Answer: 80.0

Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$

Solution: Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3,$ since $x_5x_1$ is never divisible by $3,$ now we just need to find the number of ways $x_4+x_2$ is divisible by $3.$ Note that $x_2$ and $x_4$ can be $(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),$ or $(5, 4).$ We have $2$ ways to designate $x_1$ and $x_5$ for a total of $8 \cdot 2 = 16.$ So the desired answer is $16 \cdot 5=\boxed{080}.$

Solution:
The expression $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$ has cyclic symmetry. Without the loss of generality, let $x_1=3.$ It follows that $\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.$ We have:
We construct the following table for the case $x_1=3,$ with all values in modulo $3:$ \[\begin{array}{c||c|c|c|c|c||c} & & & & & & \\ [-2.5ex] \textbf{Row} & \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \checkmark \\ 2 & 1 & 2 & 1 & 2 & 0 & \checkmark \\ 3 & 1 & 2 & 2 & 1 & 2 & \\ 4 & 2 & 1 & 1 & 2 & 1 & \\ 5 & 2 & 1 & 2 & 1 & 0 & \checkmark \\ 6 & 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}\] For Row 1, $(x_2,x_3)$ can be either $(1,4)$ or $(4,1),$ and $(x_4,x_5)$ can be either $(2,5)$ or $(5,2).$ By the Multiplication Principle, Row 1 produces $2\cdot2=4$ permutations. Similarly, Rows 2, 5, and 6 each produce $4$ permutations.
Together, we get $4\cdot4=16$ permutations for the case $x_1=3.$ By the cyclic symmetry, the cases $x_2=3, x_3=3, x_4=3,$ and $x_5=3$ all have the same count. Therefore, the total number of permutations $x_1, x_2, x_3, x_4, x_5$ is $16\cdot5=\boxed{080}.$

Solution:
WLOG, let $x_{3} = 3$ So, the terms $x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}$ are divisible by $3$
We are left with $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$ .
We need $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 \pmod{3}$ .
The only way is when They are $(+1,-1)$ or $(-1, +1) \pmod{3}$
The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1\pmod{3}$ respectively.
$+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ $= +1 \cdot +1 \cdot +1$ $\;\;\; OR \;\;\;+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ ) = $-1 \cdot -1 \cdot +1$
$-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ $= -1 \cdot -1 \cdot -1$ $\;\;\; OR \;\;\;-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ ) = $-1 \cdot +1 \cdot +1$
But, as we have just two $+1's$ and two $-1's$ .
Hence, We will have to take $+1 = +1 \cdot -1 \cdot -1$ and $-1 = -1 \cdot +1 \cdot +1$ .
Among these two, we have a $+1$ and $-1$ in common, i.e. $(x_{5}, x_{1}) = (+1, -1) or (-1, +1)$ (because $x_{1}$ and $x_{5}$ .
are common in $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$ ).
So, $(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}$ i.e. $8$ values.
For each value of $(x_{5}, x_{1})$ we get $2$ values for $(x_{2}, x_{4})$ .
Hence, in total, we have $8 \times 2 = 16$ ways.
But any of the $x_{i} 's$ can be $3$ .
So, $16 \times 5 = \boxed{080}$

Solution:
This is my first time doing a solution (feel free to edit it)
We have \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2.\]
We have $5$ numbers. Considering any $x$ as $3,$ we see that we are left with two terms that are not always divisible by $3,$ which means that already gives us 5 options.
Let's now consider $x_1 =3:$ We are left with \[3x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + 3x_4x_5 + 3x_5x_2.\] The two terms left over are \[x_2x_3x_4 + x_3x_4x_5 \equiv 0 \pmod{3}\] since we already have used $3$ the remaining numbers are $1,2,4,5.$
We now factor \begin{align*} (x_2 + x_5)(x_3x_4) &\equiv 0 \pmod{3} \\ (x_2 + x_5) &\equiv 0 \pmod{3} \end{align*} since $1,2,4,5$ are all not factors of $3.$
Now using the number $1,2,4,5,$ we take two to get a number divisible by $3$ for $(x_2 + x_5):$ \begin{align*} 1+5 &\equiv 0 \pmod{3}, \\ 4+2 &\equiv 0 \pmod{3}, \\ 4+5 &\equiv 0 \pmod{3}, \\ 1+2 &\equiv 0 \pmod{3}. \end{align*} We have $4$ possibilities from above. Since we can also have $5+1$ or $2+4,$ there are $4\cdot2=8$ possibilities in all.
Now using \[(x_2 + x_5)(x_3x_4) \equiv 0 \pmod{3},\] we have $(x_3x_4),$ which results in $8$ more possibilities of $2$ times more. So, we get $2\cdot2\cdot4=16.$
Remember that $3$ can be any of $5$ different variables. So, we multiply by $5$ to get the answer $\boxed{080}.$


