Problem A Day Marathon – Algebra

  1. Find solutions to $x + 5 + \frac 6 x = 0$.
    Solution Let $x$ be a non-zero real number. Suppose $x + 5 + \frac 6 x = 0$. Let's multiply both sides by $x$: $x^2 + 5x + 6 = 0$.
    Notice that the left-hand side equals $(x+2)(x+3)$.
    Therefore, $x = -2$ or $x = -3$ and, reciprocally, we can check that they're indeed solutions of the equation.
  2. $a$, $b$, $c$ and $d$ are nonnegative real numbers. If we have: $a^2 + b^2 = 1$, $c^2 + d^2 = 1$ and $ac - bd = 12$, find $ad + bc$.
    Solution (1) $a^2c^2 + b^2d^2 - abcd = (ac - bd)^2 = \frac 1 4$
    (2) $a^2d^2 + b^2c^2 + abcd = (ad + bc)^2$
    Therefore, adding (1) and (2): $a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 = (ad + bc)^2 + \frac 1 4$
    Notice that $a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 = (a^2 + b^2)(c^2 + d^2) = 1 \cdot 1 = 1$.
    Consequently, $(ad + bc)^2 + \frac 1 4 = 1$, so $(ad + bc)^2 = \frac 3 4$
    Finally, since $ad + bc \ge 0$, we have $ad + bc = \frac {\sqrt 3} 2.
  3. Two numbers sum to 1337 and have a product 9001. Find the two numbers.
    Solution Let $a$ and $b$ be the two numbers, with $a \le b$.
    Then we know that $a$ and $b$ are the solutions of the equation "$x^2 - (a+b)x + abx = 0$", with $a+b = 1337$ and $ab = 9001$. Using the quadratic formula, we find that
    $a = \frac {1337 - \sqrt{1751565}} {2}$ and
    $b = \frac {1337 + \sqrt{1751565}} {2}$
  4. Find the sum of all the real and nonreal roots of $x^{2001} + \left(\frac 1 2 - x\right)^{2001}$
    Solution Binomial expansion: $\left(\frac 1 2 - x\right)^{2001} = \sum_{k=0}^{2001} \left({2001} \choose {k} \right) \frac {1} {2^{2001-k}} (-1)^k x^k$
    Here, the coefficient before $x^{2001}$ equals $-1$, so that term will cancel with $x^{2001)$: the degree of this polynomial is actually 2000.
    Using Vieta's formula: the sum of the roots ($s$) of this polynomial equals $-\frac{a_{1999}} {a_{2000}}$ if we note $a_0, \cdots, a_{2000}$ the coefficients of this polynomial.
    $a_{1999} = \left({2001} \choose {1999} \right) \frac {1} {2^{2001-1999}} (-1)^{1999} = -250 \cdot 2001$
    $a_{2000} = \left({2001} \choose {2000} \right) \frac {1} {2^{2001-2000}} (-1)^{2000} = \frac {2001} {2}$
    Therefore, $s = 500$.
  5. Express $(0^3 - 350)(1^3 - 349)(2^3 - 348) \cdots (350^3 - 0)$ as concisely as possible.
    Solution Well, $7^3 = 343$ so the product is equal to $0$...
    You can also notice it because the product is $\prod_{k = 0}^{350} (k^3 + k - 350)$ and $7$ is a root of the polynomial $x^3 + x - 350$.
    For $k = 7$, we have $k^3 + k - 350 = 0$.
  6. If $a + \frac 1 a = 3$, find $a^4 + \frac {1} {a^4}$.
    Solution $\left(a + \frac 1 a \right)^2 = a^2 + \frac {1} {a^2} + 2 = 3^2 = 9$ so $a^2 + \frac {1} {a^2} = 7$.
    $\left(a^2 + \frac {1} {a^2} \right)^2 = a^4 + \frac {1} {a^4} + 2 = 7^2 = 49$ so $a^4 + \frac {1} {a^4} = 47$.
  7. Let $r$ and $s$ be roots of $x^2 - 5x + 2$. Find $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1}$.
    Solution According to Vieta's formulas, $r + s = -5$ and $rs = 2$.
    Notice that $r \neq 1$ and $s \neq 1$ (because $1$ isn't a root of this polynomial), so the expression is well defined.
    Moreover, $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1} = r^2 + r + 1 + s^2 + s + 1$ because of the identity "$a^3 - b^3 = (a-b)(a^2+ab+b^2)$"
    Therefore, $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1} = (r^2 + s^2) + (r + s) + 2$.
    Notice that $r^2 + s^2 = (r + s)^2 - 2rs = 25 - 4 = 21$.
    Therefore, $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1} = 21 - 5 + 2 = 18$.
  8. $f$ is a second degree polynomial such that $x^2 - 8x + 17 \le f(x) \le 2x^2 - 16x + 33$. If $f(10) = 55$, then find $f(22)$.
    Solution Let $g(x) = x^2 - 8x + 17$ an $h(x) = 2x^2 - 16x + 33$ for all $x$.
    Notice that $g(4) = h(4) = 1$, which is the minimum value of $g$ and of $h$. Therefore $f(4) = 1$ and it's the minimum value of $f$.
