k , by
f k ( X ) = ∏ i = 0 k − 1 ( 1 − q i X ) ( 1 − q i + 1 X − 1 ) . f_k (X) = \prod_{i=0}^{k-1} (1 - q^i X)(1 - q^{i+1} X^{-1}). f k ( X ) = i = 0 ∏ k − 1 ( 1 − q i X ) ( 1 − q i + 1 X − 1 ) .
Show that the constant term of f k ( X ) f_k (X) f k ( X ) , i.e., the coefficient of X 0 X^0 X 0 in
f k ( X ) f_k (X) f k ( X ) , is equal to
( 1 − q k + 1 ) ( 1 − q k + 2 ) ⋯ ( 1 − q 2 k ) ( 1 − q ) ( 1 − q 2 ) ⋯ ( 1 − q k ) . \frac{(1 - q^{k+1})(1 - q^{k+2}) \cdots (1 - q^{2k})}{(1 - q)(1 - q^2) \cdots (1 - q^k)}. ( 1 − q ) ( 1 − q 2 ) ⋯ ( 1 − q k ) ( 1 − q k + 1 ) ( 1 − q k + 2 ) ⋯ ( 1 − q 2 k ) .