Answer: 461.0

Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$

Solution:
If the first number is $6$ , then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If the first number is $5$ $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If the first number is $4$ , ....
4 6 _ _ _ _ $\implies$ 24 ways
4 _ 6 _ _ _ $\implies$ 24 ways
4 _ _ 6 _ _ $\implies$ 24 ways
4 _ _ _ 6 _ $\implies$ 5 must go between $4$ and $6$ , so there are $3 \cdot 3! = 18$ ways.
$24 + 24 + 24 + 18 = 90$ ways if 4 is first.
If the first number is $3$ , ....
3 6 _ _ _ _ $\implies$ 24 ways
3 _ 6 _ _ _ $\implies$ 24 ways
3 1 _ 6 _ _ $\implies$ 4 ways
3 2 _ 6 _ _ $\implies$ 4 ways
3 4 _ 6 _ _ $\implies$ 6 ways
3 5 _ 6 _ _ $\implies$ 6 ways
3 5 _ _ 6 _ $\implies$ 6 ways
3 _ 5 _ 6 _ $\implies$ 6 ways
3 _ _ 5 6 _ $\implies$ 4 ways
$24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84$ ways
If the first number is $2$ , ....
2 6 _ _ _ _ $\implies$ 24 ways
2 _ 6 _ _ _ $\implies$ 18 ways
2 3 _ 6 _ _ $\implies$ 4 ways
2 4 _ 6 _ _ $\implies$ 6 ways
2 5 _ 6 _ _ $\implies$ 6 ways
2 5 _ _ 6 _ $\implies$ 6 ways
2 _ 5 _ 6 _ $\implies$ 4 ways
2 4 _ 5 6 _ $\implies$ 2 ways
2 3 4 5 6 1 $\implies$ 1 way
$24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ ways
Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Solution:
If $6$ is the first number, then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$ , and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If $6$ is the third number, then we cannot have the following:
1 _ 6 _ _ _ $\implies$ 24 ways
2 1 6 _ _ _ $\implies$ 6 ways
$120 - 24 - 6 = 90$ ways
If $6$ is the fourth number, then we cannot have the following:
1 _ _ 6 _ _ $\implies$ 24 ways
2 1 _ 6 _ _ $\implies$ 6 ways
2 3 1 6 _ _ $\implies$ 2 ways
3 1 2 6 _ _ $\implies$ 2 ways
3 2 1 6 _ _ $\implies$ 2 ways
$120 - 24 - 6 - 2 - 2 - 2 = 84$ ways
If $6$ is the fifth number, then we cannot have the following:
_ _ _ _ 6 5 $\implies$ 24 ways
1 5 _ _ 6 _ $\implies$ 6 ways
1 _ 5 _ 6 _ $\implies$ 6 ways
2 1 5 _ 6 _ $\implies$ 2 ways
1 _ _ 5 6 _ $\implies$ 6 ways
2 1 _ 5 6 _ $\implies$ 2 ways
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 $\implies$ 3 ways
$120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71$ ways
Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Solution: Note the condition in the problem is equivalent to the following condition: for each $k$ with $1 \le k \le 5$ , the first $k$ terms is not a permutation $(1, 2, \ldots, k)$ (since it would mean there must be some integer $x$ in the first $k$ terms such that $x \not \in \{1, 2, \ldots, k\} \implies x > k$ ). Then, let $a_n$ denote the number of permutations of $(1, 2, \ldots, n)$ satisfying the condition in the problem. We use complementary counting to find $a_n$ . Notice that in order to not satisfy the condition in the problem, there are $n-1$ cases: the first $1 \le k \le n-1$ (we don't include $k = n$ since the condition in the problem only holds up to $n-1$ ) terms are a permutation of $(1, 2, \ldots, k)$ , and for all $k+1 \le i \le n-1$ , the condition in the problem still holds. Then, for each of these cases, there are $k!$ ways to arrange the first $k$ terms, and then $a_{n-k}$ ways to arrange the $k + 1$ to $n$ terms (assume by contradiction that terms from $k+1$ to $i$ is a permutation of $(k+1, k+2, \ldots, i)$ . Then, since the first $k$ terms are already a permutation of $(1, 2, \ldots, k)$ , the first $i$ terms form a permutation of $(1, 2, \ldots, i)$ . This contradicts our assumption that for all $k+1 \le i \le n-1$ , the condition still holds. Thus, we can only include the $a_{n-k}$ permutations of these terms). Now, summing the cases up and subtracting from $n!$ , we have: $a_n = n! - \sum_{k=1}^{n-1} a_{n-k} k!$ . From this recursion, we derive that $a_1 = 1$ $a_2 = 1$ $a_3 = 3$ $a_4 = 13$ $a_5 = 71$ , and finally $a_6 = \boxed{461}$

Solution:
Let $A_i$ be the set of permutations such that there is no number greater than $i$ in the first $i$ places. Note that $\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}$ for all $1\le b_0 < b_1\cdots < b_k \le 5$ and that the set of restricted permutations is $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$
We will compute the cardinality of this set with PIE. \begin{align*} &|A_1| + |A_2| + |A_3| + |A_4| + |A_5|\\ = &120 + 48 + 36 + 48 + 120 = 372\\ \\ &|A_1 \cap A_2| + |A_1 \cap A_3| + |A_1 \cap A_4| + |A_1 \cap A_5| + |A_2 \cap A_3|\\ + &|A_2 \cap A_4| + |A_2 \cap A_5| + |A_3 \cap A_4| + |A_3 \cap A_5| + |A_4 \cap A_5|\\=&24+12+12+24+12+8+12+12+12+24=152\\ \\ &|A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_2 \cap A_5| + |A_1 \cap A_3 \cap A_4| + |A_1 \cap A_3 \cap A_5|\\ +& |A_1 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_5| +  |A_2 \cap A_4 \cap A_5| + |A_3 \cap A_4 \cap A_5|\\=&6 + 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6 = 48\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4| + |A_1 \cap A_2 \cap A_3 \cap A_5| + |A_1 \cap A_2 \cap A_4 \cap A_5| + |A_1 \cap A_3 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4 \cap A_5|\\=&2 + 2 + 2 + 2 + 2 = 10\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = 1 \end{align*} To finish, $720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}$

Solution:
Define the function $f(p,q)$ as the amount of permutations with maximum digit $q$ and string length $p$ that satisfy the condition within bounds. For example, $f(4,5)$ would be the amount of ways to make a string with length $4$ with the highest digit being $5$ . We wish to obtain $f(6,6)=f(5,6)$
To generate recursion, consider how we would get to $f(p,q)$ from $f(p-1,a)$ for all $a$ such that $p\le{a}\le6$ . We could either jump from the old maximum $a$ to the new $q$ by concatenating the old string and the new digit $q$ , or one could retain the maximum, in which case $a=q$ . To retain the maximum, one would have to pick a new available digit not exceeding $q$
In the first case, there is only one way to pick the new digit, namely picking $q$ . For the second case, there are $q-p+1$ digits left to choose, because there are $q$ digits between 1 and $q$ total and there are $p-1$ digits already chosen below or equal to $q$ . Thus, $f(p,q)=[\sum^{q-1}_{n=p}f(p-1,n)] + (q-p+1)f(p-1,q)$ . Now that we have the recursive function, we can start evaluating the values of $f(p,q)$ until we get to $f(6,6)=f(5,6)$
\[f(2,3)=3, f(2,4)=5, f(2,5)=7, f(2,6)=9\] \[f(3,4)=13, f(3,5)=29, f(3,6)=51\] \[f(4,5)=71, f(4,6)=195\] \[f(5,6)=461\] Our requested answer is thus $\boxed{461}$ ~sigma

