Math Challenge - October 2020

[fresh_42]

@julian I think this was a bit long, but you are right: Arzelà-Ascoli was the crucial point and checking why it can be applied the task. Well done.

 

[Not anonymous]

Spoiler: Question 12

$4^n$ and $5^n$ are positive integers. Any positive integer $x$ can be written as the product of a positive real number between 1 (inclusive) and 10 (exclusive) and an integer (non-negative) power of 10, i.e. $x = A \times 10^k$ where $1 \leq A \lt 9$ and $k \in \mathbb{N}_{0}$. Taking the base-10 logarithm, we get $\log_{10} {x} = k + \log_{10} {A}$. It is easy to see that the first digit of $x$ is determined entirely by $A$ and hence knowing $\log_{10} {A}$ (the mantissa) means knowing the that first digit.

Hence, if $x = 5^n$ and $y = 4^n$ have the same first digit, then their respective mantissas $a, b$ too should be such that they yield the same first digit, i.e. the integer parts of $10^{a}$ and $10^{b}$ should be the same.

$\log_{10} {5^n} = n \log_{10} {5}$ (Eq. 1)
$\log_{10} {4^n} = 2n \log_{10} {2} = 2n \log_{10} {\frac {10} {5}} = 2n (1 - \log_{10} {5}) = 2n - 2n \log_{10} {5}$ (Eq. 2)

$n \log_{10} {5}$ must be a real number and can therefore be written as $n \log_{10} {5} = m + a$ (Eq. 3)

where $m \in \mathbb{N}_{0}$ and $0 \lt a \lt 1$. Here $a$ is the mantissa corresponding to $x$. Note that $a = 0$ does not arise since $5^n$ is never expressible as an integer power of 10.

Combining equations (1), (2) and (3), we find that $\log_{10} {4^n} = 2n - 2(m + a) = 2n - 2m - 2a$ (Eq. 4)

Since $\log_{10} {5} \approx 0.699$, we get $0 \lt n \log_{10} {5} \lt n \Rightarrow m \in \{0, 1, ..., n-1\}$, i.e., $m$ is a non-negative integer less than $n$, i.e. $m \leq (n - 1) \Rightarrow 2n - 2m \geq 2$. Using this property in (Eq. 4), we see that $\log_{10} {4^n}$ can be written as $p + 2 - 2a$ where $p$ is a non-negative integer given by $(2n - 2m - 2)$.

Given that $0 \leq a \lt 1$, we consider two subcases and express the mantissa of $4^n$ in terms of the mantissa of $5^n$:

1. $0 < a < 0.5 \Rightarrow 0 < 2a < 1$. In this case, $\log_{10} {4^n}$ can be expressed as $(p + 1) + b$ where $b = (1-2a) \in (0, 1)$. Here the fraction $b = (1-2a)$ is the mantissa
2. $0.5 \leq a < 1 \Rightarrow 1 \leq 2a < 2$. In this case, $\log_{10} {4^n}$ can be expressed as $p + b$ where $b = (2-2a) \in (0, 1)$. Here the fraction $b = (2-2a)$ is clearly the mantissa

Thus, if we know the mantissa corresponding to $5^n$, then the mantissa corresponding to $4^n$ is uniquely determined.

We now consider various possible values for the first digit of $5^n$ (obviously that digit is determined uniquely by $n$), find what the corresponding range of values for mantissa $a$ would be, derive the possible range of values of $b$ (mantissa of $4^n$) for the given range of values of $a$ and therefore find what are the possible values for the first digit of $4^n$. Note: The table gives only approximate values for the fractions in the range (as most of those values are irrational numbers), but the ranges are such that even with the approximation, the first digits can be determined without error.

