Properties of Numbers
Solutions

1. Suppose that a palindrome $p$ is the sum of the three consecutive integers $a-1$, $a$, $a+1$.

   In this case, $p=(a-1)+a+(a+1)=3a$, so $p$ is a multiple of $3$.

   The largest palindromes less than $200$ are $191$, $181$, $171$.

   Note that $191$ and $181$ are not divisible by $3$, but $171$ is divisible by $3$. We can easily check this using the divisibility by $3$ test. For each of these integers, the sum of their digits is $11$, $10$ and $9$, respectively. Only $9$ is divisible by $3$ and so $171$ is the only one that is divisible by $3$.

   The integer $171$ can be written as $56+57+58$, so $171$ is the largest palindrome less than $200$ that is the sum of three consecutive integers.

2. Suppose that $n$ has digits $AB$. Then $n=10A+B$. The average of the digits of $n$ is ${\displaystyle \frac{A+B}{2}}$. Putting a decimal point between the digits of $n$ is equivalent to dividing $n$ by $10$, so the resulting number is ${\displaystyle \frac{10A+B}{10}}$.

   So we want to determine $A$ and $B$ so that $$\begin{array}{rl}{\displaystyle \frac{10A+B}{10}}& ={\displaystyle \frac{A+B}{2}}\\ 10A+B& =5(A+B)\\ 5A& =4B\end{array}$$ Since $A$ and $B$ are digits such that $5A=4B$, we have $A=4$ and $B=5$ is the only possibility. Therefore, $n=45$. (We can quickly check that the average of the digits of $n$ is $4.5$, the number obtained by putting a decimal point between the digits of $n$.)

3. Solution 1

   The integer equal to ${10}^{20}$ consists of the digit $1$ followed by $20$ $0$s. The integer equal to ${10}^{20}-1$ thus consists of $20$ $9$s.

   Now, $n={10}^{20}-20$ is $19$ less than ${10}^{20}-1$ which is the integer that consists of $20$ $9$s.

   So $n={10}^{20}-20=99\cdots 980$, where this integer has $18$ $9$s.

   Therefore, the sum of the digits of $n$ is $18(9)+8+0=162+8=170$.

   Solution 2

   Since ${10}^{20}-20=10({10}^{19}-2)$ and ${10}^{19}-2=99\cdots 98$ (where this integer has 18 9s), we have ${10}^{20}-20=99\cdots 980$, where this integer has $18$ $9$s.

   Therefore, the sum of the digits of $n$ is $18(9)+8+0=162+8=170$.

4. The parity of an integer is whether it is even or odd.

   Since the Fibonacci sequence begins $1,1,2,3,5,8,13,21,\dots$, it follows that the parities of the first eight terms are Odd, Odd, Even, Odd, Odd, Even, Odd, Odd.

   In the sequence, if $x$ and $y$ are consecutive terms, then the next term is $x+y$.

   In general, suppose that $x$ and $y$ are integers. If $x$ is even and $y$ is even, then $x+y$ is even. If $x$ is even and $y$ is odd, then $x+y$ is odd. If $x$ is odd and $y$ is even, then $x+y$ is odd. If $x$ is odd and $y$ is odd, then $x+y$ is even. Therefore, the parities of two consecutive terms $x$ and $y$ in the Fibonacci sequence determine the parity of the following term $x+y$.

   Also, once there are two consecutive terms whose parities match the parities of two earlier consecutive terms in the sequence, then the parities will repeat in a cycle. In particular, the parities of the fourth and fifth terms (Odd, Odd) are the same as the parities of the first and second terms (Odd, Odd). Therefore, the parities in the sequence repeat the cycle Odd, Odd, Even. This cycle has length $3$. Therefore, the 99th term in the Fibonacci sequence ends one of these cycles, since $99$ is a multiple of $3$. In particular, the 99th term ends the 33rd cycle.

   Each cycle contains two odd terms. Therefore, the first $99$ terms in the sequence include $2\times 33=66$ odd terms.

   Finally, the 100th term in the sequence begins a new cycle, so it is odd. Therefore, the first $100$ terms include $66+1=67$ odd terms.