Solution:
We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first $k$ terms is greater than $k$ for $1\leq$ $k$ $\leq$ $5$ . We can further simplify this method by approaching it through casework on the first $k$ terms.
Case 1: None of the first one terms is greater than 1
The first term must obviously be one. Since there are five spaces left for numbers, there are a total of $5!=120$ permutations for this case.
Case 2: None of the first two terms is greater than 2
The first two terms must be 1 and 2 in some order. However, we already counted all cases starting with 1 in the previous case, so all of the permutations in this case must begin with $12\cdots$ . Since there are four spaces left, there are a total of $4!=24$ permutations for this case.
Case 3: None of the first three terms is greater than 3
The first three terms must be 1, 2, and 3 in some order. However, the cases beginning with 1__ and 21_ have already been accounted for. There are now $3!-3 = 3$ ways to order the first three numbers of the permutation, and $3!$ ways to order the last three numbers, for a total of $3\times6 = 18$ permutations.
Case 4: None of the first four terms is greater than 4
You can see where the pattern is going - the first four terms must be 1, 2, 3, and 4 in some order. All cases starting with 1 (there are $3!=6$ ), the cases starting with 21 (there are $2!=2$ ), and the 3 cases from case 3 (there are $3\times 1! = 3$ ) have been accounted for, so there are a total of $(4!-6-2-3)2!=26$ permutations for this case.
Case 5: None of the first five terms is greater than 5
This is perhaps the hardest case to work with, simply because there are so many subcases, so keeping track is crucial here. Obviously, the first five terms must be 1, 2, 3, 4, and 5, meaning there are 120 ways to order them. Now, we count the permutations we have already counted in previous cases. $4!$ start with 1, $3!$ start with 2, $3\times2!=6$ start with 3, and $13\times1!=13$ start with 4. Subtracting, we get a total of $120-24-6-6-13=71$ permutations.
Now, we subtract the total number of permutations from our cases from the total number of permutations, which is $6!$ $720 - 120 - 24 - 18 - 26 - 71 = \boxed{461}$


Answer: 16.0

Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\]

Solution: $m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}$ . Now, divide by $m^{x_0}$ to get $1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}$ . Notice that since we can choose all nonnegative $x_0,...,x_{2011}$ , we can make $x_n-x_0$ whatever we desire. WLOG, let $x_0\geq...\geq x_{2011}$ and let $a_n=x_n-x_0$ .  Notice that, also, $m^{a_{2011}}$ doesn't matter if we are able to make $m^{a_1} +m^{a_2} + .... + m^{a_{2010}}$ equal to $1-\left(\frac{1}{m}\right)^x$ for any power of $x$ . Consider $m=2$ . We can achieve a sum of $1-\left(\frac{1}{2}\right)^x$ by doing $\frac{1}{2}+\frac{1}{4}+...$ (the "simplest" sequence). If we don't have $\frac{1}{2}$ , to compensate, we need $2\cdot 1\frac{1}{4}$ 's. Now, let's try to generalize. The "simplest" sequence is having $\frac{1}{m}$ $m-1$ times, $\frac{1}{m^2}$ $m-1$ times, $\ldots$ . To make other sequences, we can split $m-1$ $\frac{1}{m^i}$ s into $m(m-1)$ $\text{ }\frac{1}{m^{i+1}}$ s since $(m-1)\cdot\frac{1}{m^{i}} = m(m-1)\cdot\frac{1}{m^{i+1}}$ . Since we want $2010$ terms, we have $\sum$ $(m-1)\cdot m^x=2010$ . However, since we can set $x$ to be anything we want (including 0), all we care about is that $(m-1)\text{ }|\text{ } 2010$ which happens $\boxed{016}$ times.

Solution:
Let \[P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}\] . The problem then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_{2011}$ are nonnegative integers. We plug in $m = 1$ and see that \[P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010\] Now, we can say that \[P(m) = (m-1)Q(m) - 2010\] for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$ $(m-1)Q(m) = 2010$ . Thus, if $P(m) = 0$ , then $m-1 | 2010$ .
Now, we need to show that \[m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}\forall m-1|2011\] We try with the first few $m$ that satisfy this.
For $m = 2$ , we see we can satisfy this if $x_0 = 2010$ $x_1 = 2009$ $x_2 = 2008$ $\cdots$ $x_{2008} = 2$ $x_{2009} = 1$ $x_{2010} = 0$ $x_{2011} = 0$ , because \[2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots\] (based on the idea $2^n + 2^n = 2^{n+1}$ , leading to a chain of substitutions of this kind) \[= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}\] . Thus $2$ is a possible value of $m$ . For other values, for example $m = 3$ , we can use the same strategy, with \[x_{2011} = x_{2010} = x_{2009} = 0\] \[x_{2008} = x_{2007} = 1\] \[x_{2006} = x_{2005} = 2\] \[\cdots\] \[x_2 = x_1 = 1004\] and \[x_0 = 1005\] , because \[3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004}\] \[=\cdots = 3^{1004} +3^{1004}+3^{1004} = 3^{1005}\] It's clearly seen we can use the same strategy for all $m-1 |2010$ . We count all positive $m$ satisfying $m-1 |2010$ , and see there are $\boxed{016}$

Solution:
One notices that $m-1 \mid 2010$ if and only if there exist non-negative integers $x_0,x_1,\ldots,x_{2011}$ such that $m^{x_0} = \sum_{k=1}^{2011}m^{x_k}$
To prove the forward case, we proceed by directly finding $x_0,x_1,\ldots,x_{2011}$ . Suppose $m$ is an integer such that $m^{x_0} = \sum_{k=1}^{2011}m^{x_k}$ . We will count how many $x_k = 0$ , how many $x_k = 1$ , etc. Suppose the number of $x_k = 0$ is non-zero. Then, there must be at least $m$ such $x_k$ since $m$ divides all the remaining terms, so $m$ must also divide the sum of all the $m^0$ terms. Thus, if we let $x_k = 0$ for $k = 1,2,\ldots,m$ , we have, \[m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.\] Well clearly, $m^{x_0}$ is greater than $m$ , so $m^2 \mid m^{x_0}$ $m^2$ will also divide every term, $m^{x_k}$ , where $x_k \geq 2$ . So, all the terms, $m^{x_k}$ , where $x_k < 2$ must sum to a multiple of $m^2$ . If there are exactly $m$ terms where $x_k = 0$ , then we must have at least $m-1$ terms where $x_k = 1$ . Suppose there are exactly $m-1$ such terms and $x_k = 1$ for $k = m+1,m+2,2m-1$ . Now, we have, \[m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.\] One can repeat this process for successive powers of $m$ until the number of terms reaches 2011. Since there are $m + j(m-1)$ terms after the $j$ th power, we will only hit exactly 2011 terms if $m-1$ is a factor of 2010. To see this, \[m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.\] Thus, when $j = 2010/(m-1) - 1$ (which is an integer since $m-1 \mid 2010$ by assumption, there are exactly 2011 terms. To see that these terms sum to a power of $m$ , we realize that the sum is a geometric series: \[1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.\] Thus, we have found a solution for the case $m-1 \mid 2010$
Now, for the reverse case, we use the formula \[x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).\] Suppose $m^{x_0} = \sum_{k=1}^{2011}m^{x^k}$ has a solution. Subtract 2011 from both sides to get \[m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).\] Now apply the formula to get \[(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],\] where $a_k$ are some integers. Rearranging this equation, we find \[(m-1)A = 2010,\] where $A = a_0 - \sum_{k=1}^{2011}a_k$ . Thus, if $m$ is a solution, then $m-1 \mid 2010$
So, there is one positive integer solution corresponding to each factor of 2010. Since $2010 = 2\cdot 3\cdot 5\cdot 67$ , the number of solutions is $2^4 = \boxed{016}$