First digit of $5^n$	Range of $a$	Range of $b$	Possible values for first digit of $4^n$
1	$(\log_{10} {1}, \log_{10} {2}) = (0, 0.301)$	$(1 - 2 \times 0.301, 1 - 2 \times 0) = (0.3979, 1)$, i.e. $b > 0.3979$	Integers less than 10 and greater than $10^{0.3979}$, i.e. $\{2, 3, 4, ..., 9\}$
2	$(\log_{10} {2}, \log_{10} {3}) = (0.301, 0.477)$	$(1 - 2 \times 0.477, 1 - 2 \times 0.301) = (0.046, 0.3979)$	$\{1, 2\}$
3	$(\log_{10} {3}, \log_{10} {4}) = (0.477, 0.602)$	$(1 - 2 \times 0.5, 1 - 2 \times 0.477)$ and $(2 - 2 \times 0.602, 2 - 2 \times 0.5)$ i.e. $(0, 0.046)$ and $(0.7959, 1)$	$\{1, 6, 7, 8, 9\}$
4	$(\log_{10} {4}, \log_{10} {5}) = (0.602, 0.699)$	$(2 - 2 \times 0.699, 2 - 2 \times 0.602) = (0.602, 0.7959)$	$\{4, 5, 6\}$
5	$(\log_{10} {5}, \log_{10} {6}) = (0.699, 0.778)$	$(2 - 2 \times 0.778, 2 - 2 \times 0.699) = (0.4437, 0.602)$	$\{2, 3, 4\}$
6	$(\log_{10} {6}, \log_{10} {7}) = (0.778, 0.845)$	$(2 - 2 \times 0.845, 2 - 2 \times 0.778) = (0.3098, 0.4437)$	$\{2\}$
7	$(\log_{10} {7}, \log_{10} {8}) = (0.845, 0.903)$	$(0.1938, 0.3098)$	$\{1, 2\}$
8	$(\log_{10} {8}, \log_{10} {9}) = (0.903, 0.954)$	$(0.0915, 0.1938)$	$\{1\}$
9	$\log_{10} {9}, \log_{10} {10}) = (0.954, 1)$	$(0, 0.0915)$	$\{1\}$

From the above table, it is seen that only in cases where the first digit of $5^n$ is either 2 or 4 can the first digit of $4^n$ too take the same value (2 and 4 respectively). For values of $n$ that lead to first digit of $5^n$ being neither 2 nor 4, the values of $4^n$ would be such that their first digit will always be different from the first digit of $5^n$. Hence the claim stated in the the question is proved.

 

[fresh_42]

Spoiler: Question 12

$4^n$ and $5^n$ are positive integers. Any positive integer $x$ can be written as the product of a positive real number between 1 (inclusive) and 10 (exclusive) and an integer (non-negative) power of 10, i.e. $x = A \times 10^k$ where $1 \leq A \lt 9$ and $k \in \mathbb{N}_{0}$. Taking the base-10 logarithm, we get $\log_{10} {x} = k + \log_{10} {A}$. It is easy to see that the first digit of $x$ is determined entirely by $A$ and hence knowing $\log_{10} {A}$ (the mantissa) means knowing the that first digit.

Hence, if $x = 5^n$ and $y = 4^n$ have the same first digit, then their respective mantissas $a, b$ too should be such that they yield the same first digit, i.e. the integer parts of $10^{a}$ and $10^{b}$ should be the same.

$\log_{10} {5^n} = n \log_{10} {5}$ (Eq. 1)
$\log_{10} {4^n} = 2n \log_{10} {2} = 2n \log_{10} {\frac {10} {5}} = 2n (1 - \log_{10} {5}) = 2n - 2n \log_{10} {5}$ (Eq. 2)

$n \log_{10} {5}$ must be a real number and can therefore be written as $n \log_{10} {5} = m + a$ (Eq. 3)

where $m \in \mathbb{N}_{0}$ and $0 \lt a \lt 1$. Here $a$ is the mantissa corresponding to $x$. Note that $a = 0$ does not arise since $5^n$ is never expressible as an integer power of 10.

Combining equations (1), (2) and (3), we find that $\log_{10} {4^n} = 2n - 2(m + a) = 2n - 2m - 2a$ (Eq. 4)

Since $\log_{10} {5} \approx 0.699$, we get $0 \lt n \log_{10} {5} \lt n \Rightarrow m \in \{0, 1, ..., n-1\}$, i.e., $m$ is a non-negative integer less than $n$, i.e. $m \leq (n - 1) \Rightarrow 2n - 2m \geq 2$. Using this property in (Eq. 4), we see that $\log_{10} {4^n}$ can be written as $p + 2 - 2a$ where $p$ is a non-negative integer given by $(2n - 2m - 2)$.

