Answer: 25.0

Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$

Solution:
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that $N \equiv a \pmod{6}$
also that $N \equiv b \pmod{5}$
Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$ $N \equiv a \pmod{6}$ $N \equiv  a \pmod{5}$ $\implies N=a \pmod{30}$ $0 \le a \le 4$
Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$
Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$ , ;
Also, we have already found which digits of $y$ will add up into the units digits of $2N$
Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work)
Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \[N \equiv 30x \pmod{36}\] \[N \equiv 30x \equiv 5x \pmod{25}\] Both of those must add up to \[2N\equiv60x \pmod{100}\]
$33 \ge x \ge 4$
Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \[N \equiv 5x \pmod{6}\] \[N \equiv 6x \equiv x \pmod{5}\] \[2N\equiv 6x \pmod{10}\]
Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x)
Then
\[N \equiv 5m+n \pmod{5}\] \[N \equiv 25m+5n \pmod{6}\] \[2N\equiv30m + 6n \pmod{10}\]
and this simplifies to
\[N \equiv n \pmod{5}\] \[N \equiv m+5n \pmod{6}\] \[2N\equiv 6n \pmod{10}\]
From careful inspection, this is true when
$n=0, m=6$
$n=1, m=6$
$n=2, m=2$
$n=3, m=2$
$n=4, m=4$
This gives you $5$ choices for $x$ , and $5$ choices for $y$ , so the answer is $5* 5 = \boxed{25}$

Solution:
Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $N$ . This means, in $100\le N\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ works because $900\equiv\dots00_5\equiv\dots00_6$
We know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\text{lcm }5,6=30$ . So we only have to care about cases of $N$ every $30$ subtracted. In each case, $2N$ subtracts $6$ /adds $4$ $N_5$ subtracts $1$ and $N_6$ adds $1$ for the $10$ 's digit.
\[\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}\]
\[\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}\]
\[\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}\]
As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\textcolor{red}{\text{red}}$ . So, thus, there are $5$ possibilities for each case, and $5\cdot 5=\boxed{25}$

Solution:
Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.
Then let $a_i$ $b_i$ denotes the digits of $N_5$ $N_6$ , respectively such that \[0\le a_i<5,0\le b_i<6\] Thus we have \[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\] \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Now we are given \[2N \equiv S \equiv N_5+N_6\pmod{100}\] \[2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}\] \[1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}\] \[50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}\] Canceling out $a_5$ left with \[50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}\]
Since $a_5$ $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$ . Thus,
\[50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}\] canceling out $10a_4$ , we have \[50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}\] \[5a_1+5a_2+5a_3 \equiv b_3\pmod{10}\] \[5(a_1+a_2+a_3) \equiv b_3\pmod{10}\]
Thus $b_3$ is a multiple of $5$
Now going back to our original equation \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Since $a_5=b_4$ \[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\] \[5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)\] \[5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]\]
Since the left side is a multiple of $5$ , then so does the right side. Thus $5\mid6(6b_1+b_2)+b_3$
Since we already know that $5\mid b_3$ , then $5\mid6(6b_1+b_2)$ , from where we also know that $5\mid6b_1+b_2$
For $b_1,b_2<6$ , there is a total of 7 ordered pairs that satisfy the condition. Namely,
\[(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)\]
Since $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$ $216b_1$ exceeds $1000$ which is not possible as $N$ is a three digit number, thus $(b_1,b_2)=(5,0)$ won't work as well.
Since we know that $a_i<5$ , for each of the ordered pairs $(b_1,b_2)$ , there is respectively one and only one solution $(a_1,a_2,a_3,a_4)$ that satisfies the equation
\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\]
Thus there are five solutions to the equation. Also since we have 5 possibilities for $a_5=b_4$ , we have a total of $5\cdot5=25$ values for $N$ $\boxed{25}$

Solution:
Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ .
Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$
If $a < \frac{25}{2}$ $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$ $2N \equiv 2a - 25 \pmod {25}$
Using the same logic for $b$ , if $b < 18$ $2N \equiv 2b \pmod {36}$ , and in the other case $2N \equiv 2b - 36 \pmod {36}$
We can do four cases:
Case 1: $a + b = 2a - 25 + 2b - 36 \implies a + b = 61$
For this case, there is trivially only one possible solution, $(a, b) = (25, 36)$ , which is equivalent to $(a, b) = (0, 0)$
Case 2: $a + b = 2a - 25 + 2b \implies a + b = 25$
Note that in this case, $a \geq 13$ must hold, and $b < 18$ must hold.
We find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.
Case 3: $a + b = 2a + 2b - 36 \implies a + b = 36$
Note that in this case, $b \geq 18$ must hold, and $a < 13$ must hold.
We find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.
Case 4: $a + b = 2a + 2b$
Trivially no solutions except $(a, b) = (0, 0)$ , which matches the solution in Case 1, which makes this an overcount.
By CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$ , our induction is complete, and our answer is $1 + 12 + 12 = \boxed{25}$


Answer: 25.0

Solution:
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that $N \equiv a \pmod{6}$
also that $N \equiv b \pmod{5}$
Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$ $N \equiv a \pmod{6}$ $N \equiv  a \pmod{5}$ $\implies N=a \pmod{30}$ $0 \le a \le 4$
Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$
Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$ , ;
Also, we have already found which digits of $y$ will add up into the units digits of $2N$
Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work)
Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \[N \equiv 30x \pmod{36}\] \[N \equiv 30x \equiv 5x \pmod{25}\] Both of those must add up to \[2N\equiv60x \pmod{100}\]
$33 \ge x \ge 4$
Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \[N \equiv 5x \pmod{6}\] \[N \equiv 6x \equiv x \pmod{5}\] \[2N\equiv 6x \pmod{10}\]
Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x)
Then
\[N \equiv 5m+n \pmod{5}\] \[N \equiv 25m+5n \pmod{6}\] \[2N\equiv30m + 6n \pmod{10}\]
and this simplifies to
\[N \equiv n \pmod{5}\] \[N \equiv m+5n \pmod{6}\] \[2N\equiv 6n \pmod{10}\]
From careful inspection, this is true when
$n=0, m=6$
$n=1, m=6$
$n=2, m=2$
$n=3, m=2$
$n=4, m=4$
This gives you $5$ choices for $x$ , and $5$ choices for $y$ , so the answer is $5* 5 = \boxed{25}$

Solution:
Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $N$ . This means, in $100\le N\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ works because $900\equiv\dots00_5\equiv\dots00_6$
We know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\text{lcm }5,6=30$ . So we only have to care about cases of $N$ every $30$ subtracted. In each case, $2N$ subtracts $6$ /adds $4$ $N_5$ subtracts $1$ and $N_6$ adds $1$ for the $10$ 's digit.
\[\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}\]
\[\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}\]
\[\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}\]
As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\textcolor{red}{\text{red}}$ . So, thus, there are $5$ possibilities for each case, and $5\cdot 5=\boxed{25}$

Solution:
Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.
Then let $a_i$ $b_i$ denotes the digits of $N_5$ $N_6$ , respectively such that \[0\le a_i<5,0\le b_i<6\] Thus we have \[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\] \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Now we are given \[2N \equiv S \equiv N_5+N_6\pmod{100}\] \[2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}\] \[1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}\] \[50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}\] Canceling out $a_5$ left with \[50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}\]
Since $a_5$ $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$ . Thus,
\[50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}\] canceling out $10a_4$ , we have \[50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}\] \[5a_1+5a_2+5a_3 \equiv b_3\pmod{10}\] \[5(a_1+a_2+a_3) \equiv b_3\pmod{10}\]
Thus $b_3$ is a multiple of $5$
Now going back to our original equation \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Since $a_5=b_4$ \[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\] \[5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)\] \[5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]\]
Since the left side is a multiple of $5$ , then so does the right side. Thus $5\mid6(6b_1+b_2)+b_3$
Since we already know that $5\mid b_3$ , then $5\mid6(6b_1+b_2)$ , from where we also know that $5\mid6b_1+b_2$
For $b_1,b_2<6$ , there is a total of 7 ordered pairs that satisfy the condition. Namely,
\[(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)\]
Since $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$ $216b_1$ exceeds $1000$ which is not possible as $N$ is a three digit number, thus $(b_1,b_2)=(5,0)$ won't work as well.
Since we know that $a_i<5$ , for each of the ordered pairs $(b_1,b_2)$ , there is respectively one and only one solution $(a_1,a_2,a_3,a_4)$ that satisfies the equation
\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\]
Thus there are five solutions to the equation. Also since we have 5 possibilities for $a_5=b_4$ , we have a total of $5\cdot5=25$ values for $N$ $\boxed{25}$