Solution: The problem is basically asking how many integers $m$ have a power that can be expressed as the sum of 2011 other powers of $m$ (not necessarily distinct). Notice that $2+2+4+8+16=32$ $3+3+3+9+9+27+27+81+81=243$ , and $4+4+4+4+16+16+16+64+64+64+256+256+256=1024$ . Thus, we can safely assume that the equation $2011 = (m-1)x + m$ must have an integer solution $x$ . To find the number of $m$ -values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant $m$ to make the equation equal a friendlier number, $2010$ , instead of the ugly prime number $2011$ $2010 = (m-1)x+(m-1)$ . Factor the equation and we get $2010 = (m-1)(x+1)$ . The number of values of $m-1$ that allow $x+1$ to be an integer is quite obviously the number of factors of $2010$ . Factoring $2010$ , we obtain $2010 = 2 \times 3 \times 5 \times 67$ , so the number of positive integers $m$ that satisfy the required condition is $2^4 = \boxed{016}$

Solution:
First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of $m$ we can multiply to get to non-negative powers.
Now we see that our problem is just a matter of m-chopping blocks.
What is meant by $m$ -chopping is taking an existing block of say $m^{k}$ and turning it into $m$ blocks of $m^{k-1}$ . This process increases the total number of blocks by $m-1$ per chop.
The problem wants us to find the number of positive integers $m$ where some number of chops will turn $1$ block into $2011$ such blocks, thus increasing the total amount by $2010= 2 \cdot 3 \cdot 5 \cdot 67$ . Thus $m-1 | 2010$ , and a cursory check on extreme cases will confirm that there are indeed $\boxed{016}$ possible $m$ s.

Solution: The problem is basically saying that we want to find 2011 powers of m that add together to equal another power of m. If we had a power of m, m^n, then to get to m^(n+1) or m*(m^n), we have to add m^n, m-1 times. Then when we are at m^(n+1), to get to m^(n+2), it is similar. So we have to have m^(some number) = m^n + (m-1)(m^n) + (m-1)(m^(n+1))...<=This expression has 1 + (m-1) + (m-1) + ... = 1 + k(m-1) terms, and the number of terms must be equal to 2011 (the problem said). Then 1+k(m-1) = 2011, and we get the equation k*(m-1) = 2010 = 2*3*5*67. 2010 has 16 factors, and setting m-1 = each of these factors makes $\boxed{016}$ values of m.


Answer: 80.0

Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution:
We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$ $5$ , and/or $6$ which are also multiples of $4$ $5$ , and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$ $5$ , and $6$ is $60$ .  We can therefore simplify the problem by finding all such numbers described above between $1$ and $60$ and multiplying the quantity of such numbers by $10$ $600$ $60$ $10$ ).
After making a simple list of the numbers between $1$ and $60$ and going through it, we see that the numbers meeting this condition are $4$ $5$ $15$ $24$ $35$ $44$ $54$ , and $55$ .  This gives us $8$ numbers. $8$ $10$ $\boxed{080}$

Solution: This is Solution 1 with a slick element included.  Solution 1 uses the concept that $60k+l$ is a solution for $n$ if $60k+l$ is a multiple of $3$ $4$ , and/or $5$ and $60k+l+1$ is a multiple of $3$ $4$ , and/or $5$ for positive integer values of $l$ and essentially any integer value of $k$ .  But keeping the same conditions in mind for $k$ and $l$ , we can also say that if $60k+l$ is a solution, then $60k-l-1$ is a solution!  Therefore, one doesn't have to go as far as determining the number of values between $1$ and $60$ and then multiplying by $10$ .  One only has to determine the number of values between $1$ and $30$ and then multiply by $20$ .  The values of $n$ that work between $1$ and $30$ are $4$ $5$ $15$ , and $24$ .  This gives us $4$ numbers. $4$ $20$ $\boxed{080}$

Solution:
1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:
1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $6$ and $n+1$ is a multiple of $5$
3. $n$ is a multiple of $5$ and $n+1$ is a multiple of $4$
4. $n$ is a multiple of $5$ and $n+1$ is a multiple of $6$
Solving for each case, we see that there are $30$ possibilities for cases 1 and 3 each, and $20$ possibilities for cases 2 and 4 each. However, we over-counted the cases where
1. $n$ is a multiple of $24$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $5$ and $n+1$ is a multiple of $24$
Each case has $10$ possibilities.
Adding all the cases and correcting for over-counting, we get $30 + 20 + 30 + 20 - 10 - 10 = \boxed{080}.$

Solution:
Here is a detailed solution for Solution 2.
Over-counted cases:
$30 + 20 + 30 + 20 - 10 - 10 = \boxed{080}$

Solution:
The problem is the same as asking how many unique sets of values of $\lfloor\frac{n}{4}\rfloor$ $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ can be produced by one and only one value of $n$ for positive integers $n$ less than or equal to 600.
Seeing that we are dealing with the unique values of the floor function, we ought to examine when it is about to change values, for instance, when $n$ is close to a multiple of 4 in $\lfloor\frac{n}{4}\rfloor$
For a particular value of $n$ , let $a$ $b$ , and $c$ be the original values of $\lfloor\frac{n}{4}\rfloor$ $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ , respectively.
Notice when $n$ $\equiv0\mod4$ and $n$ $\equiv4\mod5$ , the value of $\lfloor\frac{n-1}{4}\rfloor$ will be 1 less than the original $a$ . The value of $\lfloor\frac{n+1}{5}\rfloor$ will be 1 greater than the original value of $b$
More importantly, this means that no other value less than or greater than $n$ will be able to produce the set of original values of $a$ $b$ , and $c$ , since they make either $a$ or $b$ differ by at least 1.
Generalizing, we find that $n$ must satisfy:
Where $j$ and $k$ are pairs of distinct values of 4, 5, and 6.
Plugging in the values of $j$ and $k$ , finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are $\boxed{080}$ solutions of $n$