5. Since $900={30}^2$ and $30=2\times 3\times 5$, we have $900=2^2{3}^2{5}^2$.

   The positive divisors of $900$ are those integers of the form $d=2^a{3}^b{5}^c$, where each of $a,b,c$ is $0$, $1$ or $2$.

   For $d$ to be a perfect square, the exponent on each prime factor in the prime factorization of $d$ must be even. Thus, for $d$ to be a perfect square, each of $a,b,c$ must be $0$ or $2$.

   There are two possibilities for each of $a,b,c$ so $2\times 2\times 2=8$ possibilities for $d$.

   These are $2^0{3}^0{5}^0=1$, $2^2{3}^0{5}^0=4$, $2^0{3}^2{5}^0=9$, $2^0{3}^0{5}^2=25$, $2^2{3}^2{5}^0=36$, $2^2{3}^0{5}^2=100$, $2^0{3}^2{5}^2=225$, and $2^2{3}^2{5}^2=900$.

   Thus, $8$ of the positive divisors of $900$ are perfect squares.

6. Since each list contains $6$ consecutive positive integers and the smallest integers in the lists are $a$ and $b$, it follows that the positive integers in the first list are $a,a+1,a+2,a+3,a+4,a+5$ and the positive integers in the second list are $b,b+1,b+2,b+3,b+4,b+5$. Note that $1\le a<b$.

   We first determine the pairs $(a,b)$ for which $49$ will appear in the third list, then determine which of these pairs give a third list that contains no multiple of $64$, and then finally keep only those pairs for which there is a number in the third list larger than $75$.

   The first bullet tells us that $49$ is the product of an integer in the first list and an integer in the second list.

   Since $49=7^2$ and $7$ is prime, it follows that these integers are either $1$ and $49$ or $7$ and $7$.

   If $1$ is in one of the lists, then either $a=1$ or $b=1$. Since $1\le a<b$, it must be that $a=1$.

   If $49$ is in the second list, then one of $b,b+1,b+2,b+3,b+4,b+5$ equals $49$, and so $44\le b\le 49$.

   Therefore, for $1$ and $49$ to appear in the two lists, then $(a,b)$ must be one of $$(1,49),(1,48),(1,47),(1,46),(1,45),(1,44).$$ If $7$ appears in the first list, then one of $a,a+1,a+2,a+3,a+4,a+5$ equals $7$, so $2\le a\le 7$. Similarly, if $7$ appears in the second list, then $2\le b\le 7$.

   Therefore, for $7$ to appear in both lists, then, knowing that $a<b$, then $(a,b)$ must be one of

   $(2,3)$, $(2,4)$, $(2,5)$, $(2,6)$, $(2,7)$, $(3,4)$, $(3,5)$, $(3,6)$, $(3,7)$, $(4,5)$, $(4,6)$, $(4,7)$, $(5,6)$, $(5,7)$, $(6,7)$.
   The second bullet tells us that no pair of numbers in the first and second lists have a product that is a multiple of $64$.

   Given that the possible values of $a$ and $b$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $44$, $45$, $46$, $47$, $48$, $49$, then the possible integers in the two lists are those integers from $1$ to $12$, inclusive, and from $44$ to $54$, inclusive. (For example, if the first number in one list is $7$, then the remaining numbers in this list are $8$, $9$, $10$, $11$, $12$.)

   There is no multiple of $32$ or $64$ in these lists.

   Thus, for a pair of integers from these lists to have a product that is a multiple of $64$, one is a multiple of $4$ and the other is a multiple of $16$, or both are multiples of $8$.

   If $(a,b)=(1,48),(1,47),(1,46),(1,45),(1,44)$, then $4$ appears in the first list and $48$ appears in the second list; these have a product of $192$, which is $3\cdot 64$.

   If $(a,b)=(1,49)$, there is a multiple of $4$ but not of $8$ in the first list, and a multiple of $4$ but not of $8$ in the second list, so there is no multiple of $64$ in the third list.