Solution:
Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ .
Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$
If $a < \frac{25}{2}$ $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$ $2N \equiv 2a - 25 \pmod {25}$
Using the same logic for $b$ , if $b < 18$ $2N \equiv 2b \pmod {36}$ , and in the other case $2N \equiv 2b - 36 \pmod {36}$
We can do four cases:
Case 1: $a + b = 2a - 25 + 2b - 36 \implies a + b = 61$
For this case, there is trivially only one possible solution, $(a, b) = (25, 36)$ , which is equivalent to $(a, b) = (0, 0)$
Case 2: $a + b = 2a - 25 + 2b \implies a + b = 25$
Note that in this case, $a \geq 13$ must hold, and $b < 18$ must hold.
We find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.
Case 3: $a + b = 2a + 2b - 36 \implies a + b = 36$
Note that in this case, $b \geq 18$ must hold, and $a < 13$ must hold.
We find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.
Case 4: $a + b = 2a + 2b$
Trivially no solutions except $(a, b) = (0, 0)$ , which matches the solution in Case 1, which makes this an overcount.
By CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$ , our induction is complete, and our answer is $1 + 12 + 12 = \boxed{25}$


Answer: 26.0

Bertha has 6 daughters and no sons.  Some of her daughters have 6 daughters, and the rest have none.  Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters.  How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$

Solution: Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters.  Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$

Solution: Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters.  Adding them together gives the answer of $\boxed{26}$


Answer: 26.0

Solution: Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$


Answer: 26.0

Solution: Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters.  Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$

Solution: Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters.  Adding them together gives the answer of $\boxed{26}$


Answer: 26.0

Solution: Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$


Answer: 40.0

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

Solution:
Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that $5\cdot 3\cdot 4=60$ plates could have been made.
If two letters are added to the second set, then $5\cdot 5\cdot 4=100$ plates can be made. If one letter is added to each of the second and third sets, then $5\cdot 4\cdot 5=100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100-60=\boxed{40}$


Answer: 40.0

Solution:
There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \cdot 3 \cdot 4 = 60$ license plates.
Adding $2$ letters to the start gives $7\cdot 3 \cdot 4 = 84$ plates.
Adding $2$ letters to the middle gives $5 \cdot 5 \cdot 4 = 100$ plates.
Adding $2$ letters to the end gives $5 \cdot 3 \cdot 6 = 90$ plates.
Adding a letter to the start and middle gives $6 \cdot 4 \cdot 4 = 96$ plates.
Adding a letter to the start and end gives $6 \cdot 3 \cdot 5 = 90$ plates.
Adding a letter to the middle and end gives $5 \cdot 4 \cdot 5 = 100$ plates.
You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\boxed{40}$

Solution:
Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters.
In general, when three numbers have the same fixed sum, their product will be maximal when they are as close together as possible.  This is a 3D analogue of the fact that a rectangle with fixed perimeter maximizes its area when the sides are equal (ie when it becomes a square).  In this case, no matter where you add the letters, there will be $5 + 3 + 4 + 2 = 14$ letters in total.  If you divide them as evenly as possible among the three groups, you get $5, 5, 4$ , which is a possible situation.
As before, the answer is $5 \cdot 5 \cdot 4 - 5 \cdot 3 \cdot 4 = 40$ , and the correct choice is $\boxed{40}$


Answer: 32.0

Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution:
There are $5$ days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats $n$ bananas on May 5, and $n-4(6)=n-24$ bananas on May 1. The sum of this arithmetic sequence is equal to $100$
\begin{align*} \frac{n+n-24}{2} \cdot 5 &= 100\\ n-12&=20\\ n&=\boxed{32}

Solution: Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply $\frac{100}{5}=20$ . This means that the number of bananas the ape ate on May 5th is just $20+6*2=32$ . Select $\boxed{32}$


Answer: 10.0

Blake and Jenny each took four $100$ -point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?
$\mathrm{(A)}\  10  \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40$

Solution:
Blake scored a total of $4 \times 78=312$ points. Jenny scored $10-10+20+20=40$ points higher than Blake, so her average is $\frac{312+40}{4}=88$ .
the difference is $88-78=\boxed{10}$

Solution: The total point difference between Blake's and Jenny's scores is $10-10+20+20=40$ . The average of it is $\frac{40}{4}=\boxed{10}$


Answer: 1.0

Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$

Solution:
Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas ).
We now have that the sum of the two roots is $63$ while the product is $k$ . Since both roots are primes, one must be $2$ , otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$ . Hence, our answer is $\boxed{1}$

Solution: By Vieta's you have \[a+b=63\] \[ab=k\] Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\boxed{1}$ solution.


Answer: 1.0

Solution:
Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas ).
We now have that the sum of the two roots is $63$ while the product is $k$ . Since both roots are primes, one must be $2$ , otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$ . Hence, our answer is $\boxed{1}$

Solution: By Vieta's you have \[a+b=63\] \[ab=k\] Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\boxed{1}$ solution.


Answer: 350.0

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points.  They first meet after Brenda has run 100 meters.  They next meet after Sally has run 150 meters past their first meeting point.  Each girl runs at a constant speed.  What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ }  400\qquad \mathrm{(E) \ } 500$

Solution: Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350\Longrightarrow\boxed{350}$

Solution: The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{350}$


Answer: 350.0

Solution: Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350\Longrightarrow\boxed{350}$

Solution: The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{350}$


Answer: 5.0

Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month.  If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution:
Call this month "Month 0".  Make a table of the fish that Brent and Gretel have each month.
$\text{Month / Brent / Gretel}$
$\text{0 / 4 / 128}$
$\text{1 / 16 / 256}$
$\text{2 / 64 / 512}$
$\text{3 / 256 / 1024}$
$\text{4 / 1024 / 2048}$
$\text{5 / 4096 / 4096}$
You could create a similar table without doing all of the calculations by converting all the goldfish into powers of $2$ .  In this table, you could increase Brent's goldfish by two powers of $2$ , while increasing Gretel's fish by one power of $2$
Either way, in $5$ months they will have the same number of fish, giving an answer of $\boxed{5}$

Solution:
Create two equations.
Brent starts with $4$ goldfish and they quadruple each month, giving us $y=4\cdot 4^{x}$ or $y=2^{2x+2}$
Gretel starts with $128$ goldfish and they double each month, giving us $y=128 \cdot 2^{x}$ or $y=2^{x+7}$
Using substitution, $2^{2x+2}=2^{x+7}$ . Thus $2x+2=x+7$ giving us $x=5$ , which is $\boxed{5}$


Answer: 31.0

Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$ . What is the sum of the possible values of $w$
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$

Solution:
Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$ .
This means $a+b+c=9$ $a,b,c$ must be in the set ${1,3,4,5,6}$ .
The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$ $a+b$ , and $b+c$ then must be the remaining two numbers which are $4$ and $6$ .
The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$
\begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*}
\begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*}
The sum of the two w's is $15+16=31$ $\boxed{31}$


Answer: 31.0

Solution:
Let the four numbers be $z$ $z+a$ $z+b$ , and $z+c$ . We know that $c$ must be $9$ because that's the greatest difference. So we have $z$ $z+a$ $z+b$ , and $z+9$ . The 6 possible differences are $a$ $b$ $9$ $b-a$ $9-a$ , and $9-b$ . We are given that the differences are 1, 3, 4, 5, 6, 9. $a$ and $9-a$ and $b$ and $9-b$ add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves $1$ . That means $b-a$ has to equal $1$ . So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have $z$ $z+3$ $z+4$ , and $z+9$ $4z+16=44$ $4z=28$ $z=7$ $w=7+9=16$ . Now for 4,5, notice that it doesn't work. The differences are 4, 5, 9, 1, 4, 5. We are missing 6 and 3. For 5,6, it's $z$ $z+5$ $z+6$ , and $z+9$ . Check that the differences work; they do. We have $4z+20=44$ $4z=24$ $z=6$ $w=6+9=15$ . Therefore our answer is $15+16=\boxed{31}$ .
~MC413551