Solution:
By Chinese Remainder Theorem , the general solution of systems of $2$ linear congruences is:
$lcm[4, 5, 6] = 60$ , we solve the number of values for $n \le 60$ , then multiply by $10$ to get the number of values for $n \le 600$ . We are going to solve the following $6$ systems of linear congruences:
$n \in \{ 4, 5, 15, 24, 35, 44, 54, 55 \}$ , there are $8$ values for $n \le 60$ . For $n \le 600$ , the answer is $8 \cdot 10 = \boxed{080}$


Answer: 496.0

Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$

Solution:
We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational.
Let $x = a + \frac{b}{c}$ where $a,b,c$ are nonnegative integers and $0 \le b < c$ (essentially, $x$ is a mixed number). Then, \[n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}\]
Here it is sufficient for $\frac{ab}{c}$ to be an integer. We can use casework to find values of $n$ based on the value of $a$
$a = 0 \implies$ nothing because n is positive
$a = 1 \implies \frac{b}{c} = \frac{0}{1}$
$a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}$
$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}$
The pattern continues up to $a = 31$ . Note that if $a = 32$ , then $n > 1000$ . However if $a = 31$ , the largest possible $x$ is $31 + \frac{30}{31}$ , in which $n$ is still less than $1000$ . Therefore the number of positive integers for $n$ is equal to $1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$

Solution: Notice that $x\lfloor x\rfloor$ is continuous over the region $x \in [k, k+1)$ for any integer $k$ . Therefore, it takes all values in the range $[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)$ over that interval. Note that if $k>32$ then $k^2 > 1000$ and if $k=31$ , the maximum value attained is $31*32 < 1000$ . It follows that the answer is $\sum_{k=1}^{31} (k+1)k-k^2  = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$

Solution: Bounding gives $x^2\le n<x^2+x$ . Thus there are a total of $x$ possible values for $n$ , for each value of $x^2$ . Checking, we see $31^2+31=992<1000$ , so there are \[1+2+3+...+31= \boxed{496}\] such values for $n$

Solution:
After a bit of experimenting, we let $n=l^2+s, s < 2n+1$ . We claim that I (the integer part of $x$ ) = $l$ . (Prove it yourself using contradiction !) so now we get that $x=l+\frac{s}{l}$ . This implies that solutions exist iff $s<l$ , or for all natural numbers of the form $l^2+s$ where $s<l$ .
Hence, 1 solution exists for $l=1$ ! 2 for $l=2$ and so on. Therefore our final answer is $31+30+\dots+1= \boxed{496}$


Answer: 134.0

Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer.

Solution: We start with the last two terms of the polynomial $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}$ , which are $\frac{n^5}{5!}+\frac{n^6}{6!}$ . This can simplify to $\frac{6n^5+n^6}{720}$ , which can further simplify to $\frac{n^5(6+n)}{720}$ . Notice that the prime factorization of $720$ is $5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2$ . In order for $\frac{n^5(6+n)}{720}$ to be an integer, one of the parts must divide $5, 3$ , and $2$ . Thus, one of the parts must be a multiple of $5, 3$ , and $2$ , and the LCM  of these three numbers is $30$ . This means \[n^5 \equiv 0\pmod{30}\] or \[6+n \equiv 0\pmod{30}\] Thus, we can see that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$ . Note that as long as we satisfy $\frac{6n^5+n^6}{720}$ $2!, 3!$ , and $4!$ will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. $4! = 2\cdot2\cdot2\cdot2\cdot3$ , and this will be divisible by $2^4\cdot3^4\cdot5^4$ . Now, since we know that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$ in order for the polynomial to be an integer, $n\equiv0, 24\pmod{30}$ . To find how many integers fulfill the equation and are $<2017$ , we take $\left \lfloor\frac{2017}{30} \right \rfloor$ and multiply it by $2$ . Thus, we get $67\cdot2=\boxed{134}$

Solution:
Taking out the $1+n$ part of the expression and writing the remaining terms under a common denominator, we get $\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)$ . Therefore the expression $n^6+6n^5+30n^4+120n^3+360n^2$ must equal $720m$ for some positive integer $m$ .
Taking both sides mod $2$ , the result is $n^6 \equiv 0 \pmod{2}$ . Therefore $n$ must be even. If $n$ is even, that means $n$ can be written in the form $2a$ where $a$ is a positive integer. Replacing $n$ with $2a$ in the expression, $64a^6+192a^5+480a^4+960a^3+1440a^2$ is divisible by $16$ because each coefficient is divisible by $16$ . Therefore, if $n$ is even, $n^6+6n^5+30n^4+120n^3+360n^2$ is divisible by $16$
Taking the equation $n^6+6n^5+30n^4+120n^3+360n^2=720m$ mod $3$ , the result is $n^6 \equiv 0 \pmod{3}$ . Therefore $n$ must be a multiple of $3$ . If $n$ is a multiple of three, that means $n$ can be written in the form $3b$ where $b$ is a positive integer. Replacing $n$ with $3b$ in the expression, $729b^6+1458b^5+2430b^4+3240b^3+3240b^2$ is divisible by $9$ because each coefficient is divisible by $9$ . Therefore, if $n$ is a multiple of $3$ $n^6+6n^5+30n^4+120n^3+360n^2$ is divisibly by $9$
Taking the equation $n^6+6n^5+30n^4+120n^3+360n^2=720m$ mod $5$ , the result is $n^6+n^5 \equiv 0 \pmod{5}$ . The only values of $n (\text{mod }5)$ that satisfy the equation are $n\equiv0(\text{mod }5)$ and $n\equiv4(\text{mod }5)$ . Therefore if $n$ is $0$ or $4$ mod $5$ $n^6+6n^5+30n^4+120n^3+360n^2$ will be a multiple of $5$
The only way to get the expression $n^6+6n^5+30n^4+120n^3+360n^2$ to be divisible by $720=16 \cdot 9 \cdot 5$ is to have $n \equiv 0 \pmod{2}$ $n \equiv 0 \pmod{3}$ , and $n \equiv 0 \text{ or } 4 \pmod{5}$ . By the Chinese Remainder Theorem or simple guessing and checking, we see $n\equiv0,24 \pmod{30}$ . Because no numbers between $2011$ and $2017$ are equivalent to $0$ or $24$ mod $30$ , the answer is $\frac{2010}{30}\times2=\boxed{134}$

