## Bowling Pins

After rolling the first ball of a frame in a game of 10-pin bowling, how many different pin configurations can remain (assuming all configurations are physically possible)?
Source: NCTM Mathematics Teacher, September 2006

Solution
A pin configuration is made up of ten pins each of which can be up or down. The number of possible configurations equals $2^{10}=1024$.

Answer: $1024$

Alternative solution
Let $n$ represent the number of pins left standing after a roll. In how many ways can they be left standing?
$\binom{10}{0}=1$ way to leave zero pin (bowl a strike)
$\binom{10}{1}=10$ ways to leave a single pin
$\binom{10}{2}=45$ ways to leave two pins
$\cdots$
$\binom{10}{10}=1$ way to leave all ten pins (gutter ball)
Total number of possible ways
$\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}=1+10+45+120+210+252+210+120+45+10+1=1024$

## Spring Garden

Each spring, a $12$ meter by $12$ meter garden has its length increased by $2$ meters but its with decreased by $50$ centimeters. What will be the maximum attainable area of the garden?
Source: NCTM Mathematics Teacher, October 2006

Solution
Area in square meters per each year starting from year $0$
$0\!:\!(12)(12)$
$1\!:\!(12+2)(12-.5)=(12+1(2)(12-1(.5)$
$2\!:\!(12+2+2)(12-.5-.5)=(12+2(2))(12-2(.5)$
$3\!:\!(12+2+2+2)(12-.5-.5-.5)=(12+3(2))(12-3(.5))$
$\cdots$
In general in year $x$, the area equals
$(12+x(2))(12-x(.5))=-x^2+18x+144$
The graph of this quadratic expresion is shown below

The area reaches a maximum of $225$ square meters in year $x=9$.

Answer: $225$ square meters

## Divided by 5

Find the reminder when $3^{98}$ is divided by $5$?
Source: NCTM Mathematics Teacher, October 2006

Solution
$3^{98}$ is such a large number that it is impractical to use modulo $5$ arithmetic to find the remainder. So we are going to find the remainders of a few powers of $3$ and hope to see a pattern emerge.
Remainders when $3^x$ are divided by $5$
$3^0\qquad\qquad 1$
$3^1\qquad\qquad 3$
$3^2\qquad\qquad 4$
$3^3\qquad\qquad 2$
$3^4\qquad\qquad 1$
$3^5\qquad\qquad 3$
$3^6\qquad\qquad 4$
$3^7\qquad\qquad 2$
$3^8\qquad\qquad 1$
$3^9\qquad\qquad 3$
$3^{10}\qquad\quad\:\:\, 4$
$3^{11}\qquad\quad\:\:\, 2$
$\cdots$
The pattern of remainders is $1,3,4,2,1,3,4,2,\cdots$. The remainder equals $1$ when the even exponent is the product of $2$ and an even number, for example, the remainder of $3^8$ equals $1$ because $8=2\times 4$. The remainder equals $4$ when the even exponent is the product of $2$ and an odd number, for example, the remainder of $3^{10}$ equals $4$ because $10=2\times 5$.
Since $98=2\times 49$, the remainder of $3^{98}$ divided by $5$ equals $4$.

Answer: $4$

Alternative solution
We can group the powers $3^x$ according to their remainders as follows:
$1\!: 3^0,3^4,3^8,3^{12},\cdots$
$3\!: 3^1,3^5,3^9,3^{13},\cdots$
$4:\! 3^2,3^6,3^{10},3^{14},\cdots$
$2:\! 3^3,3^7,3^{11},3^{15},\cdots$
This is exactly what modulo $4$ arithmetic does, divide the whole numbers $0,1,2,3,4,\cdots$ into four groups
$0,4,8,12,\cdots$
$1,5,9,13,\cdots$
$2,6,10,14,\cdots$
$3,7,11,15,\cdots$
Since $98\equiv 2\bmod 4, 3^{98}$ divided by $5$ will have the same remainder as $3^2$ divided by $5$.

Each day Chris is chided for not cleaning up his room, so he picks up approximately $10$ percent of the items on the floor in the morning. (He always rounds off to the nearest whole number if his calculations result in fraction.) Each day, ten new items somehow end up on the floor. If he has a clean floor on Sunday morning, how many items will be on the floor on Saturday night?
Source: NCTM Mathematics Teacher, October 2006

Solution
Sunday morning starts with a clean floor but in the night $10$ items appear.
Monday he cleans up $10\%$ but ends up with $10-(.1)10+10=19$.
Tuesday $19-(.1)19+10=19-2+10=27$
Wednesday $27-(.1)27+10=27-3+10=34$
Thursday $34-(.1)34+10=34-3+10=41$
Friday $41-(.1)41+10=41-4+10=47$
Saturday $47-(.1)47+10=47-5+10=52$