Answer: 31.0

Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$

Solution:
The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences can't be right next to each other because they would make a difference of $8,$ which isn't given as a possibility in the problem. This means $1$ must be the difference between $y$ and $x.$ We can express the possible configurations as the lines.
If we look at the first number line, you can express $x$ as $w-5,$ $y$ as $w-6,$ and $z$ as $w-9.$ Since the sum of all these integers equal $44$ \begin{align*} w+w-5+w-6+w-9&=44\\ 4w&=64\\ w&=16 \end{align*} You can do something similar to this with the second number line to find the other possible value of $w.$ \begin{align*} w+w-3+w-4+w-9&=44\\ 4w&=60\\ w&=15 \end{align*} The sum of the possible values of $w$ is $16+15 = \boxed{31}$

Solution:
First, like Solution 1, we know that $w-z=9 \ \text{(1)}$ , because no two numbers could have a larger difference. Next, we find the sum of all the differences; since $w$ is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes $3w$ . Continuing in this way, we find that \[3w+x-y-3z=28 \ \text{(2)}\] . Now, we can subtract $3w-3z=27$ from (2) to get $x-y=1 \ \text{(3)}$ . Also, adding (2) with $w+x+y+z=44$ gives $4w+2x-2z=72$ , or $2w+x-z=36$ . Subtracting (1) from this gives $w+x=27$ . Since we know $w-z$ and $x-y$ , we find that \[(w-z)+(x-y)=(w-y)+(x-z)=9+1=10\] . This means that $w-y$ and $x-z$ must be 4 and 6, in some order. If $w-y=6$ , then subtracting this from (3) gives $(w-y)-(x-y)=6-1=5$ , so $w-x=5$ . This means that $(w-x)+(w+x)=2w=27+5=32$ , so $w=16$ . Similarly, $w$ can also equal $15$
Now if you are in a rush, you most likely would have answered $16+15=\boxed{31}$ . But we do have to check if these work. In fact, they do, giving solutions $(w,x,y,z)=(16, 11, 10, 7)$ and $(w,x,y,z)=(15, 12, 11, 6)$

Solution:
Let $w - x = a$ $w - y = b$ $w - z = c$ . As above, we know that $c = 9$ . Thus, $a < b < c$ .
So, we have $w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44$ . This means $a + b + 9$ is a multiple of $4$ . Testing values of $a$ and $b$ , we find $(a, b, c) = (1, 6, 9), (3, 4, 9),$ and $(5, 6, 9)$ all satisfy this relation. The corresponding $(w, x, y, z)$ sets are $(15, 14, 9, 6), (15, 12, 11, 6),$ and $(16, 11, 10, 7)$ . The first set does not satisfy the given conditions, but the other two do. Thus, $w = 15$ and $w = 16$ are both possible solutions so the answer is $16+15=\boxed{31}$

Solution: From the problem, we know that $w+x+y+z=44$ . Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are $(w-x)$ $(w-y)$ $(w-z)$ $(x-y)$ $(x-z)$ $(y-z)$ , the sum of $(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)$ is equal to the sum of 1, 3, 4, 5, 6, 9, so $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ . Simplifying, we get $3w-3z+x-y=28$ . Adding $w+x+y+z=44$ and $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ , we get $4w+2x-2z=72$ , and simplifying we get $2w+x-z=36$ . Since $x-z$ is one of our positive differences, we can start guessing values for $w$ , and if the equation simplifies to one of our numerical positive differences, that value of $w$ should work. We can start at $w=18$ and keep going down, because our sum has to be positive. For $w=18$ $x-z=0$ , which is not one of our sums. For $w=17$ $x-z=2$ ,which is not one of our sums. For $w=16$ $x-z=4$ , which is one of our sums, so 16 works. For $w=15$ $x-z=6$ , which is one of our sums, so 15 works. For $w=14$ $x-z=8$ , which is not one of our sums. If we keep going, $x-z$ will soon exceed 10 and exceed all our sums, so any value below $w=14$ will not work. Therefore, our only solutions for $w$ are 15 and 16, which means our sum is $\boxed{31}$ . You can check that 15 and 16 work by forming a string of 4 numbers as shown above.

Solution: Because we know that $w>x>y>z$ and that the positive differences are $1, 3, 4, 5, 6, 9$ , we can immediately come to the conclusion that $w-z= 9$ (because w is the largest integer and z is the smallest integer, so their difference must be the greatest). With this we have $3$ equations, we have that $x+y+z+w=44$ (from the problem), $w-z=9$ and because we can add up all the possible possible differences (as shown in the previous solutions), we get that $3w+x-y-3z=28$ . With these equations, we eventually manipulate these equations by doing the first equation minus 3 times the second to get $x-y=1$ . We can also add the first and third equation and subtract the second equation to get $w+x=27$ thus we know that $w-x$ can be $3, 4, 5,$ and $6$ (note: 1 and 9 are not possible because we know that $x-y=1$ and $w-z=9$ and the remaining differences can only be taken by 1 pair, so $w-x$ cannot be equal to 1 or 9). However, in order to get an integer value for x and z, we find that $w-z$ can only be equal to 3 and 5. Thus, by solving these, we see that $w= 15$ and $w=16$ $15+16=\boxed{31}$


Answer: 25.0

Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qquad\textbf{(D) }\ \frac23 \qquad\textbf{(E) }\ \frac45$

Solution: Let $m =$ Brianna's money.  We have $\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})$ .  Thus, the money left over is $m-\frac35m = \frac25m$ , so the answer is $\boxed{25}$


Answer: 25.0

Solution: Let $m =$ Brianna's money.  We have $\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})$ .  Thus, the money left over is $m-\frac35m = \frac25m$ , so the answer is $\boxed{25}$


Answer: 900.0

Bricklayer Brenda takes $9$ hours to build a chimney alone, and bricklayer Brandon takes $10$ hours to build it alone. When they work together, they talk a lot, and their combined output decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?
$\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900$

Solution:
Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.
Using $w = rt$ , we get $x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)$ . Solving for $x$ , we get $\boxed{900}$


Answer: 52.0

Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $\textdollar 0.75$ for her to make. In dollars, what is her profit for the day?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution:
She first sells one-half of her $48$ loaves, or $\frac{48}{2}=24$ loaves. Each loaf sells for $\textdollar 2.50$ , so her total earnings in the morning is equal to \[24\cdot \textdollar 2.50 = \textdollar 60\]
This leaves 24 loaves left, and Bridget will sell $\dfrac{2}{3}\times 24=16$ of them for a price of $\textdollar\frac{2.50}{2}=\textdollar 1.25$ . Thus, her total earnings for the afternoon is \[16\cdot \textdollar 1.25  = \textdollar 20\]
Finally, Bridget will sell the remaining $24-16=8$ loaves for a dollar each. This is a total of $\textdollar 1\cdot 8 = \textdollar 8$
The total amount of money she makes is equal to $60+20+8=\textdollar 88$
However, since Bridget spends $\textdollar 0.75$ making each loaf of bread, the total cost to make the bread is equal to $\textdollar 0.75\cdot48=\textdollar 36$
Her total profit is the amount of money she spent subtracted from the amount of money she made, which is \[88-36=52\implies\boxed{52}\]


Answer: 10.0

Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 9\qquad\text{(D)}\ 10\qquad\text{(E)}\ 11$

Solution:
Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - $a:b$ $a$ is for hrs. and $b$ is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minutes from Houston, so you are delighted at the first bus you have passed.
At 1:15 pm, the 9:00 am Dallas bus meets you again, being 4:15 away from Dallas and therefore 45 minutes from Houston. After a while you might notice that the buses meet you in 30 minute intervals after 12:45 pm; indeed, the bus that left an hour later than the bus most recently has traveled 30 minutes closer, and so have you, for a total of an hour closer, and when you passed the most recent bus, the bus after it had indeed left an hour later. You decide to predict how many buses you pass by finding all valid times that are 30 minute intervals from 12:45. They are
12:45, 1:15, 1:45, 2:15, 2:45, 3:15, 3:45, 4:15, 4:45, 5:15, and 5:30, your arrival time.
Hence, you will pass $\boxed{10}$ Dallas-bound buses.


