Length of Side of Square

If the numerical value of the area of a square plus two times the numerical value of its perimeter is equal to $20$, what is the length of one side of the square?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $a$ represent the length of one side of a square. The following equation is true
$a^2+2(4a)=20$
$a^2+8a-20=0$
$(a-2)(a+10)=0$
$a=2$ or $a=-10$ (extraneous solution)
The length of one side of the square equals $2$ units.

Answer: $2$ units

John has a lemonade stand. He sells a small lemonade for $50$ cents and a large lemonade for $\1$. A small serving contains $1$ cup of lemonade; a large contains $1.5$ cups. At the end of the day, John has made $\9$ and sold $15.5$ cups of lemonade. How many small and large lemonades has he sold?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $x$ represent the number of small lemonades and $y$ the number of large lemonades sold. We have the following system of equations
$x(.50)+y(1)=9\qquad\qquad\; (1)$
$x(1)+y(1.5)=15.5\qquad\quad\, (2)$
Multiply Eqs. $(1)$ and $(2)$ by $2$
$x+2y=18\qquad\quad\:\: (3)$
$2x+3y=31\qquad\quad (4)$
Multiply Eq. $(3)$ by $-2$ and add to Eq. $(4)$
$-2x-4y=-36$
$2x+3y=31$
————————-
$-y=-5$
$y=5$
Substitute the value of $y=5$ into Eq. $(3)$
$x+2(5)=18$
$x=8$
John has sold $8$ small and $5$ large lemonades.

Answer: $8$ small and $5$ large

Red and Yellow Plums

One store sold red plums at $4$ for $\1$ and yellow plums at $3$ for $\1$. A second store sold red plums at $4$ for $\1$ and yellow plums at $6$ for $\1$. You bought $m$ red plums and $n$ yellow plums from each store, spending a total of $\10$. How many plums in all did you buy?
Source: NCTM Mathematics Teacher, September 2006

Solution
In dollars the following equation depicts how you spent $\10$ at the two stores
$m(1/4)+n(1/3)+m(1/4)+n(1/6)=10$
Collect like terms and simplify
$m(1/2)+n(1/2)=10$
$m+n=20$
At each store you purchase $20$ plums; in all you bought $40$ plums.

Answer: $40$

Value of Sum

What is the value of the following sum?
$2006[(-1)^1+(-1)^2+(-1)^3+(-1)^4+\cdots+(-1)^{2006}]$
Source: NCTM Mathematics Teacher, September 2006

Solution
The square bracket in the sum expression contains 1003 pairs of consecutive powers $(-1)^n+(-1)^{n+1}$ which pairwise sum to zero
$(-1)^1+(-1)^2=-1+1=0$
$(-1)^3+(1-)^4=-1+1=0$
$(-1)^5+(-1)^6=-1+1=0$
$\cdots$
$(-1)^{2005}+(-1)^{2006}=-1+1=0$
$2006[(-1)^1+(-1)^2+(-1)^3+\cdots+(-1)^{2006}]=2006[0+0+0+\cdots+0]=0$

Answer: $0$

Biggest Pizza Slice

An 8-inch pizza is cut into $3$ equal slices. A 10-inch pizza is cut into $4$ equal slices. A 12-inch pizza is cut into $6$ equal slices. A 14-inch pizza is cut into $8$ equal slices. From which pizza would you take a slice if you want as much pizza as possible?
Source: NCTM Mathematics Teacher, September 2006

Solution
Areas of pizza slices
8-inch: $\dfrac{\pi\times 4^2}{3}=\dfrac{\pi\times 16}{3}=\dfrac{\pi\times 128}{24}$
10-inch: $\dfrac{\pi\times 5^2}{4}=\dfrac{\pi\times 25}{4}=\dfrac{\pi\times 150}{24}$
12-inch: $\dfrac{\pi\times 6^2}{6}=\dfrac{\pi\times 6^2}{6}=\dfrac{\pi\times 144}{24}$
14-inch: $\dfrac{\pi\times 7^2}{8}=\dfrac{\pi\times 49}{8}=\dfrac{\pi\times 147}{24}$
Since $\dfrac{150}{24}$ is the largest fraction, you want to take a slice from the 10-inch pizza.

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: $210$