GLaw’s challenge problems

Jul 2015
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Ilford
I’m back! (Smile)

Solution to Problem #3:
An isomorphism is given by $\phi:\mathbb Z[x] \to \mathbb Q^+$; $\phi(0)=1$ and if $f = a_1 + a_2x + \cdots + a_nx^{n-1} \not\equiv 0$, $\phi(f) = 2^{a_1}3^{a_2}\cdots p_n^{a_n}$ where $p_n$ is the $n$th prime.

Challenge Problem #4:

Find all integers $a$ such that

\(\displaystyle \frac{2 \cdot 1000^k - a}{27}\)​

is an integer for all integers $k\geqslant0$. Similarly find all integers $b$ and $c$ such that

\(\displaystyle \frac{2 \cdot 10^{3k-1} - b}{27}\ \text{and}\ \frac{2 \cdot 10^{3k-2} - c}{27}\)​

are integers for all integers $k\geqslant1$.
 
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Dec 2012
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A solution using congruence mod 27:
For any integer $x$, ${x\over27}$ an integer is the same as saying $x\equiv 0\pmod {27}$ . Easily, the multiplicative order of 10 mod 27 is 3. So
1. $2\cdot 1000^k-a\equiv 0\pmod {27}$ if and only if $2(10^{3})^k-a\equiv 0\pmod {27}$ if and only if $a\equiv 2\pmod {27}$ or $a=2+27q\text{ for some integer }q$
2. $2\cdot 10^{3k-1}\equiv b\pmod {27}$ iff $2\cdot 10^{-1}\equiv b\pmod {27}$. Here $10^{-1}=19$ is the multiplicative inverse of 10. So $b\equiv 11\pmod {27}$
3. $2\cdot 10^{3k-2}\equiv c\pmod {27}$ iff $2\cdot 10^{-2}\equiv c\pmod {27}$ iff $c\equiv 20 \pmod {27}$
 
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I’m back! (Smile)

Solution to Problem #3:
An isomorphism is given by $\phi:\mathbb Z[x] \to \mathbb Q^+$; $\phi(0)=1$ and if $f = a_1 + a_2x + \cdots + a_nx^{n-1} \not\equiv 0$, $\phi(f) = 2^{a_1}3^{a_2}\cdots p_n^{a_n}$ where $p_n$ is the $n$th prime.

Challenge Problem #4:

Find all integers $a$ such that

\(\displaystyle \frac{2 \cdot 1000^k - a}{27}\)​

is an integer for all integers $k\geqslant0$. Similarly find all integers $b$ and $c$ such that

\(\displaystyle \frac{2 \cdot 10^{3k-1} - b}{27}\ \text{and}\ \frac{2 \cdot 10^{3k-2} - c}{27}\)​

are integers for all integers $k\geqslant1$.
An answer equivalent to johng's using induction.

$\dfrac{2 * 1000^0 + 25}{27} = \dfrac{2 + 25}{27} = \dfrac{27}{27} = 1 \in \mathbb Z.$

$Let\ k\ be\ any\ integer \ge 0\ such\ that\ \dfrac{2 * 1000^k + 25}{27} \in \mathbb Z.$

There is at least one such integer, namely 0.

$Let\ u = \dfrac{2 * 1000^k + 25}{27} \implies u \in \mathbb Z.$

$Let\ v = 74 * 1000^k,\ which\ obviously \in \mathbb Z.$

$\dfrac{2 * 1000^{(k+1)}}{27} - u = \dfrac{2 * 1000^{(k+1)} + 25 - (2 * 1000^k + 25)}{27} = \dfrac{2 * 1000^{(k+1)} - 2 * 1000^k}{27} \implies$

$\dfrac{2 * 1000^k * (1000 - 1)}{27} = 1000^k * \dfrac{2 * 999}{27} = v.$

$So\ \dfrac{2 * 1000^{(k+1)}}{27} - u = v \implies \dfrac{2 * 1000^{(k+1)}}{27} = u + v \in \mathbb Z.$

$therefore, n \in \mathbb\ Z\ and\ n \ge 0 \implies \dfrac{2 * 1000^n + 25}{27} \in \mathbb Z.$

$Let\ n = any\ non-negative\ integer\ and\ x = \dfrac{2 * 1000^n + 25}{27}.$

$x \in \mathbb Z.$

$Let\ m\ be\ any\ integer\ and\ a = 27m - 25 \implies a \in \mathbb Z.$

$\dfrac{2 * 1000^n - a}{27} = \dfrac{2 * 1000^n + 25 - 27m}{27} = \dfrac{2 * 1000^n + 25}{27} - \dfrac{27m}{27} = x - m \in \mathbb Z.$

So the answer is that a must be of the form a = 27m - 25, where m is an integer.