Solution: Note that $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}$ will have a denominator that divides $5!$ . Therefore, for the expression to be an integer, $\frac{n^6}{6!}$ must have a denominator that divides $5!$ . Thus, $6\mid n^6$ , and $6\mid n$ . Let $n=6m$ . Substituting gives $1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}$ . Note that the first $5$ terms are integers, so it suffices for $\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}$ to be an integer. This simplifies to $\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)$ . It follows that $5\mid m^5(m+1)$ . Therefore, $m$ is either $0$ or $4$ modulo $5$ . However, we seek the number of $n$ , and $n=6m$ . By CRT, $n$ is either $0$ or $24$ modulo $30$ , and the answer is $67+67=\boxed{134}$

Solution: Clearly $1+n$ is an integer. The part we need to verify as an integer is, upon common denominator, $\frac{360n^2+120n^3+30n^4+6n^5+n^6}{720}$ . Clearly, the numerator must be even for the fraction to be an integer. Therefore, $n^6$ is even and n is even, aka $n=2k$ for some integer $k$ . Then, we can substitute $n=2k$ and see that $\frac{n^2}{2}$ is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get $\frac{60k^3+30k^4+12k^5+4k^6}{45}$ . It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the $4k^6$ , and we see that $k=3b$ for some integer $b$ . From there we now know that $n=6b$ . If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that $n^5(6n+1) \equiv 0 \pmod5$ , so combining with divisibility by 6, $n$ is $24$ or $0\pmod{30}$ . There are $67$ cases for each, hence the answer $\boxed{134}$


Answer: 50.0

Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$

Solution:
We want to find the number of positive integers $n<1000$ which can be written in the form $n = 2^a - 2^b$ for some non-negative integers $a > b \ge 0$ (note that if $a=b$ , then $2^a-2^b = 0$ ). We first observe $a$ must be at most 10; if $a \ge 11$ , then $2^a - 2^b \ge 2^{10} > 1000$ . As $2^{10} = 1024 \approx 1000$ , we can first choose two different numbers $a > b$ from the set $\{0,1,2,\ldots,10\}$ in $\binom{11}{2}=55$ ways. This includes $(a,b) = (10,0)$ $(10,1)$ $(10,2)$ $(10,3)$ $(10,4)$ which are invalid as $2^a - 2^b > 1000$ in this case. For all other choices $a$ and $b$ , the value of $2^a - 2^b$ is less than 1000.
We claim that for all other choices of $a$ and $b$ , the values of $2^a - 2^b$ are pairwise distinct. More specifically, if $(a_1,b_1) \neq (a_2,b_2)$ where $10 \ge a_1 > b_1 \ge 0$ and $10 \ge a_2 > b_2 \ge 0$ , we must show that $2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}$ . Suppose otherwise for sake of contradiction; rearranging yields $2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}$ . We use the fact that every positive integer has a unique binary representation:
If $a_1 \neq b_2$ then $\{a_1,b_2\} = \{a_2,b_1\}$ ; from here we can deduce either $a_1=a_2$ and $b_1=b_2$ (contradicting the assumption that $(a_1,b_1) \neq (a_2,b_2)$ , or $a_1=b_1$ and $a_2=b_2$ (contradicting the assumption $a_1>b_1$ and $a_2>b_2$ ).
If $a_1 = b_2$ then $2^{a_1}+2^{b_2} = 2 \times 2^{a_1}$ , and it follows that $a_1=a_2=b_1=b_2$ , also contradicting the assumption $(a_1,b_1) \neq (a_2,b_2)$ . Hence we obtain contradiction.*
Then there are $\binom{11}{2}-5$ choices for $(a,b)$ for which $2^a - 2^b$ is a positive integer less than 1000; by the above claim, each choice of $(a,b)$ results in a different positive integer $n$ . Then there are $55-5 = \boxed{050}$ integers which can be expressed as a difference of two powers of 2.

Solution:
Case 1: When our answer is in the form $2^n-2^i$ , where $i$ is an integer such that $0\le i\le 4$
We start with the subcase where it is $2^n-2^0$ , for some integer $n$ where $n>0$ (this is because the case where $n=0$ yields $2^0-2^0=0$ , which doesn't work because it must be a positive integer.)
Note that $2^{10}=1024$ , and $2^9=512$ . Our answer needs to be less than $1000$ , so the maximum possible result (in this case) is $2^9-2^0$ . Our lowest result is $2^1-2^0$ . All the positive powers of two less than $1024$ work, so we have $9$ possibilities for this subcase. For subcases $i=1, i=2, i=3,$ and $i=4$ , we have $8, 7, 6,$ and $5$ possibilities, respectively.
Case 2: When our answer is in the form of $2^n-2^j$ , where $j$ is an integer such that $5\le j\le 9$
We can start with the subcase where $j=5$ . We notice that $2^5=32$ , and $2^{10}-2^5=992$ which is less than $1000$ , so the greatest result in this subcase is actually $2^{10}-2^5$ , and the lowest is $2^6-2^5$ . Thus, we have $5$ possibilities. For the other four subcases, we have $4, 3, 2,$ and $1$ possibilities, respectively.
Answer: We note that these are our only cases, as numbers in the form of $2^n-2^{10}$ and beyond are greater than $1000$
Thus, our result is $9+8+7+6+5+5+4+3+2+1=(9+8+7+6+5+4+3+2+1)+5=\boxed{50}$ . ~jehu26

Solution:
We look for all positive integers of the form $2^a-2^b<1000,$ where $0\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \[\begin{array}{c|c} & \\ [-2.25ex] \boldsymbol{a} & \boldsymbol{b} \\  \hline  & \\ [-2ex] 1 & 0 \\   2 & 0,1 \\ 3 & 0,1,2 \\ 4 & 0,1,2,3 \\ 5 & 0,1,2,3,4 \\ 6 & 0,1,2,3,4,5 \\ 7 & 0,1,2,3,4,5,6 \\ 8 & 0,1,2,3,4,5,6,7 \\ 9 & 0,1,2,3,4,5,6,7,8 \\ 10 & \xcancel{0},\xcancel{1},\xcancel{2},\xcancel{3},\xcancel{4},5,6,7,8,9 \\ [0.5ex] \end{array}\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid.
Note that each of the remaining ordered pairs generates one unique desired positive integer.
We prove this statement as follows:
Together, we have justified our claim in bold. The answer is \[1+2+3+4+5+6+7+8+9+5=\boxed{050}.\]

