## Difference of Sequences

For the sequence $1,-2,3,-4,5,-6,7,\cdots$ , what is the difference between the mean of the sequence’s first $400$ terms and the mean of its first $200$ terms?
Source: NCTM Mathematics Teacher February 2008

Solution
If we add the first $400$ terms by pairs $1+(-2),3+(-4),5+(-6),\cdots$ we end up with $200(-1)=-200$. The mean of the first $400$ terms is $-200/400=-1/2$.
Likewise, if we add the first $200$ terms by pairs, we get $100(-1)=-100$. The mean of the first $200$ terms is $-100/200=-1/2$.
The difference between the two means is $-1/2-(-1/2)=0$.

Answer: $0$

## Sums of Reciprocals

The sum of the positive divisors of $480$ is $1512$. Find the sum of the reciprocals of the positive divisors of $480$.
Source: NCTM Mathematics Teacher, February 2008

Solution
Given that $480=2^5\cdot 3^1\cdot 5^1$, the number of divisors is $(5+1)(1+1)(1+1)=24$. The first twelve divisors are small and easy to guess: $1,2,3,4,5,6,8,10,12,15,16$, and $20$ and as a bonus we get the last twelve by dividing $480$ by the first twelve divisors. For example, $480/1=480,480/2=240,480/3=160$, etc. : $480,240,160,120,96,80,60,48,40,32,30,24$.
When we add the reciprocals of the divisors, the divisors appear as denominators in a set of 24 fractions
$\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{240}+\dfrac{1}{480}$
When we reduce the fractions to $480$ (the least common denominator), the $24$ divisors appear as numerators
$\dfrac{480}{480}+\dfrac{240}{480}+\cdots+\dfrac{2}{480}+\dfrac{1}{480}=\dfrac{480+240+\cdots+2+1}{480}=\dfrac{1512}{480}$

Answer: $1512/480$

## Divisible by Units Digit

The number $64$ is divisible by its units digit $(4)$. How many whole numbers less than $64$ are divisible by their respective units digit?
Source: NCTM Mathematics Teacher, February 2008

Solution
Nine numbers: $1,2,3,4,5,6,7,8,9$ divisible by themselves
Six numbers: $11,21,31,41,51,61$ divisible by $1$
Six numbers: $12,22,32,42,52,62$ divisible by $2$
Five numbers: $15,25,35,45,55$ divisible by $5$
Six numbers: $33,63,24,44,36,48$ divisible by their respective units digit
Total $9+6+6+5+6=32$

Answer: $32$

## Sum of 7’s

In the sum of the expression $7+77+777+7777+\cdots+7,\!777,\!777,\!777,\!777,\!777,\!777$, what digit will be in the tens place?
Source: NCTM Mathematics Teacher, February 2008

Solution
We are adding nineteen numbers from the smallest $7$ to the largest $7,\!777,\!777,\!777,\!777,\!777,\!777$. Note that there are nineteen $7\mathrm{'s}$ in the ones place and eighteen $7\mathrm{'s}$ in the tens place. First, we add the nineteen $7\mathrm{'s}$ in the ones place and get $19\times 7=133$. Consider the number $133$. The $3$ in the ones place of $133$ will be the ones place digit of the final sum. The $3$ in the tens place of $133$ means $30$ and if we regroup $30$ with the eighteen $70\mathrm{'s}$, we get $30+18\times 70=1290$. The digit in the tens place of the final sum is $9$.

Answer: $9$

## Numerical Puzzle

The five-square numerical puzzle below requires the vertical column to be filled with the digits from a three-digit integral power of $5$ and the horizontal row to be filled with the digits from a three-digit integral power of $2$. What digit will be in the shaded square?

Source: NCTM Mathematics Teacher, February 2008

Solution
$5^2=25\quad 5^3=125\quad 5^4=625\quad 5^5=3125$
$2^6=64\quad 2^7=128\quad 2^8=256\quad 2^9=512\quad 2^{10}=1024$

Answer: $6$

## 4-Digit Numbers Divisible by 7

Given the digits $1,3,6$, and $9$, find the probability that a four-digit number formed by using each of them only once is divisible by $7$.
Source: NCTM Mathematics Teacher, February 2008

Solution
There are $4\times 3\times 2\times 1=24$ possible four-digit numbers using $1,3,6,9$ each only once
$1369,1396,1639,1693,1936,1963$
$3169,3196,\underline{3619},3691,3916,3961$
$\underline{6139},6193,6319,\underline{6391},6913,6931$
$9136,\underline{9163},9316,9361,9613,9631$
Numbers divisible by $7$ are $3619,6139,6391,9163$
Probability = $4/24=1/6$.

Answer: $1/6$

## Two-mile Long Train

How long will it take a two-mile long train traveling $12$ miles per hour to travel completely through a mile-long tunnel?
Source: NCTM Mathematics Teacher, February 2008

Solution

The diagram shows that for the train to clear the tunnel, point $C$ must travel three mile-long segments each of which takes $1/12$ hour or $5$ minutes to finish. The train will clear the tunnel in $3\times 5=15$ minutes.