Answer: 39.6

Business is a little slow at Lou's Fine Shoes, so Lou decides to have a
sale. On Friday, Lou increases all of Thursday's prices by $10$ percent. Over the
weekend, Lou advertises the sale: "Ten percent off the listed price. Sale
starts Monday." How much does a pair of shoes cost on Monday that
cost $40$ dollars on Thursday?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44$

Solution: On Friday, the shoes would cost $40 \cdot 1.1= 44$ dollars. Then on Monday, the shoes would cost $44- \frac{44}{10}=44-4.4=\boxed{39.60}$


Answer: 279.0

Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$

Solution:
When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile:
\[\frac{a}{6} + \frac{n-a}{4} = \frac{n-a}{6} + \frac{2a}{5} \rightarrow  a = \frac{5}{19}n.\]
The smallest possible integral value of $n$ is $19$ , so we plug in $n = 19$ and $a = 5$ and get $t = \frac{13}{3}$ hours, or $260$ minutes. So our answer is $19 + 260 = \boxed{279}$


Answer: 5.0

Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution:
Looking at the answer choices, you see that you can list them out.
Doing this gets you:
$UUDDUD$
$UDUDUD$
$UUUDDD$
$UDUUDD$
$UUDUDD$
Counting all the paths listed above gets you $\boxed{5}$

Solution: Any combination can be written as some re-arrangement of $UUUDDD$ . Clearly we must end going down, and start going up, so we need the number of ways to insert 2 $U$ 's and 2 $D$ 's into $U\, \_ \, \_ \, \_ \, \_ \, D$ . There are ${4\choose 2}=6$ ways, but we have to remove the case $UDDUUD$ , giving us $\boxed{5}$

Solution:
These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is $\boxed{5}$ .
~andliu766

Solution:
First step is U, last step is D.
After third step we can get only positions 3 or 1.
In the first case there is only one way UUUDDD.
In the second case we have two way to get this position UDU and UUD.
Similarly, we have two way return to position 0 (UDD and DUD).
Therefore, we have $1 + 2 \cdot 2 = \boxed{5}$

Solution:
We can find the total cases then deduct the ones that don't work.
Let $U$ represent "Up" and $D$ represent "Down". We know that in order to land back at the bottom of the stairs, we must have an equal number of $U$ 's and $D$ 's, therefore six hops means $3$ of each.
The number of ways to arrange $3$ $U$ 's and $3$ $D$ 's is $\dfrac{6!}{(3!)^2}=\dfrac{720}{36}=20$
Case $1$ : Start with $D$
Case $2$ : Start with $UDD$
Case $3$ : Start with $UUDDD$
Case $4$ : Start with $UDUDD$
Case $1$ is asking us how many ways there are to arrange $3$ $U$ 's and $2$ $D$ 's, which is $\dfrac{5!}{3!2!}=\dfrac{120}{12}=10$
Case $2$ is asking us how many ways there are to arrange $2$ $U$ 's and $1$ $D$ , which is $\dfrac{3!}{2!1!}=\dfrac{6}{2}=3$
Case $3$ is asking us how many ways there are to arrange $1$ $U$ , which is $1$
Case $4$ is asking us the same thing as Case $3$ , giving us $1$
Therefore, deducting all cases from $20$ gives $20-10-3-1-1=\boxed{5}$


Answer: 182.0

By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

Solution:
Let $p(n)$ denote the product of the distinct proper divisors of $n$ . A number $n$ is nice in one of two instances:
We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form $n = pqr$ (with $p,q$ prime and $r > 1$ ) or $n = p^e$ (with $e \neq 3$ ).
In the former case, it suffices to note that $p(n) \ge (pr) \cdot (qr) = pqr^2 > pqr = n$
In the latter case, then $p(n) = p \cdot p^2 \cdots p^{(e-1)} = p^{(e-1)e/2}$
For $p(n) = n$ , we need $p^{(e-1)e/2} = p^e \Longrightarrow e^2 - e = 2e \Longrightarrow$ $e = 0$ or $e = 3$
Since $e \neq 3$ , in the case $e = 0 \Longrightarrow n = 1$ does not work.
Thus, listing out the first ten numbers to fit this form, $2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,$ $\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,$ $\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,$ $\ 3^3 = 27,\ 3 \cdot 11 = 33$ Summing these yields $\boxed{182}$


Answer: 4.0

By inserting parentheses, it is possible to give the expression \[2\times3 + 4\times5\] several values.  How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$

Solution: The three operations can be performed on any of $3! = 6$ orders.  However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained.  It is easy to check that the values of the four expressions \begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + 4\times5) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*} are in fact all distinct.  So the answer is $\boxed{4}$


Answer: 4.0

Solution: The three operations can be performed on any of $3! = 6$ orders.  However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained.  It is easy to check that the values of the four expressions \begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + 4\times5) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*} are in fact all distinct.  So the answer is $\boxed{4}$


Answer: 20.0

Solution:
Case $1$ : The numbers are separated by $1$
We this case with $a=0, b=1,$ and $c=2$ . Following this logic, the last set we can get is $a=7, b=8,$ and $c=9$ . We have $8$ sets of numbers in this case.
Case $2$ : The numbers are separated by $2$
This case starts with $a=0, b=2,$ and $c=2$ . It ends with $a=5, b=7,$ and $c=9$ . There are $6$ sets of numbers in this case.
Case $3$ : The numbers start with $a=0, b=3,$ and $c=6$ . It ends with $a=3, b=6,$ and $c=9$ . This case has $4$ sets of numbers.
It's pretty clear that there's a pattern: $8$ sets, $6$ sets, $4$ sets. The amount of sets per case decreases by $2$ , so it's obvious Case $4$ has $2$ sets. The total amount of possible five-digit numbers is $8+6+4+2=\boxed{20}$


Answer: 25.0

Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$

Solution: Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\boxed{25}$ cupcakes.

Solution: In $300$ seconds ( $5$ minutes), Cagney will frost $\dfrac{300}{20} = 15$ cupcakes, and Lacey will frost $\dfrac{300}{30} = 10$ cupcakes. Therefore, working together they will frost $15 + 10 = \boxed{25}$ cupcakes.

Solution: Since Cagney frosts $3$ cupcakes a minute, and Lacey frosts $2$ cupcakes a minute, they together frost $3+2=5$ cupcakes a minute. Therefore, in $5$ minutes, they frost $5\times5 = 25 \Rightarrow \boxed{25}$


Answer: 25.0

Solution: Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\boxed{25}$ cupcakes.

Solution: In $300$ seconds ( $5$ minutes), Cagney will frost $\dfrac{300}{20} = 15$ cupcakes, and Lacey will frost $\dfrac{300}{30} = 10$ cupcakes. Therefore, working together they will frost $15 + 10 = \boxed{25}$ cupcakes.

Solution: Since Cagney frosts $3$ cupcakes a minute, and Lacey frosts $2$ cupcakes a minute, they together frost $3+2=5$ cupcakes a minute. Therefore, in $5$ minutes, they frost $5\times5 = 25 \Rightarrow \boxed{25}$


Answer: 840.0

Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.