Answer: $52$

## Divisible by 7

How many four digit positive integers divisible by $7$ have the property that, when the first and last digits are interchanged, the result is a (not necessarily four-digit) positive integer divisible by $7$?
Source: NCTM Mathematics Teacher, October 2006

Solution
Is $8673$ divisible by $7$? One way to find out is by determining if $8673$ “behaves” like $7$ or $14$ or some other small multiples of $7$. Modulo $7$ arithmetic will help us do that. In its most basic definition, modulo arithmetic is the arithmetic of remainder. We say “$8$ is congruent to $15$ modulo $7$” and write $8\equiv 15\bmod 7$, because $8$ and $15$ yield the same remainder when divided by $7$.
Using $8673=8(1000)+6(100)+7(10)+3(1)$, we will do the same calculation but with congruent numbers mod $7$. For ease in presentation we drop the modulo $7$ notation.
Since $1000\equiv 6,100\equiv 2,10\equiv 3,8\equiv 1,7\equiv 0$,
$8(1000)+6(100)+7(10)+3(1)\equiv 1(6)+6(2)+0(3)+3(1)$
$\equiv 6+12+0+3\equiv 21$
$8673$ is divisible by $7$.

Let $axy\,b$ be a 4-digit positive integer where the digits are $a,x,y$, and $b$. Suppose $7$ divides $axy\,b$ and $bxya$. Show that $7$ divides $(a-b)$.
Since $7$ divides $axy\,b$ and $bxya$, there exist integers $k_1$ and $k_2$ such that
$a(1000)+x(100)+y(10)+b=7k_1\qquad\qquad (1)$
$b(1000)+x(100)+y(10)+a=7k_2\qquad\qquad (2)$
Subtract Eq. $(2)$ from Eq. $(1)$
$(a-b)1000+(b-a)=7k_3$ where $k_3=k_1-k_2$
$1000a-1000b+b-a=7k_3$
$999(a-b)=7k_3$
In other words, $7$ divides $999(a-b)$. Since $7$ does not divide $999$, $7$ must divide $(a-b)$.

Show that if $7$ divides $axy\,b$, then $7$ divides $10x+y$.
Given $a(1000)+x(100)+y(10)+b=7k_4$ for some integer $k_4$,
$1000a+100x+10y+b=7k_4$
$1001a-a+100x+10y+b=7k_4$
$1001a-(a-b)+100x+10y=7k_4$
Since $7$ divides $1001a$ and $(a-b)$, $7$ must divide $100x+10y$. Hence, $7$ divides $10x+y$.

We have two cases to consider: $a=b$ and $a\neq b$.
Case 1: $a=b$
The integers are of the form $axy\,a$ where the $xy$ numbers are divisible by $7$. The $15$ possible $xy$ numbers are: $00,07,14,21,28,35,42,49,56,63,70,77,84,91,98$. The $9$ possible $a$ digits are: $1,2,3,4,5,6,7,8,9$. There are $9\times 15=135$ possible $axy\,a$.
Cases 2: $a\neq b$
The integers are of the form $axy\,b$ where digits $a$ and $b$ are congruent. Interchanging congruent digits like $0$ and $7$, $1$ and $8$, $2$ and $9$ has no bearing on the remainder. The $5$ possible $axy\,b$ are: $1xy8,8xy1,2xy9,9xy2,7xy0$. There are $5\times 15=75$ possible $axy\,b$.
Total number of integers equals $135+75=210$.
List of some of the integers
$1001,1008,1071,1078,1141,1148,1211,1218,1281,1288,1351,1358,1421,1428,1491,1498$
$\cdots$
$7000,7007,7070,7077,7140,7147,7210,7217,7280,7287,7350,7357,7420,7427,7490,7497$
$\cdots$
$9002,9009,9072,9079,9142,9149,9212,9219,9282,9289,9352,9359,9422,9429,9492,9499$

Answer: $210$

## Tangent Lines

Suppose to co-planar circles $C_1$ and $C_2$ have no points in common. Determine how many lines, if any, are tangent to both $C_1$ and $C_2$?
Source: NCTM Mathematics Teacher, October 2006

Solution

When one circle is inside the other, there are no tangent lines to both circles. When the two circles are side by side, there are exactly four tangent lines to both circles.