Solution:
Because the difference is less than $1000$ , we can simply list out all numbers that satisfy $2^n < 1000$ . We get $0 \le n < 10$ , where n is an integer. Because the sequence $2^n$ is geometric, the difference of any two terms will be unique. $\binom{10}{2}$ will be the number of differences for $0\le n < 10$ . However, we also need to consider the case in which $n=10$ . With simple counting, we find that $5$ numbers: $(32, 64, 128, 256, 512)$ could be subtracted from $1024$ , which makes another 5 cases. There is no need to check for higher exponents since the lowest difference would be $2^{11} - 2^{10} = 1024$ , which exceeds $1000$
Thus, the final answer is $\binom{10}{2} + 5 = \boxed{050}.$


Answer: 222.0

Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$

Solution: The base- $3$ representation of $2017_{10}$ is $2202201_3$ . Because any $7$ -digit base- $3$ number that starts with $22$ and has no digit equal to $0$ must be greater than $2017_{10}$ , all $7$ -digit numbers that have no digit equal to $0$ must start with $21$ or $1$ in base $3$ . Of the base- $3$ numbers that have no digit equal to $0$ , there are $2^5$ $7$ -digit numbers that start with $21$ $2^6$ $7$ -digit numbers that start with $1$ $2^6$ $6$ -digit numbers, $2^5$ $5$ -digit numbers, $2^4$ $4$ -digit numbers, $2^3$ $3$ -digit numbers, $2^2$ $2$ -digit numbers, and $2^1$ $1$ -digit numbers. Summing these up, we find that the answer is $2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}$

Solution:
Note that $2017=2202201_{3}$ , and $2187=3^7=10000000_{3}$ . There can be a $1,2,...,7$ digit number less than $2187$ , and each digit can either be $1$ or $2$ . So $2^1$ one digit numbers and so on up to $2^7$ $7$ digit.
Now we have to subtract out numbers from $2018$ to $2187$
Then either the number must begin $221...$ or $222...$ with  four more digits at the end
Using $1$ s and $2$ s there are $2^4$ options for each so:
$2+4+8+16+32+64+128-2*16=256-2-32=\boxed{222}$

Solution:
Since the greatest power of $3$ that can be used is $3^6$ , we can do these cases.
Coefficient of $3^6=0$ : Then if the number has only $3^0$ , it has 2 choices (1 or 2). Likewise if the number has both a $3^1$ and $3^0$ term, there are 4 choices, and so on until $3^5$ , so the sum is $2+4+...+64=127-1=126$
Coefficient of $3^6=1$ : Any combination of $1$ or $2$ for the remaining coefficients works, so $2^6=64$ . Why is this less than $126$ ? Because the first time around, leading zeroes didn't count. But this time, all coefficients $3^n$ of $n<6$ need 1 and 2.
Coefficient of $3^6=2$ : Look at $3^5$ coefficient. If 1, all of them work because $3^7=2187-3^5=243<<2017$ . That's 32 cases. Now of this coefficient is 2, then at the coefficient of $3^4=81$ is at least 1. However, $3^6*2+3^5*2+3^4>>2017$ , so our answer is $126+64+32=\boxed{222}$


Answer: 435.0

Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$

Solution:
$10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors
$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.
$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.
Now, we use the Principle of Inclusion-Exclusion .  We have $121 + 64 + 276$ total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers.  Thus, we must subtract off the divisors of their pair-wise greatest common divisors
$\gcd(10^{10},15^7) = 5^7$ which has 8 divisors.
$\gcd(15^7, 18^{11}) = 3^7$ which has 8 divisors.
$\gcd(18^{11}, 10^{10}) = 2^{10}$ which has 11 divisors.
So now we have $121 + 64 + 276 - 8 -8 -11$ potential divisors.  However, we've now undercounted those factors which divide all three of our numbers.  Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of $121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}$

Solution:
We can rewrite the three numbers as $10^{10} = 2^{10}\cdot 5^{10}$ $15^7 = 3^7\cdot5^7$ , and $18^{11} = 2^{11}\cdot3^{22}$ .
Assume that $n$ (a positive integer) is a divisor of one of the numbers. Therefore, $n$ can be expressed as ${p_1}^{e_1}$ or as ${p_2}^{e_2}{p_3}^{e_3}$ where $p_1$ $p_2$ are in $\{2,3,5\}$ and $e_1$ $e_2$ are positive integers.
If $n$ is the power of a single prime, then there are 11 possibilities ( $2^1$ to $2^{11}$ ) for $p_1=2$ , 22 possibilities ( $3^1$ to $3^{22}$ ) for $p_1=3$ , 10 possibilities ( $5^1$ to $5^{10}$ ) for $p_1=5$ , and 1 possibility if $n=1$ . From this case, there are $11+22+10+1=44$ possibilities.
If $n$ is the product of the powers of two primes, then we can just multiply the exponents of each rewritten product to get the number of possibilities, since each exponent of the product must be greater than 0. From this case, there are $10*10+11*22+7*7=100+242+49=391$ possibilities.
Adding up the two cases, there are $44+391=\boxed{435}$ positive integers.


Answer: 40.0

Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$

Solution: A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$ , there are two possible values for $a$ and $c$ , since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$ , and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$ . There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{040}$

Solution: A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$ . If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$ . If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$ . For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{040}$ numbers.

Solution: For this number to fit the requirements $bc$ and $ba$ must be divisible by 4. So $bc = 00, 04, 08, 12, 16, ... , 92, 96$ and so must $ba$ for each two digits of $bc$ . There are two possibilities for $ba$ if $b$ is odd and three possibilities if $b$ is even. So there are $2^{2} \cdot 5+3^{2} \cdot 4 = 65$ possibilities but this overcounts when $a$ or $c = 0$ . So when $bc = 00, 20, 40, 60, 80$ and the corresponding $ba$ should be removed, so $65 - 5 \cdot 3 = 50$ . But we are still overcounting when $b$ is even because then $a$ can be 0. So the answer is $50 - 10 \cdot 1 = \boxed{040}$


Answer: 200.0

Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: Let the numerator and denominator $x,y$ with $\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\gcd(x,y) = 1$ then $\gcd(x,y) =\gcd(x,1000-x)= \gcd(x,1000-x-(-1)x)=\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\frac{x}{y}<1$ and note by Euclidean algorithm $\gcd(1000,1000-x)=1$ , so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $\boxed{200}$ desired fractions.