   If $(a,b)=(3,4),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7),(5,6),(5,7),(6,7)$, then $8$ appears in both lists, so $64$ appears in the third list.

   If $(a,b)=(2,3),(2,4),(2,5),(2,6),(2,7)$, then there is no multiple of $8$ or $16$ in the first list and no multiple of $16$ in the second list, so there is no multiple of $64$ in the third list.

   Therefore, after considering the first two bullets, the possible pairs $(a,b)$ are $(1,49),(2,3),(2,4),(2,5),(2,6),(2,7)$.

   The third bullet tells us that there is at least one number in the third list that is larger than $75$.

   Given the possible pairs $(a,b)$ are $(1,49),(2,3),(2,4),(2,5),(2,6),(2,7)$, the corresponding pairs of largest integers in the lists are $(6,54),(7,8),(7,9),(7,10),(7,11),(7,12)$.

   The corresponding largest integers in the third list are the products of the largest integers in the two lists; these products are $324$, $56$, $63$, $70$, $77$, $84$, respectively.

   Therefore, the remaining pairs $(a,b)$ are $(1,49),(2,6),(2,7)$.

   Having considered the three conditions, the possible pairs $(a,b)$ are $(1,49),(2,6),(2,7)$.

7. First, we determine the perfect squares between $1300$ and $1400$ and between $1400$ and $1500$. Since $\sqrt{1300}\approx 36.06$, the first perfect square larger than $1300$ is ${37}^2=1369$. The next perfect squares are ${38}^2=1444$ and ${39}^2=1521$.

   Since Charles was born between 1300 and 1400 in a year that was a perfect square, Charles must have been born in 1369.

   Since Louis was born between 1400 and 1500 in a year that was a perfect square, Louis must have been born in 1444.

   Suppose that on April 7 in some year, Charles was $m^2$ years old and Louis was $n^2$ years old for some positive integers $m$ and $n$. Thus, Charles was $m^2$ years old in the year $1369+m^2$ and Louis was $n^2$ years old in the year $1444+n^2$.

   Since these expressions represent the same years, we have that $1369+m^2=1444+n^2$, or $m^2-n^2=1444-1369=75$. In other words, we want to find two perfect squares less than $110$ (since their ages are less than $110$) whose difference is $75$.

   The perfect squares less than $110$ are $1,4,9,16,25,36,49,64,81,100$. The two that differ by $75$ are $100$ and $25$. Thus, $m^2=100$ and $n^2=25$.

   This means that the year in which the age of each of Charles and Louis was a perfect square was the year $1369+100=1469$.

8. Since ${10}^y\ne 0$, the equation ${\displaystyle \frac{1}{32}}={\displaystyle \frac{x}{{10}^y}}$ is equivalent to ${10}^y=32x$.

   So the given question is equivalent to asking for the smallest positive integer $x$ for which $32x$ equals a positive integer power of $10$.

   Now $32=2^5$ and so $32x=2^{5}x$. For $32x$ to equal a power of $10$, each factor of $2$ must be matched with a factor of $5$. Therefore, $x$ must be divisible by $5^5$ (that is, $x$ must include at least $5$ powers of $5$), and so $x\ge 5^5=3125$.

   But $32(5^5)=2^5{5}^5={10}^5$, and so if $x=5^5=3125$, then $32x$ is indeed a power of 10, namely ${10}^5$. This tells us that the smallest positive integer $x$ for which ${\displaystyle \frac{1}{32}}={\displaystyle \frac{x}{{10}^y}}$ for some positive integer $y$ is $x=5^5=3125$.

9. Since the average of three consecutive multiples of $3$ is $a$, $a$ is the middle of these three integers, and so the integers are $a-3,a,a+3$.

   Since the average of four consecutive multiples of $4$ is $a+27$, we have that $a+27$ is halfway in between the second and third of these multiples (which differ by $4$), so the second and third of the multiples are $(a+27)-2=a+25$ and $(a+27)+2=a+29$. Therefore, the four integers are $a+21,a+25,a+29,a+33$.

   The smallest of these seven integers is $a-3$ and the largest is $a+33$.