Answer: $0$ or exactly $4$

## Wins and Losses

A person starting with $\256$ makes eight bets and wins exactly four times. The wins and losses occur in random order. If each wager is for half the money she has at the time of the bet, how much money will she win or lose?
Source: NCTM Mathematics Teacher, October 2006

Solution
Suppose the eight bets result in the following arbitrary sequence of wins and losses
$1\quad 256/2=128\quad\:\:\textrm{win}\quad 256+128=384$
$2\quad 384/2=192\quad\:\:\textrm{loss}\quad 384-192=192$
$3\quad 192/2=96\quad\:\:\:\:\textrm{win}\quad 192+96=288$
$4\quad 288/2=144\quad\:\:\textrm{loss}\quad 288-144=144$
$5\quad 144/2=72\quad\:\:\:\:\textrm{win}\quad 144+72=216$
$6\quad 216/2=108\quad\:\:\textrm{loss}\quad 216-108=108$
$7\quad 108/2=54\quad\:\:\:\:\textrm{loss}\quad 108-54=54$
$8\quad 54/2=27\quad\:\:\:\:\:\:\textrm{win}\quad 54+27=81$
The gambler loses $256-81=\175$.
What if we loaded all the four wins consecutively in the beginning so that she wins big and loses little money?
$1\quad 256/2=128\quad\:\:\textrm{win}\quad 256+128=384$
$2\quad 384/2=192\quad\:\:\textrm{win}\quad 384+192=576$
$3\quad 576/2=288\quad\:\:\textrm{win}\quad 576+288=864$
$4\quad 864/2=432\quad\:\:\textrm{win}\quad 864+432=1296$
$5\quad 1296/2=648\quad\textrm{loss}\quad 1296-648=648$
$6\quad 648/2=324\quad\:\:\textrm{loss}\quad 648-324=324$
$7\quad 324/2=162\quad\:\:\textrm{loss}\quad 324-162=162$
$8\quad 162/2=81\quad\:\:\:\:\textrm{loss}\quad 162-81=81$
Alas! The wins are big and so are the losses. She still loses $\175$, but why lose the same amount in the two different scenarios?
Let $x$ = money available before a bet. After a win she will have more money, $x+x/2=x(3/2)$, that is $3/2$ as much as before; after a loss she will have less money $x-x/2=x(1/2)$, that is $1/2$ as much. We track the ups and downs of money using this multiplicative rule as follows
$1\quad x/2\qquad\quad\:\:\,\textrm{win}\:\:\:\:\: x(3/2)$
$2\quad x(3/4)\quad\:\:\:\:\:\,\textrm{loss}\quad\, x(3/2)(1/2)$
$3\quad x(3/8)\quad\:\:\:\:\:\,\textrm{win}\quad\, x(3/2)^2(1/2)$
$4\quad x(9/16)\quad\:\:\:\:\textrm{loss}\quad x(3/2)^2(1/2)^2$
$5\quad x(9/32)\quad\:\:\:\:\textrm{win}\quad x(3/2)^3(1/2)^2$
$6\quad x(27/64)\quad\:\:\textrm{loss}\quad x(3/2)^3(1/2)^3$
$7\quad x(27/128)\quad\textrm{loss}\quad x(3/2)^3(1/2)^4$
$8\quad x(27/256)\quad\textrm{win}\quad x(3/2)^4(1/2)^4$
She starts out with $x=256$ and ends up with
$x(3/2)^4(1/2)^4=x(81/256)$
$=256(81/256)$
$=\81$
Though the wins and losses occur in random order, she always loses $256-81=\175$.

Answer: Will lose $\175$

## Nested Square Roots 2

Find the value of $\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}$
Source: NCTM Mathematics Teacher, October 2006

Solution
Let $x=\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}$
By substitution
$x=\sqrt{16+x}$
Squaring both sides
$x^2=16+x$
$x^2-x-16=0$
Using the quadratic formula to solve for $x$
$x=\dfrac{1\pm\sqrt{65}}{2}$
$\dfrac{1-\sqrt{65}}{2}$ is an extraneous solution because the expression cannot be negative.
$\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}=\dfrac{1+\sqrt{65}}{2}$

Answer: $\dfrac{1+\sqrt{65}}{2}$

## Digital Clock

The display on a digital clock reads $6\!:\!38$. What will the clock display twenty-seven digit changes later?
Source: NCTM Mathematics Teacher, October 2006

Solution

The figure shows there are four digit changes from $6\!:\!38$ to $6\!:\!42$ corresponding to four minute intervals. Thus, $27$ changes later $27$ minutes will have passed. The clock will display $6\!:\!38+0\!:\!27=7\!:\!05$.

Answer: $7\!:\!05$

## Waiting in Line

In a movie theater line, $x$ people are behind Mark, who is $y$ places in front of Sam. If there are $z$ people in front of Sam, how many people are in line?
Source: NCTM Mathematics Teacher, October 2006

Solution

The figure illustrates the specifics of the line. The dots represent the people waiting in line starting from the left and Mark and Sam are randomly placed in the line.
$x$ = number of people behind Mark
$y$ = Mark is $y$ places in front of Sam
$z$ = number of people in front of Sam
Let $u$ represent Sam and the number of people behind him.
Number of people in line = $z+u$
$u=x-y+1$
$z+u=z+x-y+1$

Answer: $z+x-y+1$

Alternative solution
$z$ = number of people in front of Sam excluding Sam
$x-y$ = number of people behind Sam excluding Sam
Total number of people waiting in line = $z+x-y+1$.