Solution: Assume that the largest geometric number starts with a $9$ . We know that the common ratio must be a rational of the form $k/3$ for some integer $k$ , because a whole number should be attained for the 3rd term as well. When $k = 1$ , the number is $931$ . When $k = 2$ , the number is $964$ . When $k = 3$ , we get $999$ , but the integers must be distinct. By the same logic, the smallest geometric number is $124$ . The largest geometric number is $964$ and the smallest is $124$ . Thus the difference is $964 - 124 = \boxed{840}$

Solution:
Consider the three-digit number $abc$ . If its digits form a geometric progression, we must have that ${a \over b} = {b \over c}$ , that is, $b^2 = ac$
The minimum and maximum geometric numbers occur when $a$ is minimized and maximized, respectively. The minimum occurs when $a = 1$ ; letting $b = 2$ and $c = 4$ achieves this, so the smallest possible geometric number is 124.
For the maximum, we have that $b^2 = 9c$ $b$ is maximized when $9c$ is the greatest possible perfect square; this happens when $c = 4$ , yielding $b = 6$ . Thus, the largest possible geometric number is 964.
Our answer is thus $964 - 124 = \boxed{840}$

Solution: The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $\boxed{840}.$


Answer: 19990.0

Call a $7$ -digit telephone number $d_1d_2d_3-d_4d_5d_6d_7$ memorable if the prefix sequence $d_1d_2d_3$ is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digits $0,1,2, \ldots, 9$ , the number of different memorable telephone numbers is
$\mathrm{(A)}\ 19,810 \qquad\mathrm{(B)}\ 19,910 \qquad\mathrm{(C)}\ 19,990 \qquad\mathrm{(D)}\ 20,000 \qquad\mathrm{(E)}\ 20,100$

Solution: In this problem, we only need to consider the digits $\overline{d_4d_5d_6d_7}$ . Each possibility of $\overline{d_4d_5d_6d_7}$ gives $2$ possibilities for $\overline{d_1d_2d_3}$ , which are $\overline{d_1d_2d_3}=\overline{d_4d_5d_6}$ and $\overline{d_1d_2d_3}=\overline{d_5d_6d_7}$ with the exception of the case of $d_4=d_5=d_6=d_7$ , which only gives one sequence. After accounting for overcounting, the answer is $(10 \times 10 \times 10 \times 10) \times 2 - 10=19990 \Rightarrow \boxed{19,990}$


Answer: 11.0

Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\  10 \qquad\textbf{(C)}\  11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

Solution:
The special fractions are \[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\] We rewrite them in the simplest form: \[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$
For the set $\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$
For the set $\left\{\frac{1}{4},2\frac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$
Together, there are $\boxed{11}$ distinct integers that can be written as the sum of two, not necessarily different, special fractions: \[1,2,3,4,6,7,8,13,16,18,28.\] ~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution:
Let $a=15-b,$ so the special fraction is \[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\] We can ignore the $-1$ part and only focus on $\frac{15}{b}.$
The integers are $\frac{15}{1},\frac{15}{3},\frac{15}{5},$ which are $15,5,3,$ respectively. We get $30,20,18,10,8,6$ from this group of numbers.
The halves are $\frac{15}{2},\frac{15}{6},\frac{15}{10},$ which are $7\frac12,2\frac12,1\frac12,$ respectively. We get $15,10,9,5,4,3$ from this group of numbers.
The quarters are $\frac{15}{4},\frac{15}{12},$ which are $3\frac34,1\frac14,$ respectively. We get $5$ from this group of numbers.
Note that $10$ and $5$ each appear twice. Therefore, the answer is $\boxed{11}.$

Solution:
We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are $\frac{10}{5}$ (or $2$ ), $\frac{12}{3}$ (or $4$ ), and $\frac{14}{1}$ (or $14$ ). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Case 2: fraction + fraction
The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$ ), $\frac{9}{6}$ (or $\frac{3}{2}$ ), and $\frac{13}{2}$ . Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Thus, $6+6=12$
So now you would just go ahead and innocently choose $\textbf{(D) }12$ , right? No! We overcounted $8$ , as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$ . Therefore, the correct answer is actually $12-1=\boxed{11}$


Answer: 11.0

Solution:
The special fractions are \[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\] We rewrite them in the simplest form: \[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$
For the set $\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$
For the set $\left\{\frac{1}{4},2\frac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$
Together, there are $\boxed{11}$ distinct integers that can be written as the sum of two, not necessarily different, special fractions: \[1,2,3,4,6,7,8,13,16,18,28.\] ~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution:
Let $a=15-b,$ so the special fraction is \[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\] We can ignore the $-1$ part and only focus on $\frac{15}{b}.$
The integers are $\frac{15}{1},\frac{15}{3},\frac{15}{5},$ which are $15,5,3,$ respectively. We get $30,20,18,10,8,6$ from this group of numbers.
The halves are $\frac{15}{2},\frac{15}{6},\frac{15}{10},$ which are $7\frac12,2\frac12,1\frac12,$ respectively. We get $15,10,9,5,4,3$ from this group of numbers.
The quarters are $\frac{15}{4},\frac{15}{12},$ which are $3\frac34,1\frac14,$ respectively. We get $5$ from this group of numbers.
Note that $10$ and $5$ each appear twice. Therefore, the answer is $\boxed{11}.$

Solution:
We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are $\frac{10}{5}$ (or $2$ ), $\frac{12}{3}$ (or $4$ ), and $\frac{14}{1}$ (or $14$ ). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Case 2: fraction + fraction
The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$ ), $\frac{9}{6}$ (or $\frac{3}{2}$ ), and $\frac{13}{2}$ . Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Thus, $6+6=12$
So now you would just go ahead and innocently choose $\textbf{(D) }12$ , right? No! We overcounted $8$ , as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$ . Therefore, the correct answer is actually $12-1=\boxed{11}$


Answer: 486.0

Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$

Solution:
The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$ , before $n - 2$ , and at the very end.
EXAMPLE:
Putting 4 into the string 123:
4 can go before the 2: 1423,
Before the 3: 1243,
And at the very end: 1234.
Only the addition of the next number, n, will change anything.
Thus the number of permutations with n elements is three times the number of permutations with $n-1$ elements.
Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$
Thus for $n=7$ there are $2*3^5=\boxed{486}$ permutations.


Answer: 315.0

Call a positive integer $N$ 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$ . For example, $51$ is a 7-10 double because its base- $7$ representation is $102$ . What is the largest 7-10 double?

Solution:
Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.
Given this is an AIME problem, $A<1000$ . If we look at $B$ in base $10$ , it must be equal to $2A$ , so $B<2000$ when $B$ is looked at in base $10.$
If $B$ in base $10$ is less than $2000$ , then $B$ as a number in base $7$ must be less than $2*7^3=686$
$686$ is non-existent in base $7$ , so we're gonna have to bump that down to $666_7$
This suggests that $A$ is less than $\frac{666}{2}=333$
Guess and check shows that $A<320$ , and checking values in that range produces $\boxed{315}$

Solution: Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3.  Let the number be \[abc\] in base 7. Then the number in expanded form is \[49a+7b+c\] in base 7 and \[100a+10b+c\] in base 10.  Since the number in base 7 is half the number in base 10, we get the following equation. \[98a+14b+2c=100a+10b+c\] which simplifies to \[2a=4b+c.\] The largest possible value of a is 6 because the number is in base 7.  Then to maximize the number, $b$ is $3$ and $c$ is $0$ .  Therefore, the largest 7-10 double is 630 in base 7, or $\boxed{315}$ in base 10.


Answer: 472.0

Call a positive integer $n$ $k$ pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$

Solution:
Every 20-pretty integer can be written in form $n = 2^a 5^b k$ , where $a \ge 2$ $b \ge 1$ $\gcd(k,10) = 1$ , and $d(n) = 20$ , where $d(n)$ is the number of divisors of $n$ . Thus, we have $20 = (a+1)(b+1)d(k)$ , using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check.
If $a+1 = 4$ , then $b+1 = 5$ . But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$
If $a+1 = 5$ , then $b+1 = 2$ or $4$ . The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$ . The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 7440$ . In the second case $b+1 = 4$ and $d(k) = 1$ , and there is one solution $n = 2^4 \cdot 5^3 = 2000$
If $a+1 = 10$ , then $b+1 = 2$ , but this gives $2^9 \cdot 5^1 > 2019$ . No other values for $a+1$ work.
Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$

Solution:
For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$ where $p,q,r$ are primes. No number that is a multiple of $20$ can be expressed in the first form because 20 has two primes in its prime factorization, while the first form has only one , and the only integer divisible by $20$ that has the second form is $2^{9}5$ , which is 2560, greater than $2019$
For the third form, the only $20$ -pretty numbers are $2^45^3=2000$ and $2^35^4=5000$ , and only $2000$ is small enough.
For the fourth form, any number of the form $2^45r$ where $r$ is a prime other than $2$ or $5$ will satisfy the $20$ -pretty requirement. Since $n=80r<2019$ $r\le 25$ . Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$
Thus, $\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}$