Answer: $15$ minutes

Alternative solution
When the head of the train enters the tunnel, the end of the train is $3$ miles away from the exit. The rear of the train will have to travel $3$ miles at $12$ miles per hour.

## Triangular Array of Numbers

If you continue the triangular array of numbers shown in the figure, what number would be directly below $122$?
$|\qquad\qquad\qquad\:\: 1$
$|\quad\quad\quad\quad\:\: 2\quad\:\: 3\quad\:\: 4$
$|\quad\quad\:\: 5\quad\:\:\: 6\quad\:\: 7\quad\:\: 8\quad\:\: 9$
$|\:10\quad 11\quad12\quad 13\quad 14\quad 15\quad 16$
Source: NCTM Mathematics Teacher, February 2008

Solution
$|\qquad\qquad\qquad\qquad\:\: 1$
$|\qquad\quad\quad\quad\quad\:\: 2\quad\:\: 3\quad\:\: 4$
$|\qquad\quad\quad\: 5\quad\:\:\: 6\quad\:\: 7\quad\:\: 8\quad\:\: 9$
$|\qquad\: 10\quad 11\quad12\quad 13\quad 14\quad 15\quad 16$
$|\:17\quad 18\quad 19\quad 20\quad 21\quad 22\quad 23\quad 24\quad 25$
$|\:\cdots$
Note that the leading numbers increase by $1,3,5,7$, etc. If we follow this pattern, the leading numbers of the first $13$ rows are $1,2,5,10,17,26,37,50,65,82,101,122,145$, which means that the number directly below $122$ is $146$.

Answer: $146$

Alternative solution 1
Note that the ending number of each row is a perfect square $1,4,9,16,25$, etc. If we follow this pattern, $121=11^2$ is the ending number of row $11$ and $144=12^2$ is the ending number of row $12$.

Alternative solution 2
Let $x=0,1,2,3,\cdots$ represent the row number. We want to find a function $f(x)$ that relates row $x$ to the leading number of that row. We already know a few values of $f(x)$
$x\quad f(x)$
$0\quad 1$
$1\quad 2$
$2\quad 5$
$3\quad 10$
$4\quad 17$
Is $f(x)$ a linear or quadratic function or neither? The first differences (FD) and second differences (SD) in the values of $f(x)$ will tell us what it is
$x\quad f(x)\:\mathrm{FD}\:\mathrm{SD}$
$0\quad\:\: 1$
$1\quad\:\: 2\quad\: 1$
$2\quad\:\: 5\quad\: 3\quad\: 2$
$3\quad 10\quad\: 5\quad\: 2$
$4\quad 17\quad\: 7\quad\: 2$
$f(x)$ is a quadratic function because the second differences are a constant $2$
$f(x)=ax^2+bx+c$
$a=\mathrm{SD}/2=2/2=1$
$c=f(0)=1$
$b=0$ because the term $bx$ does not contribute anything to the function
$f(x)=x^2+1$
$f(11)=11^2+1=122$
$f(12)=12^2+1=145$

## Smaller Factor

Given a three-digit number, interchange the units digit and the hundreds digit. The product of the original number and the new number is $65125$. What is the smaller three-digit factor?
Source: NCTM Mathematics Teacher, February 2008

Solution
Since the product of the original number and the new number $65125$ is odd, the two numbers must be odd. Furthermore, their units digits must multiply to $5$. These restrictions limit the possibilities to $1\!-\!5,3\!-\!5,5\!-\!7$, and $5\!-\!9$. Let’s examine a few numbers to see if we can narrow down the choices starting with $5\!-\!9$
$519\times 915=474885$ – not possible, too big a product
$517\times 715=369655$ – not possible
$315\times 513=161595$ – not possible
$115\times 511=58765$
$125\times 521=65125$
The smaller three-digit factor is $125$.

Answer: $125$

Alternative solution
The prime factorization of $65125=5^3\times 521$.
$2\times 521=1042$
$3\times 521=1563$
$\cdots$
$521$ is a prime number and any multiple of $521$ is a $4$-digit number, thus the two factors must be $5^3=125$ and $521$.

## Non-Congruent Triangles

Among the triangles of the previous problem (repeated below), how many distinct (non-congruent) triangles exist? Previous Problem: How many distinct triangles can be constructed by choosing three vertices from among the corners of a unit cube?
Source: NCTM Mathematics Teacher, February 2008

Solution

$24$ triangles on the faces of the cube have side lengths of $1,1,\sqrt 2$.

$24$ triangles formed by two diagonals and one side have side lengths of $1,\sqrt 2,\sqrt 3$.

$8$ triangles that are bases of tetrahedrons formed by connecting three corners of the cube have side lengths of $\sqrt 2,\sqrt 2,\sqrt 2$.

There are $3$ distinct (non-congruent) triangles.

Answer: $3$