Given that $0 \leq a \lt 1$, we consider two subcases and express the mantissa of $4^n$ in terms of the mantissa of $5^n$:

1. $0 < a < 0.5 \Rightarrow 0 < 2a < 1$. In this case, $\log_{10} {4^n}$ can be expressed as $(p + 1) + b$ where $b = (1-2a) \in (0, 1)$. Here the fraction $b = (1-2a)$ is the mantissa
2. $0.5 \leq a < 1 \Rightarrow 1 \leq 2a < 2$. In this case, $\log_{10} {4^n}$ can be expressed as $p + b$ where $b = (2-2a) \in (0, 1)$. Here the fraction $b = (2-2a)$ is clearly the mantissa

Thus, if we know the mantissa corresponding to $5^n$, then the mantissa corresponding to $4^n$ is uniquely determined.

We now consider various possible values for the first digit of $5^n$ (obviously that digit is determined uniquely by $n$), find what the corresponding range of values for mantissa $a$ would be, derive the possible range of values of $b$ (mantissa of $4^n$) for the given range of values of $a$ and therefore find what are the possible values for the first digit of $4^n$. Note: The table gives only approximate values for the fractions in the range (as most of those values are irrational numbers), but the ranges are such that even with the approximation, the first digits can be determined without error.

First digit of $5^n$	Range of $a$	Range of $b$	Possible values for first digit of $4^n$
1	$(\log_{10} {1}, \log_{10} {2}) = (0, 0.301)$	$(1 - 2 \times 0.301, 1 - 2 \times 0) = (0.3979, 1)$, i.e. $b > 0.3979$	Integers less than 10 and greater than $10^{0.3979}$, i.e. $\{2, 3, 4, ..., 9\}$
2	$(\log_{10} {2}, \log_{10} {3}) = (0.301, 0.477)$	$(1 - 2 \times 0.477, 1 - 2 \times 0.301) = (0.046, 0.3979)$	$\{1, 2\}$
3	$(\log_{10} {3}, \log_{10} {4}) = (0.477, 0.602)$	$(1 - 2 \times 0.5, 1 - 2 \times 0.477)$ and $(2 - 2 \times 0.602, 2 - 2 \times 0.5)$ i.e. $(0, 0.046)$ and $(0.7959, 1)$	$\{1, 6, 7, 8, 9\}$
4	$(\log_{10} {4}, \log_{10} {5}) = (0.602, 0.699)$	$(2 - 2 \times 0.699, 2 - 2 \times 0.602) = (0.602, 0.7959)$	$\{4, 5, 6\}$
5	$(\log_{10} {5}, \log_{10} {6}) = (0.699, 0.778)$	$(2 - 2 \times 0.778, 2 - 2 \times 0.699) = (0.4437, 0.602)$	$\{2, 3, 4\}$
6	$(\log_{10} {6}, \log_{10} {7}) = (0.778, 0.845)$	$(2 - 2 \times 0.845, 2 - 2 \times 0.778) = (0.3098, 0.4437)$	$\{2\}$
7	$(\log_{10} {7}, \log_{10} {8}) = (0.845, 0.903)$	$(0.1938, 0.3098)$	$\{1, 2\}$
8	$(\log_{10} {8}, \log_{10} {9}) = (0.903, 0.954)$	$(0.0915, 0.1938)$	$\{1\}$
9	$\log_{10} {9}, \log_{10} {10}) = (0.954, 1)$	$(0, 0.0915)$	$\{1\}$

From the above table, it is seen that only in cases where the first digit of $5^n$ is either 2 or 4 can the first digit of $4^n$ too take the same value (2 and 4 respectively). For values of $n$ that lead to first digit of $5^n$ being neither 2 nor 4, the values of $4^n$ would be such that their first digit will always be different from the first digit of $5^n$. Hence the claim stated in the the question is proved.