   The average of these two integers is ${\displaystyle \frac{1}{2}}(a-3+a+33)={\displaystyle \frac{1}{2}}(2a+30)=a+15$.

   Since $a+15=42$, we have $a=27$.

10. First, we factor the left side of the given equation to obtain $a(a^2+2b)=2013$.

    Next, we factor the integer $2013$ as $2013=3\times 671=3\times 11\times 61$. Note that each of $3$, $11$ and $61$ is prime, so we can factor $2013$ no further. (We can find the factors of $3$ and $11$ using tests for divisibility by $3$ and $11$, or by systematic trial and error.)

    Since $2013=3\times 11\times 61$, the positive divisors of $2013$ are $$1,3,11,33,61,183,671,2013$$ Since $a$ and $b$ are positive integers, $a$ and $a^2+2b$ are both positive integers.

    Since $a$ and $b$ are positive integers, we have $a^2\ge a$ and $2b>0$, so $a^2+2b>a$.

    Since $a(a^2+2b)=2013$, $a$ and $a^2+2b$ must be a divisor pair of $2013$ (that is, a pair of positive integers whose product is $2013$) with $a<a^2+2b$.

    We make a table of the possibilities:

    $a$	$a^2+2b$	$2b$	$b$
    $1$	$2013$	$2012$	$1006$
    $3$	$671$	$662$	$331$
    $11$	$183$	$62$	$31$
    $33$	$61$	$-1028$	N/A

    Note that the last case is not possible, since $b$ must be positive. Therefore, the three pairs of positive integers that satisfy the equation are $(1,1006)$, $(3,331)$, $(11,31)$. (We can verify by substitution that each is a solution of the original equation.)

11. Suppose that the auditorium with these properties has $r$ rows and $c$ columns of chairs. Then there are $rc$ chairs in total.

    Each chair is empty, is occupied by a teacher, or is occupied by a student.

    Since there are $14$ teachers in each row, there are $14r$ chairs occupied by teachers. Since there are $10$ students in each column, there are $10c$ chairs occupied by students. Since there are exactly $3$ empty chairs, the total number of chairs can also be written as $14r+10c+3$. Therefore, $rc=14r+10c+3$.

    We proceed to find all pairs of positive integers $r$ and $c$ that satisfy this equation. We note that since there are $14$ teachers in each row, there must be at least $14$ columns (that is, $c\ge 14$) and since there are $10$ students in each column, there must be at least $10$ rows (that is, $r\ge 10$).

    Manipulating the equation, $$\begin{array}{rl}rc& =14r+10c+3\\ rc-14r& =10c+3\\ r(c-14)& =10c+3\\ r& ={\displaystyle \frac{10c+3}{c-14}}\\ r& ={\displaystyle \frac{10c-140+143}{c-14}}\\ r& ={\displaystyle \frac{10c-140}{c-14}}+{\displaystyle \frac{143}{c-14}}\\ r& =10+{\displaystyle \frac{143}{c-14}}\end{array}$$ Since $r$ is an integer, $10+{\displaystyle \frac{143}{c-14}}$ is an integer, and so ${\displaystyle \frac{143}{c-14}}$ must be an integer.

    Therefore, $c-14$ is a divisor of 143. Since $c\ge 14$, we have $c-14\ge 0$, and so $c-14$ is a positive divisor of 143.

    Since $143=11\times 13$, its positive divisors are $1$, $11$, $13$, $143$.

    We make a table of the possible values of $c-14$ along with the resulting values of $c$, $r$ (calculated using $r=10+{\displaystyle \frac{143}{c-14}}$) and $rc$:

    $c-14$	$c$	$r$	$rc$
    $1$	$15$	$153$	$2295$
    $11$	$25$	$23$	$575$
    $13$	$27$	$21$	$567$
    $143$	$157$	$11$	$1727$

    Therefore, the four possible values for $rc$ are $567$, $575$, $1727$, $2295$. That is, the smallest possible number of chairs in the auditorium is $567$. (Can you create a grid with $27$ columns and $21$ rows that has the required properties?)