Answer: 49.0

Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$

Solution:
$n$ can either be $0$ or $1$ (mod $2$ ).
Case 1: $n \equiv 0 \pmod{2}$
Then, $n \equiv 2 \pmod{4}$ , which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$ , and therefore $n \equiv 3 \pmod{5}$ . Using CRT , we obtain $n \equiv 58 \pmod{60}$ , which gives $16$ values for $n$
Case 2: $n \equiv 1 \pmod{2}$
$n$ is then $3 \pmod{4}$ . If $n \equiv 0 \pmod{3}$ $n \equiv 3 \pmod{6}$ , a contradiction. Thus, $n \equiv 2 \pmod{3}$ , which implies $n \equiv 5 \pmod{6}$ $n$ can either be $0 \pmod{5}$ , which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$ , which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.
The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$

Solution:
Because the LCM of all of the numbers we are dividing by is $60$ , we know that all of the remainders are $0$ again at $60$ , meaning that we have a cycle that repeats itself every $60$ numbers.
After listing all of the remainders up to $60$ , we find that $35$ $58$ , and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\cdot16 = 960$ and $3\cdot16 = 48$ , so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$ , so we have $1$ other extra-distinct number ( $35$ ).
$48 + 1 =  \boxed{049}$

Solution:
$\textbf{Case 0:  } {\rm Rem} \ \left( n, 6 \right) = 0$
We have ${\rm Rem} \ \left( n, 2 \right) = 0$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 1:  } {\rm Rem} \ \left( n, 6 \right) = 1$
We have ${\rm Rem} \ \left( n, 2 \right) = 1$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 2:  } {\rm Rem} \ \left( n, 6 \right) = 2$
We have ${\rm Rem} \ \left( n, 3 \right) = 2$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 3:  } {\rm Rem} \ \left( n, 6 \right) = 3$
The condition ${\rm Rem} \ \left( n, 6 \right) = 3$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ ${\rm Rem} \ \left( n, 3 \right) = 0$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 \right\}$ is a permutation of $\left\{ 0, 1 ,2 \right\}$ .
Thus, ${\rm Rem} \ \left( n, 4 \right) = 2$
However, ${\rm Rem} \ \left( n, 4 \right) = 2$ conflicts ${\rm Rem} \ \left( n, 2 \right) = 1$ .
Therefore, this case has no solution.
$\textbf{Case 4:  } {\rm Rem} \ \left( n, 6 \right) = 4$
The condition ${\rm Rem} \ \left( n, 6 \right) = 4$ implies ${\rm Rem} \ \left( n, 2 \right) = 0$ and ${\rm Rem} \ \left( n, 3 \right) = 1$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 , 5 \right\}$ is a permutation of $\left\{ 0, 1 ,2 , 3 \right\}$
Because ${\rm Rem} \ \left( n, 2 \right) = 0$ , we must have ${\rm Rem} \ \left( n, 4 \right) = 2$ . Hence, ${\rm Rem} \ \left( n, 5 \right) = 3$
Hence, $n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}$ .
Hence, $n \equiv - 2 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this case is 16.
$\textbf{Case 5:  } {\rm Rem} \ \left( n, 6 \right) = 5$
The condition ${\rm Rem} \ \left( n, 6 \right) = 5$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ and ${\rm Rem} \ \left( n, 3 \right) = 2$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, 4 \right)$ and ${\rm Rem} \ \left( n, 5 \right)$ are two distinct numbers in $\left\{ 0, 3, 4 \right\}$ .
Because ${\rm Rem} \ \left( n, 4 \right) \leq 3$ and $n$ is odd, we have ${\rm Rem} \ \left( n, 4 \right) = 3$ .
Hence, ${\rm Rem} \ \left( n, 5 \right) = 0$ or 4.
$\textbf{Case 5.1:  } {\rm Rem} \ \left( n, 6 \right) = 5$ ${\rm Rem} \ \left( n, 4 \right) = 3$ ${\rm Rem} \ \left( n, 5 \right) = 0$
We have $n \equiv 35 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this subcase is 17.
$\textbf{Case 5.2:  } {\rm Rem} \ \left( n, 6 \right) = 5$ ${\rm Rem} \ \left( n, 4 \right) = 3$ ${\rm Rem} \ \left( n, 5 \right) = 4$
$n \equiv - 1 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this subcase is 16.
Putting all cases together, the total number of extra-distinct $n$ is $16 + 17 + 16 = \boxed{049}$


Answer: 6.0

Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$

Solution: The divisibility rule of $15$ is that the number must be congruent to $0$ mod $3$ and congruent to $0$ mod $5$ . Being divisible by $5$ means that it must end with a $5$ or a $0$ . We can rule out the case when the number ends with a $0$ immediately because the only integer that is uphill and ends with a $0$ is $0$ which is not positive. So now we know that the number ends with a $5$ . Looking at the answer choices, the answer choices are all pretty small, so we can generate all of the numbers that are uphill and are divisible by $3$ . These numbers are $15, 45, 135, 345, 1245, 12345$ , or $\boxed{6}$ numbers.

Solution:
First, note how the number must end in either $5$ or $0$ in order to satisfying being divisible by $15$ . However, the number can't end in $0$ because it's not strictly greater than the previous digits. Thus, our number must end in $5$ . We do casework on the number of digits.
Case 1: $1$ digit. No numbers work, so $0$ numbers.
Case 2: $2$ digits. We have the numbers $15, 45,$ and $75$ , but $75$ isn't an uphill number, so $2$ numbers
Case 3: $3$ digits. We have the numbers $135, 345$ , so $2$ numbers.
Case 4: $4$ digits. We have the numbers $1235, 1245$ and $2345$ , but only $1245$ satisfies this condition, so $1$ number.
Case 5: $5$ digits. We have only $12345$ , so $1$ number.
Adding these up, we have $2+2+1+1=\boxed{6}$

Solution:
Like solution 2, we can proceed by using casework. A number is divisible by $15$ if is divisible by $3$ and $5.$ In this case, the units digit must be $5,$ otherwise no number can be formed.
Case 1: sum of digits = 6
There is only one number, $15.$
Case 2: sum of digits = 9
There are two numbers: $45$ and $135.$
Case 3: sum of digits = 12
There are two numbers: $345$ and $1245.$
Case 4: sum of digits = 15
There is only one number, $12345.$
We can see that we have exhausted all cases, because in order to have a larger sum of digits, then a number greater than $5$ needs to be used, breaking the conditions of the problem. The answer is $\boxed{6}.$

Solution:
An integer is divisible by $15$ if it is divisible by $3$ and $5$ . Divisibility by $5$ means ending in $0$ or $5$ , but since no digit is less than $0$ , the only uphill integer ending in $0$ could be $0$ , which is not positive. This means the integer must end in $5$
All uphill integers ending in $5$ are formed by picking any subset of the sequence $(1,2,3,4)$ of digits (keeping their order), then appending a $5$ . Divisibility by $3$ means the sum of the digits is a multiple of $3$ , so our choice of digits must add to $0$ modulo $3$
$5 \equiv -1 \pmod{3}$ , so the other digits we pick must add to $1$ modulo $3$ . Since $(1,2,3,4) \equiv (1,-1,0,1) \pmod{3}$ , we can pick either nothing, or one residue $1$ (from $1$ or $4$ ) and one residue $-1$ (from $2$ ), and we can then optionally add a residue $0$ (from $3$ ). This gives $(1+2\cdot1)\cdot2 = \boxed{6}$ possibilities.


Answer: 1524.0

Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution:
Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\sum_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{1524}$ monotonous numbers.

Solution:
Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{1524}$

Solution:
Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$ . So, this case gives us $511$
Onto the second case, if there are no $0$ s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$ , there are $2$ solutions. This gives \[2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022\] possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$ ).
All in all, that gives $511 + 1022 - 9 = \boxed{1524}$ monotonous numbers.

Solution:
Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$ , we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$ . When $n=1$ , there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.
The answer is:
$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n}  -  \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) -  \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{1524}$

Solution:
We have 2 cases.
Case 1: Ascending order
We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ $0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number of ascending integers. So, $2^9=512$ . However, the empty set ( $\emptyset$ ) is not an integer, so we must subtract 1. Thus, $512-1=511$
Case 2: Descending order
Similarly, any subset of the digits $0$ to $9$ can be rearranged into a descending monotonous number. So, $2^{10}=1024$ . However, $\emptyset$ and $0$ are not positive integers, so we must subtract 2. Thus, $1024-2=1022$
We have covered all the cases. We add $511$ to $1022$ , giving us $1533$ . So now we just innocently go ahead and choose $\textbf{(C) } 1533$ as our answer, right? No! We overcounted the $9$ single-digit integers . The answer is actually $1533-9=\boxed{1524}$


Answer: 1524.0

Solution:
Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\sum_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{1524}$ monotonous numbers.

