Solution of Problems 1-5

Problem #1   $x$ and $y$ are positive integers. If $x+y=7$ and $x>y>0$, find $x^3-y^3+2y(y^2+x^2)+xy(x+3y)$.

Solution

$$\begin{array}{cc}& =x^3-y^3+2y(y^2+x^2)+xy(x+3y)\\ & =x^3-y^3+2y^3+2x^{2}y+x^{2}y+3x{y}^2\\ & =x^3+3x^{2}y+3x{y}^2+y^3\\ & =(x+y{)}^3\\ & =7^3\\ & =343\end{array}$$

Problem #2   Find a positive integer $k$ whose product of digits is equal to $\frac{11k}{4}-199$.

Solution

Let $k=\overline{a_n{a}_{n-1}\cdots a_0}$. From the question we have $$a_n{a}_{n-1}\cdots a_0=\frac{11k}{4}-199.$$ Considering $a_n\cdot 9^n\ge a_n{a}_{n-1}\cdots a_0$ and $k\ge a_n\cdot {10}^n$, we have $$\begin{array}{cc}{a}_n\cdot 9^n& \ge a_n{a}_{n-1}\cdots a_0\\ & =\frac{11k}{4}-199\\ & \ge \frac{11a_n\cdot {10}^n}{4}-199,\end{array}$$

i.e. $$\begin{array}{cc}{a}_n\cdot 9^n& \ge \frac{11a_n\cdot {10}^n}{4}-199.\\ {\left(\frac{9}{10}\right)}^n& \ge \frac{11}{4}-\frac{199}{a_n\cdot {10}^n}\end{array}$$

If $k<10$, then $n=0$. The given question cannot be satisfied because $\frac{11k}{4}-199$ is negative.

If $k\ge 200$, then $a_n\cdot {10}^n\ge 200$. We have $\frac{11}{4}-\frac{199}{a_n\cdot {10}^n}>\frac{7}{4}$, so the inequality cannot be satisfied.

$\therefore 10\le k\le 199$. We have either $k=\overline{a_1{a}_0}$ or $k=\overline{1a_1{a}_0}$. In both cases, the given question becomes $$a_1{a}_0=\frac{11k}{4}-199.$$

Consider $0\le a_1\le 9$ and $0\le a_0\le 8$ ($a_0$ must be an even number), we have

$$\begin{array}{c}0\le \frac{11k}{4}-199\le 72\\ 199\le \frac{11k}{4}\le 271\\ 72\frac{4}{11}\le k\le 98\frac{6}{11}\end{array}$$

From the original equation we know that $k$ is divisible by $4$. So the numbers remaining are $76,80,\cdots ,96$. Testing each number, we obtain the unique solution $84$.

(Note: the solution above was to find ALL solutions instead of just one solution.)

Problem #3   For positive integer $n$, let $f\left(n\right)$ denote the unit digit of $1+2+3+\cdots +n$. Find the value of $f\left(1\right)+f\left(2\right)+\cdots +f\left(2014\right)$. (The unit digit of $456$ is $6$, and of $759$ is $9$, etc.)

Solution The values of the $f\left(1\right)$ to $f\left(20\right)$ are $$1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0.$$ And we have $f(n+20)=f\left(n\right)$ because the same digits are added again.

Now we have $$\begin{array}{cc}& =f\left(1\right)+f\left(2\right)+\cdots +f\left(2014\right)\\ & =100\left[f\right(1)+f(2)+\cdots +f(20\left)\right]\\ & +f\left(1\right)+f\left(2\right)+\cdots +f\left(14\right)\\ & =100\left(70\right)+60\\ & =7060\end{array}$$

Problem #4   In pentagon $ABCDE$, $AB=BC=CD=DE$, $\angle B={96}^{\circ }$ and $\angle C=\angle D={108}^{\circ }$. Find $\angle E$.

Solution Join $BD$ and $CE$, and name their intersection $P$. Join $AP$.

The size of an interior angle of a regular pentagon is $\frac{5-2}{5}\times {180}^{\circ }={108}^{\circ }$.

$\begin{array}{cc}\angle PCD& =\angle PDC=\angle PBC=\angle PED\\ & =\frac{{180}^{\circ }-{108}^{\circ }}{2}& & \left(\angle \text{sum of}△\right)\\ & ={36}^{\circ }\\ \\ \angle BCP& ={108}^{\circ }-\angle PCD\\ & ={108}^{\circ }-{36}^{\circ }\\ & ={72}^{\circ }\\ \\ \angle BPC& =\angle DPE\\ & =\angle PCD+\angle PDC& & \left(\text{ext.}\angle \text{of}△\right)\\ & ={36}^{\circ }+{36}^{\circ }\\ & ={72}^{\circ }\\ & =\angle BCP\\ \\ \therefore BP& =BC& & \left(\text{sides opp. equal}\angle \text{s}\right)\\ & =AB\end{array}$

Similarly, we have $PE=DE$.

$\begin{array}{cc}\angle PBA& ={96}^{\circ }-\angle PBC\\ & ={96}^{\circ }-{36}^{\circ }\\ & ={60}^{\circ }\\ \\ \angle BAP& =\angle BPA\\ & =\frac{{180}^{\circ }-\angle PBA}{2}& & \left(\angle \text{sum of}△\right)\\ & =\frac{{180}^{\circ }-{60}^{\circ }}{2}\\ & ={60}^{\circ }\end{array}$

$\therefore △ABP$ is an equilateral triangle. Hence we have $PA=PE$.

$\begin{array}{cccc}\angle APE& ={180}^{\circ }-\angle BPA-\angle DPE& & \left(\text{adj.}\angle \text{s on st. line}\right)\\ & ={180}^{\circ }-{60}^{\circ }-{72}^{\circ }\\ & ={48}^{\circ }\end{array}$

$\begin{array}{cccc}\angle PEA& =\frac{{180}^{\circ }-\angle APE}{2}& & \left(\angle \text{sum of}△\right)\\ & =\frac{{180}^{\circ }-{48}^{\circ }}{2}\\ & ={66}^{\circ }\\ \\ \angle AED& =\angle PED+\angle PEA\\ & ={36}^{\circ }+{66}^{\circ }\\ & ={102}^{\circ }\end{array}$

Problem #5   Let $p$ and $q$ be positive integers such that $\frac{72}{487}<\frac{p}{q}<\frac{18}{121}$. Find the smallest possible value of $p$.

Solution

The required fraction $\frac{p}{q}$ can be found by continued fraction. A continued fraction is defined as:

$$[a_0;a_1,a_2,a_3,\dots ]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots }}}$$

We have $\frac{18}{121}=[0;6,1,2,1,1,2]$ and $\frac{72}{487}=[0;6,1,3,4,4]$. Therefore

$$\frac{p}{q}=[0;6,1,3]=0+\frac{1}{6+\frac{1}{1+\frac{1}{3}}}=\frac{4}{27}$$

And the smallest possible value of $p$ is $4$.

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