I don't have time to do the follow ons right now. But thanks: this was fun.
 
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Jul 2015
217
116
Ilford
Solution to Problem #4:
Define

\(\displaystyle \begin{array}{rcl}s_1 &=& 1 \\ s_2 &=& 11 \\ s_3 &=& 111 \\ &\vdots& \end{array}\)​

\(\displaystyle \therefore\ s_n\ \equiv\ 1\pmod5\ \forall n=1,2,\ldots\)

\(\displaystyle \Rightarrow\ s_n\ \equiv\ 1,6,11\pmod{15}\)

\(\displaystyle \text{Also}\ s_n\ \equiv\ \text{sum of digits}\ =\ n\pmod3\)

\(\displaystyle \therefore\ s_n\ \equiv\ 1,11,6\pmod{15}\ \text{if}\ n\equiv1,2,3\pmod3\ \text{respectively}\)

\(\displaystyle \Rightarrow\ 10^n\ =\ 9s_n+1\ \equiv\ 10,100,55\pmod{135}\ \text{if}\ n\equiv1,2,3\pmod3\)

\(\displaystyle \Rightarrow\ 2\cdot10^{n-1}\ \equiv\ 2,20,11\pmod{27}\ \text{if}\ n\equiv1,2,3\pmod3\)

\(\displaystyle \Rightarrow\ a\equiv2\pmod{27},\ b\equiv11\pmod{27},\ c\equiv20\pmod{27}\)

Challenge Problem #5:

Let $a,b,c$ be non-negative real numbers. Prove that

\(\displaystyle (a+b)\sqrt{ab}+(b+c)\sqrt{bc}+(c+a)\sqrt{ca}\ \leqslant\ \frac{a^2+b^2+c^2+3(ab+bc+ca)}2\)​
 
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Feb 2015
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Ottawa Ontario
Hopefully your last one this year :)
 
Feb 2014
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$a,\ b,\ c,\ x,\ y\ are\ all\ real\ and \ge 0.$

$STEP\ I:$

$0 \le \sqrt{xy} \le \dfrac{x + y}{2} \implies 0 \le \dfrac{x + y}{2} - \sqrt{xy} \implies$

$0 \le \left(\dfrac{x + y}{2} - \sqrt{xy}\right)^2 = \left(\dfrac{x + y}{2}\right)^2 - 2\sqrt{xy}\left(\dfrac{x + y}{2}\right) + xy \implies$

$0 \le \left(\dfrac{x + y}{2}\right)^2 - (x + y)\sqrt{xy}+ xy \implies$

$(x + y)\sqrt{xy} \le \left(\dfrac{x + y}{2}\right)^2 + xy.$

$STEP\ II:$

$\dfrac{a^2 + b^2 + c^2 + 3(ab + bc + ac)}{2} = \dfrac{2a^2 + 2b^2 + 2c^2 + 6(ab + bc + ac)}{4} =$

$\dfrac{(a^2 + 2ab + b^2) + (a^2 + 2ac + c^2) + (b^2 + 2bc + c^2) + 4(ab + bc + ac)}{4} =$

$\left\{\left(\dfrac{a + b}{2}\right)^2 + ab\right\} + \left\{\left(\dfrac{a + c}{2}\right)^2 + ac\right\} +\left\{\left(\dfrac{b + c}{2}\right)^2 + bc\right\}.$

$STEP\ III:$

$By\ STEP\ I:$

$(a + b)\sqrt{ab} \le \left(\dfrac{a + b}{2}\right)^2 + ab,\ and$

$(a + c)\sqrt{ac} \le \left(\dfrac{a + c}{2}\right)^2 + ac,\ and$

$(b + c)\sqrt{bc} \le \left(\dfrac{b + c}{2}\right)^2 + bc.$

$STEP\ IV:$

$Combining\ STEPS\ II\ and\ III:$

$ (a + b)\sqrt{ab} + (a + c)\sqrt{ac} + (b + c)\sqrt{bc} \le \dfrac{a^2 + b^2 + c^2 + 3(ab + bc + ac)}{2}.$
I had the general idea right away, but finding the specific route took a while. Thanks for the challenge.

I still think you should put a new challenge into a new thread.
 
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Jul 2015
217
116
Ilford
Thanks, Jeff.

I still think you should put a new challenge into a new thread.
I’m out of challenge problems for now but if I have a new one I’ll start a new thread. (Wink)

Solution to Problem #5:
Expand $\left(\sqrt a-\sqrt b\right)^4+\left(\sqrt b-\sqrt c\right)^4+\left(\sqrt c-\sqrt a\right)^4\geq0$.
 
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Feb 2015
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You can borrow this one:
why is number 18 a good pick for me right now?