Solution:
Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{1524}$

Solution:
Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$ . So, this case gives us $511$
Onto the second case, if there are no $0$ s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$ , there are $2$ solutions. This gives \[2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022\] possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$ ).
All in all, that gives $511 + 1022 - 9 = \boxed{1524}$ monotonous numbers.

Solution:
Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$ , we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$ . When $n=1$ , there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.
The answer is:
$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n}  -  \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) -  \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{1524}$

Solution:
We have 2 cases.
Case 1: Ascending order
We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ $0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number of ascending integers. So, $2^9=512$ . However, the empty set ( $\emptyset$ ) is not an integer, so we must subtract 1. Thus, $512-1=511$
Case 2: Descending order
Similarly, any subset of the digits $0$ to $9$ can be rearranged into a descending monotonous number. So, $2^{10}=1024$ . However, $\emptyset$ and $0$ are not positive integers, so we must subtract 2. Thus, $1024-2=1022$
We have covered all the cases. We add $511$ to $1022$ , giving us $1533$ . So now we just innocently go ahead and choose $\textbf{(C) } 1533$ as our answer, right? No! We overcounted the $9$ single-digit integers . The answer is actually $1533-9=\boxed{1524}$


Answer: 8.0

Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$ . For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\\ \textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.}$

Solution: Define a super-special number to be a number whose decimal expansion only consists of $0$ 's and $1$ 's. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\frac{1}{7}=0.142857142857\hdots$ . This can be done in $8$ numbers if we take \[0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots\] Now assume for sake of contradiction that we can do this with strictly less than $8$ super-special numbers (in particular, less than $10$ .) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a $1$ in that place. This means that in order to obtain the $8$ in $0.1428\hdots$ , there must be $8$ super-special numbers, so the answer is $\boxed{8}$


Answer: 252.0

Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$

Solution:
We shall solve this problem by doing casework on the lowest element of the subset. Note that the number $1$ cannot be in the subset because $1*1=1$ . Let $S$ be a product-free set. If the lowest element of $S$ is $2$ , we consider the set $\{3, 6, 9\}$ . We see that 5 of these subsets can be a subset of $S$ $\{3\}$ $\{6\}$ $\{9\}$ $\{6, 9\}$ , and the empty set). Now consider the set $\{5, 10\}$ . We see that 3 of these subsets can be a subset of $S$ $\{5\}$ $\{10\}$ , and the empty set). Note that $4$ cannot be an element of $S$ , because $2$ is. Now consider the set $\{7, 8\}$ . All four of these subsets can be a subset of $S$ . So if the smallest element of $S$ is $2$ , there are $5*3*4=60$ possible such sets.
If the smallest element of $S$ is $3$ , the only restriction we have is that $9$ is not in $S$ . This leaves us $2^6=64$ such sets.
If the smallest element of $S$ is not $2$ or $3$ , then $S$ can be any subset of $\{4, 5, 6, 7, 8, 9, 10\}$ , including the empty set. This gives us $2^7=128$ such subsets.
So our answer is $60+64+128=\boxed{252}$

Solution: We will consider the $2^9 = 512$ subsets that do not contain 1. A subset is product-free if and only if it does not contain one of the groups $\{2, 4\}, \{3, 9\}, \{2, 3, 6\},$ or $\{2, 5, 10\}$ . There are $2^7$ subsets that contain 2 and 4 since each of the seven elements other than 2 and 4 can either be in the subset or not. Similarly, there are $2^7$ subsets that contain 3 and 9, $2^6$ subsets that contain 2, 3, and 6, and $2^6$ subsets that contain 2, 5, and 10. The number of sets that contain one of the groups is: \[2^7 + 2^7 + 2^6 + 2^6 = 384\] For sets that contain two of the groups, we have: \[2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4 = 160\] For sets that contain three of the groups, we have: \[2^4 + 2^3 + 2^3 + 2^3 = 40\] For sets that contain all of the groups, we have: \[2^2 = 4\] By the principle of inclusion and exclusion, the number of product-free subsets is \[512 - 384 + 160 - 40 + 4 = \boxed{252}\]

Solution:
Let $X$ be a product-free subset, and note that 1 is not in $x$ . We consider four cases:
1.) both 2 and 3 are not in $X$ . Then there are $2^7=128$ possible subsets for this case.
2.) 2 is in $X$ , but 3 is not. Then 4 in not in $X$ , so there are $2^6=64$ subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$ , so there are $64\cdot \frac{3}{4}=48$ subsets for this case.
3.) 2 is not in $X$ , but 3 is. Then, 9 is not in $X$ , so there are $2^6=64$ subsets for this case.
4.) 2 and 3 are both in $X$ . Then, 4, 6, and 9 are not in $X$ , so there are $2^4=16$ total subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$ , so there are $16\cdot \frac{3}{4}=12$ subsets for this case.
Hence our answer is $128+48+64+12=\boxed{252}$ . -Stormersyle

Solution:
Note that if any element $s\geq6$ makes an invalid set, it must be $c$ of $ab=c$ . Otherwise, $ab\geq12>10$ so no $ab=c$ will suffice. Therefore, whether or not an element $s\geq6$ depends only on the previous elements $1\leq\omega\leq5$
First, if a set is product-free, it must never contain an element $\omega=1$ or $1\cdot1=1$
We do casework on the subset of $S\in{1,2,3,4,5}=s_1$ to determine $S\in{6,7,8,9,10}=s_2$ (There are only 12 to cover so we just list them out):
When it is empty, there are clearly no restrictions for $s_2$ so there are $2^5=32$ cases.
$s_1={2}, s_1={4}, s_1={5}$ has no restrictions $s_2$ so we get $2^5\cdot3=96$ total cases.
$s_1=3$ only cannot contain 9 so there are $2^4=16$ cases.
$s_1={2, 5}; s_1={3, 4}; s_1={3, 5}$ each have one restriction in $s_2$ so there are $2^4\cdot3=48$ ways.
$s_1={2, 3}$ has 2 restrictions (6 and 9) so there are $2^3=8$ ways.
$s_1={4, 5}$ has no restrictions so there are $2^5=32$ ways.
$s_1={2, 3, 5}$ has 3 restrictions (6, 9, and 10) so there are $2^2=4$ ways.
$s_1={3, 4, 5}$ has 1 restriction so there are $2^4=16$ cases.
Summing the numbers, we get $32+96+16+48+8+32+4+16=\boxed{252}$


Answer: 129.0

Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$

Solution:
Let $S_{n}$ denote the number of spacy subsets of $\{ 1, 2, ... n \}$ .  We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$
The spacy subsets of $S_{n + 1}$ can be divided into two groups:
Hence,
From this recursion , we find that
And so the answer is $\boxed{129}$