This looks like the published solution, only that you made your life artificially difficult by dealing with logarithms and real numbers. You can do the same by setting
\begin{align*}
z\cdot 10^r &\leq 4^n=2^{2n}< (z+1)\cdot 10^r\\
z\cdot 10^s &\leq 5^n < (z+1)\cdot 10^s
\end{align*}
Then square the second and multiply both.

 

[Not anonymous]

Spoiler: Question 15

$a+b+c+2 = abc \Rightarrow c = \dfrac{a+b+2}{ab-1}$ (Eq. 1)

$(a+1)(b+1)(c+1)$ can therefore be expressed as a function $f$ of variables $a, b$ alone.

$(a+1)(b+1)(c+1) \equiv f(a, b) = (a+1)(b+1)\left(\dfrac{a+b+2}{ab-1} + 1\right)$
$= (a+1)(b+1)\dfrac{(a+1)(b+1)}{ab-1} = \dfrac{(a+1)^2(b+1)^2}{ab-1}$ (Eq. 2)

Since $a, b > 0$, it follows that the numerator of (Eq. 1) is positive. Since we also need $c > 0$, we also need denominator in that equation to be positive and hence we must have $ab > 1$.

To find the minimum value of $f(a, b)$ for a fixed value of $b$, find the partial derivative of this function w.r.t $a$ and equate to 0.$\dfrac {\partial f} {\partial a} = 0 \equiv \dfrac{(b+1)^2 (a+1) (ab - 2 - b)}{(ab-1)^2} = 0 \Rightarrow a = \dfrac{2+b}{b}$ (Eq. 3)

That the above is a minimum, not a maximum, can be proven by looking at the sign of $\dfrac{\partial^2 f} {\partial a^2}$ at $a = \frac{2+b}{b}$.
$\dfrac{\partial^2 f} {\partial a^2} = \dfrac{(b+1)^2}{(ab-1)^2} \left( \dfrac{-2b(a+1)(ab - 2 - b)}{ab - 1} + (2ab - 2) \right)$, which when evaluated at $a = \dfrac{2+b}{b}$ is equal to
$\dfrac{1}{b+1} \left( -2(2+2b)(2+b-2-b) + 2(b+1)^2 \right) = 2(b+1)$. Since this value is positive for any positive value of $b$ (in fact it is positive for $b > -1$), $a = \frac{2+b}{b}$ must correspond to a local minimum, not a maximum.
$f(\frac{2+b}{b}, b)$ is therefore minimum value of $f(a, b)$ for a specific value of $b$, and for variable $b$, these minima can be expressed as a function of $b$.

$g(b) \equiv f(\frac{2+b}{b}, b) = \dfrac{4(b+1)^3}{b^2}$.
The minimum possible value of $(a+1)(b+1)(c+1)$ is the minimum value of $f(a, b)$ across all allowed values of $a, b$ and this is $\min_{a>0, b>0} f(a, b) = \min_{b>0} f(\frac{2+b}{b}, b) = \min_{b>0} g(b)$

The minimum value of $g(b)$ for $b > 0$ can be found by finding the local minima of $g(b)$.
$g'(b) = 0 \Rightarrow \dfrac{-8(b+1)^3}{b^3} + \dfrac{12(b+1)^2}{b^2} \Rightarrow \dfrac{(b+1)^2}{b^3}(4b-8) = 0 \Rightarrow b=2$ (Eq. 4)

That the above is a local minimum, not a maximum, can be seen by evaluating $g''(b)$ at $b=2$ and noting that $g''(b=2)$ is a positive value ($\frac{9}{2} = 4.5$ if I calculated correctly).

Thus, the minimum value of $(a+1)(b+1)(c+1)$ that meets the conditions stated in the question is $g(2) = \dfrac{4(2+1)^3}{2^2} = 3^3 = 27$. This proves that $(a+1)(b+1)(c+1) \geq 27$ for the stated conditions and that the minimum value of 27 is achieved when $b=2$ whereby using (Eq. 2) and (Eq. 1) we get the corresponding values of $a, c$ both to be 2, i.e. $a=b=c=2$ gives the minimum value for the expression.