    Let's note $f(x) = ax^2 + bx + c$. We have the following equalities:
    (1) $100a + 10b + c = 55$   because $f(10) = 55$
    (2) $16a + 4b + c = 1$   because $f(4) = 1$
    (3) $-\frac {b} {2a} = 4$   because the minimum value of $f$ occurs at $x = 4$
    (1) - (2) gives $84a + 6b = 54$, that is, $14a + b = 9$.
    (3) gives $b = -8a$.
    Therefore, $14a - 8a = 9$, so $a = \frac 3 2$. So we can deduce that $b = -12$ and $c = 25$.
    Finally, $f(x) = \frac 3 2 x^2 - 12x + 25$ and $f(22) = 487$.
  9. If $\frac 4 5 x = y$ and $x^y = y^x$, then $x - y$ can be expressed as a reduced fraction $\frac a b$. Find the value of $a + b$.
    Solution We assume that $x$ and $y$ are positive numbers.
    $x^{\frac 4 5 x} = \left(\frac 4 5 x\right)^x$ so $\left(x^{\frac 4 5}\right)^x = \left(\frac 4 5 x\right)^x$.
    Therefore, $x^{\frac 4 5} = \frac 4 5 x$ and $x = \left(\frac 5 4\right)^5$.
    Consequently, $y = \frac 4 5 x = \left(\frac 5 4\right)^4$.
    $x - y = \left(\frac 5 4\right)^5 - \left(\frac 5 4\right)^4 = \frac {5^4} {4^5}$ (which is a reduced fraction) so $a + b = 5^4 + 4^5 = 1649$.
  10. Find $(1 + i)^{12345678987654321}$
    Solution $1 + i = \sqrt 2 e^{i \frac {\pi} {4}}$ and $12345678987654321 \equiv 1$ (mod 8)
    Therefore, $(1 + i)^{12345678987654321} = (\sqrt 2)^{12345678987654321} e^{i \frac {\pi} {4}}$
  11. Find $\exp \left({\ln (x+1)+ \ln (x^2+1)+\ln (x^4+1)+\cdots+\ln (x^{128}+1)}\right)$ fully simplified and without any parentheses in the final result.
    Solution Let's note that number $P(x)$.
    Using properties of exp and ln, we can find that, in fact, $P(x) = \prod_{k=0}^{7} (x^{2^k} + 1)$.
    Let $x \neq 1$. Then, for all $k$, $\left(x^{2^k} + 1\right)\left(x^{2^{k+1}} + 1\right) = x^{2^k + 2^{k+1}} + x^{2^k} + x^{2^{k+1}} + 1 = \left(x^{2^k}\right)^3 + \left(x^{2^k}\right)^2 + x^{2^k} + 1 = \frac {\left(x^{2^k}\right)^4 - 1} {x^{2^k} - 1}$ (formula for a geometric sum).
    Therefore, $P\left(x\right) = \left(\frac {x^{4} - 1} {x - 1}\right) \left(\frac {x^{16} - 1} {x^{4} - 1}\right)\left(\frac {x^{64} - 1} {x^{16} - 1}\right)\left(\frac {x^{256} - 1} {x^{64} - 1}\right)$
    There is a lot of simplifications, leaving $P\left(x\right) = \frac {x^{256} - 1} {x - 1}$
    Hence, $P(x) = x^{255} + \cdots + x + 1$, which is also true if $x = 1$.
  12. Let $F_0 = 0$ and $F_1 = 1$ and for $n \ge 2$ we have $F_{n-2} + F_{n-1} = F_n$. Find $\sum_{k=0}^{\infty} \frac {F_k} {7^k}$.
    Solution The series converges: we can prove by induction that $F_k \le 2^k$ for all nonnegative integer $k$.
    So $0 \le \frac {F_k} {7^k} \le \left(\frac 2 7\right)^k$ and, consequently, the series converge accodring to the comparison test (as $\sum \left(\frac 2 7\right)^k$ converges).
    Let $l$ be the value of the sum.
    Let $u_k = \frac {F_k} {7^k}$ for all nonnegative integer $k$.
    For all integer $k \ge 2$, $u_k = \frac {F_k} {7^k} = \frac {F_{k-1}} {7^k} + \frac {F_{k-2}} {7^k} = \frac 1 7 u_{k-1} + \frac 1 {49} u_{k-2}$
    Let $N \ge 2$ be an integer. $\sum_{k=0}^{N} u_k = u_0 + u_1 + \sum_{k=2}^{N} (\frac 1 7 u_{k-1} + \frac 1 {49} u_{k-2}) = \frac 1 7 + \frac 1 7 \sum_{k=1}^{N-1} u_k + \frac 1 {49} \sum_{k=0}^{N-2} u_k$
    Taking the limit ($N \to \infty$) of both sides: $l = \frac 1 7 + \frac 1 7 (l - u_0) + \frac 1 {49} l = \frac 8 {49} l + 1$ ; so $l = \frac {7} {41}$.
  13. The sum of the real coefficients of the terms of the expanded form of $(1 + i)^{1337}$ can be expressed as $2n$ where $n$ is an integer. Find $n$.
    Solution $1 + i = \sqrt 2 e^{i \frac {\pi} 4}$ and $1337 \equiv 1$ (mod 8)
    Therefore, $1 + i = \sqrt 2 ^ {1337} e^{i \frac {\pi} 4} = 2^{668} \sqrt 2 (\frac {\sqrt 2} 2 + i\frac {\sqrt 2} 2) = 2^{668} + 2^{668}i$ so $n = 668$.