Solution:
Let us consider each size of subset individually. Since each integer in the subset must be at least $3$ away from any other integer in the subset, the largest spacy subset contains $4$ elements.
First, it is clear that there is $1$ spacy set with $0$ elements in it, the empty set. Next, there are $12$ spacy subsets with $1$ element in them, one for each integer $1$ through $12$
Now, let us consider the spacy subsets with $2$ elements in them. If the smaller integer is $1$ , the larger integer is any of the $9$ integers from $4$ to $12$ . If the smaller integer is $2$ , the larger integer is any of the $8$ integers from $5$ to $12$ . This continues, up to a smaller integer of $9$ and $1$ choice for the larger integer, $12$ . This means that there are $9 + 8 + \cdots + 1 = 45$ spacy subsets with $2$ elements.
For spacy subsets with $3$ elements, we first consider the middle integer. The smallest such integer is $4$ , and it allows for $1$ possible value for the smaller integer ( $1$ ) and possible $6$ values for the larger integer ( $7$ through $12$ ), for a total of $1 \cdot 6 = 6$ possible subsets. The next middle integer, $5$ , allows for $2$ smaller integers and $5$ larger integers, and this pattern continues up until the middle integer of $9$ , which has $6$ values for the smaller integer and $1$ value for the larger integer. This means that there are $1 \cdot 6 + 2 \cdot 5 + \cdots + 6 \cdot 1 = 56$ spacy subsets with $3$ elements.
Lastly, there are $3$ main categories for spacy subsets with $4$ elements, defined by the difference between their smallest and largest values. The difference ranges from $9$ to $11$ . If it is $9$ , there is only $1$ set of places to put the two middle values ( $n + 3$ and $n + 6$ , where $n$ is the smallest value). Since there are $3$ possible sets of smallest and largest values, there are $1 \cdot 3 = 3$ sets in this category. If the difference is $10$ , there are now $3$ sets of places to put the two middle values ( $n + 3$ and $n + 6$ or $7$ , and $n + 4$ and $n + 7$ ). There are $2$ possible sets of smallest and largest values, so there are $3 \cdot 2 = 6$ sets in this category. Finally, if the difference is $11$ , there are $6$ possible sets of places to put the two middle values ( $n + 3$ and $n + 6$ $7$ , or $8$ $n + 4$ and $n + 7$ or $8$ , and $n + 5$ and $n + 8$ ) and one possible set of smallest and largest values, meaning that there are $6 \cdot 1 = 6$ sets in this category. Adding them up, there are $3 + 6 + 6 = 15$ spacy subsets with $4$ elements.
Adding these all up, we have a total of $1 + 12 + 45 + 56 + 15 = \boxed{129}$ spacy subsets. ~ emerald_block


Answer: 129.0

Solution:
Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.
From the set $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents ${1,4,7,10}$
For subsets of size $k$ there must be $2(k - 1)$ dividers between the balls, leaving $12 - k - 2(k - 1) = 12 - 3k + 2$ dividers to be be placed in $k + 1$ spots between the balls. The number of way this can be done is $\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k$
Therefore, the number of spacy subsets is $\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = \boxed{129}$

Solution: A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most $4$ elements. Given any arrangment, we subract $2i-2$ from the $i-th$ element in our subset, when the elements are arranged in increasing order. This creates a bijection with the number of size $k$ subsets of the set of the first $14-2k$ positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for $k$ ${14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}$

Solution:
We build a modified version of Markov's chain to solve this problem. First, start off by representing each element of the chosen subset with a $1$ , and every other element $0$ . The possible states are as follows; if the restriction was broken previously or not. For the sake of not over counting, we let the former be its own state (call it "C"), and the latter will be split up into more states. The previous two digits can either be $00$ $10$ , or $01$ . If a previous digit doesn't exist, it is $0$ for obvious reasons. We know that we start off at state " $00$ ". We have two options; include the first digit ( $1$ ) or not ( $0$ ). Thus, state $00$ transitions to state $01$ in $1$ case and state $00$ in $1$ case. State $01$ transitions to $10$ in $1$ case and the "C" state in $1$ case. State $10$ transitions to $00$ in $1$ case and state "C" in $1$ case. Since it doesn't matter where or how many times the restriction is broken for the subset to fail, state "C" can transition to itself in $2$ cases.
Notice how this corresponds to a transition matrix of $\begin{bmatrix}  	1 & 1 & 0 & 0\\ 	0 & 0 & 1 & 1\\ 	1 & 0 & 0 & 1 \\         0 & 0 & 0 & 2 \\ \end{bmatrix}$ . Since we start off on state $00$ , our starting state matrix is $\begin{bmatrix}  	1 & 0 & 0 & 0\\ \end{bmatrix}$ . We wish to compute the value $a+b+c = 4096-d$ when given $f^{12}(\begin{bmatrix}  	1 & 0 & 0 & 0\\ \end{bmatrix})=\begin{bmatrix}  	a & b & c & d\\ \end{bmatrix}$ , where $f(x)$ is $x\times\begin{bmatrix}  	1 & 1 & 0 & 0\\ 	0 & 0 & 1 & 1\\ 	1 & 0 & 0 & 1 \\         0 & 0 & 0 & 2 \\ \end{bmatrix}$ Through repeated applications of matrix multiplication, we obtain the output matrix of $\begin{bmatrix}  	60 & 41 & 28 & 3967\\ \end{bmatrix}$ . Our answer is thus $60+41+28=\boxed{129}$ .
Remark: Markov's Chain is more commonly used to calculate probabilities, but it also can be very useful for counting problems with a uniform requirement.
-SigmaPiE


Answer: 31.0

Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

Solution:
Let the terms be $a-b$ $a$ , and $a+b$ . Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$ , or $3a^2+2b^2=ab^2$ . Rearranging, we get $b^2=\frac{3a^2}{a-2}$ . Simplifying further, $b^2=3a+6+\frac{12}{a-2}$ . Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\pm1, 2, 3, 4, 6, 12$ . Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$ . After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=\boxed{031}$ .
~JHawk0224

Solution:
Let the common difference be $d$ and let the middle term be $x$ . Then, we have that the sequence is \[x-d,~x,~x+d.\] This means that the sum of the squares of the 3 terms of the sequence is \[(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.\] We know that this must be equal to $xd^2,$ so we can write that \[3x^2+2d^2=xd^2,\] and it follows that \[3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.\]
Now, we can treat $d$ as a constant and use the quadratic formula to get \[x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.\] We can factor pull $d^2$ out of the square root to get \[x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.\] Here, it is easy to figure out the values of $d$ . Let $\sqrt{d^2-24} = k$ , then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ note that $d$ $k$ are integers. Examining the parity, we find that $d+k$ and $d-k$ are of the same parity. Now, we solve by factoring. We can find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\sqrt{d^2-24}$ a positive integer. $^{*}$ $d=5$ gives $x=5$ and $x=\frac{10}{3}$ , but we can ignore the latter. $d=7$ gives $x=14$ , as well as a fraction which we can ignore.
Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\boxed{031}$ . -BorealBear, minor edit by Kinglogic

Solution:
As in Solution 1, write the three integers in the sequence as $a-d$ $a$ , and $a+d$
Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$
Setting this equal to the middle term times the common difference squared, which is $ad^2$
and solving for $d^2$ we get:
$3a^2+2d^2 = ad^2 \implies ad^2-2d^2 = 3a^2 \implies d^2(a-2) = 3a^2 \implies d^2 = \frac{3a^2}{a-2}$
The numerator has to be positive, so the denominator has to be positive too for the sequence
to be strictly increasing; that is, $a>2$
For $\frac{3a^2}{a-2}$ to be a perfect square, $\frac{3}{a-2}$ must be a perfect square as well.
This means that $a-2$ is divisible by 3, and whatever left over is a perfect square.
We can express this as an equation: let the perfect square left over be $n^2$ . Then:
$3n^2 = a-2$ . Now when you divide the numerator and denominator by 3, you are left with
$d^2 = \frac{a^2}{n^2} \implies d = \frac{a}{n}$ . Because the sequence is of integers, d must also be an
integer, which means that $n$ must divide $a$
Taking the above equation we can solve for $a$ $3n^2 = a-2 \implies a = 3n^2+2$
This means that $3n^2+2$ is divisible by $n$ $3n^2$ is automatically divisible by $n$ , so
$2$ must be divisible by $n$ . Then $n$ must be either of $\{1,2\}$ . Plugging back into the
equation,
$n = 1 \implies a = 5 \implies d = 5$ , so $a+d = 5+5 = 10$
$n = 2 \implies a = 14 \implies d = 7$ , so $a+d = 14+7 = 21$
Finally, $10+21 = \boxed{031}$

Solution: Following from previous solutions, we derive $3x^2+2a^2=xa^2.$ We divide both sides to get $3\left(\frac{x}{a}\right)^2+2=x.$ Since $x$ is an integer, $\frac{x}{a}$ must also be an integer, so we have $x=pa$ , for some factor $p$ . We then get $3p^2+2=pa.$ We then take this to modulo $p$ , getting $2\equiv 0 \pmod p.$ The only possibilities for $p$ are therefore 1 and 2. We plug these into $3p^2+2=pa$ , for $a=5$ and $x=5$ , giving us the sequence $0,5,10$ , or $2a=14$ and $x=14$ , for the sequence $7,14,21.$ $10+21 = \boxed{031}.$