Solution: If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of using $\frac{n}{m}$ as the fraction to use the Euclidean Algorithm on, we can rewrite this as $\frac{500-x}{500+x}$ or \[\gcd(500+x,500-x)=\gcd((500+x)+(500-x),500-x)=\gcd(1000,500-x).\] Thus, we want $\gcd(1000,500-x)=1$ . You can either proceed as Solution $1$ , or consider that no even numbers work, limiting us to $250$ choices of numbers and restricting $x$ to be odd. If $x$ is odd, $500-x$ is odd, so the only possible common factors $1000$ and $500-x$ can share are multiples of $5$ . Thus, we want to avoid these. There are $50$ odd multiples of $5$ less than $500$ , so the answer is $250-50=\boxed{200}$

Solution: Say $r=\frac{d}{1000-d}$ ; then $1\leq d\leq499$ . If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$ . Since $1000=2^3\cdot5^3$ , that modulus can only be $2$ or $5$ . This implies that if $d\mid2$ or $d\mid5$ , the fraction is reducible. There are $249$ cases where $d\mid2$ $99$ where $d\mid5$ , and $49$ where $d\mid(2\cdot5=10)$ , so by PIE, the number of fails is $299$ , so our answer is $\boxed{200}$

Solution: We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of $5$ because the denominator would also be a multiple of $5$ . Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have $499$ fractions to start with, and $250$ with odd numerators. Subtract $50$ to account for the multiples of $5$ , and we get $\boxed{200}$

Solution:
We know that the set of these rational numbers is from $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.
We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ .
Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreducible, and $\dfrac{1000}{m}$ is irreducible if $m$ is not divisible by $2$ or $5$ . Thus, the answer to the question is the number of integers between $501$ and $999$ inclusive that are not divisible by $2$ or $5$
We note there are $499$ numbers between $501$ and $999$ , and
Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$ , so our answer is $\boxed{200}$

Solution: We notice that there are a total of $400$ fractions that are in simplest form where the numerator and denominator add up to $1000$ . Because the numerator and denominator have to be relatively prime, there are $\varphi(1000)=400$ fractions. Half of these are greater than $1$ , so the answer is $400\div2=\boxed{200}$

Solution: Our fraction can be written in the form $\frac{1000 - a}{a} = \frac{1000}{a} - 1.$ Thus the fraction is reducible when $a$ divides $1000.$ We also want $500 < a < 1000.$ By PIE , the total values of $a$ that make the fraction reducible is, \[249 + 99 - 49 = 299.\] By complementary counting , the answer we want is $499 - 299 = \boxed{200}.$

Solution:
Suppose our fraction is $\frac{a}{b}$ . The given condition means $a+b=1000$ . Now, if $a$ and $b$ share a common factor greater than $1$ , then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even.
There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$ , which means the answer is $250-50=\boxed{200}$


Answer: 315.0

Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy]

Solution:
By Furaken
There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1, and those whose sides are not (Case 2).
For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangle's sides coincide with these segments as shown in the diagram.  We use inclusion-exclusion for this. There are 30 valid rectangles contained in $A_1A_5A_7A_{11}$ , as well as 30 in $A_2A_4A_8A_{10}$ . However, the 9 rectangles contained in $B_1B_2B_3B_4$ have been counted twice, so we subtract 9 and we have 51 rectangles in the diagram. Multiplying by 3, we get 153 rectangles for Case 1.
For Case 2, we have this diagram. To be honest, you can count the rectangles here in whatever way you like.  There are 36 rectangles contained within $A_2A_5A_8A_{11}$ , and 18 that use points outside $A_2A_5A_8A_{11}$ . So we get a total of $3(36+18)=162$ rectangles for Case 2.
Adding the two cases together, we get the answer $\boxed{315}$

Solution:
Using the same diagram as Solution 1, we can get the number of rectangles from Case 1 by adding the number of rectangles of $A_2$ $A_8$ $A_8$ $A_{10}$ and $A_1$ $A_5$ $A_7$ $A_{11}$ and then subtracting the overlaps,
\[\binom{5}{2}\binom{3}{2} + \binom{5}{2}\binom{3}{2} - \binom{3}{2}\binom{3}{2}\] \[=51\]
We multiply this by 3 to get the total number of rectangles for Case 1, which is 153.
For Case 2, we can first get the total number of rectangles from $A_2A_3A_4A_5A_8A_9A_{10}A_{11}$ then add $A_1A_6A_7A_{12}$ and subtract by the overlaps, \[\binom{4}{2}\binom{4}{2} + \binom{6}{2} - \binom{4}{2} + \binom{6}{2} - \binom{4}{2}\] \[= 54\] Multiply that by 3 and add it to Case 1 to get $\boxed{315}$

Solution:
We put the dodecagon in the right position that there exists a side whose slope is 0.
Note that finding a rectangle is equivalent to finding two pairs of lines, such that two lines in each pair are parallel and lines from different pairs are perpendicular.
Now, we use this property to count the number of rectangles.
Because two pairs of lines that form a rectangle are perpendicular, we only need to use the slope of one pair, denoted as $k$ , to determine the direction of the rectangle.
The slope of the other pair is thus $- \frac{1}{k}$ . To avoid overcounting, we do casework analysis by defining each case in term of $0 \leq k < \infty$ only (we make a convention that if $k = 0$ , then $- \frac{1}{k} = \infty$ ).
In our counting, we will frequently quantify the distance between two vertices of the regular dodecagon.
To characterize this in a straightforward way, we simply measure the number of vertices (on the minor arc) between our measured two vertices. For instance, two vertices on a side has distance 0. Distances between two vertices that are diagonals can be 1, 2, 3, 4, 5.
Case 1: $k = 0, \tan 30^\circ, \tan 60^\circ$
We only count for $k = 0$ . The number of solutions for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Consider $k = 0$ .
We need to find a pair of horizontal segments and a pair of vertical segments to form a rectangle.
For $k = 0$ , the length of each horizontal segment can only be 0, 2, 4.
Denote by $2i$ the shorter length of two parallel horizontal segments.
Given $i$ , the number of pairs of two parallel horizontal segments is $1 + 2 \left( 4 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 2}{2}$
Therefore, for $k = 0$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^2 \left( 1 + 2 \left( 4 - 2 i \right) \right)
\binom{2i + 2}{2}
& = 54 .
\end{align*}
The number of rectangles for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Therefore, the total number of rectangles in this case is $54 \cdot 3 = 162$
Case 2: $k = \tan 15^\circ$ $\tan 45^\circ$ $\tan 75^\circ$
The number of rectangles under all these $k$ s are the same.
So we only count for $k = \tan 15^\circ$
For $k = \tan 15^\circ$ , the length of each segment can only be 1, 3, 5.
However, there is only one segment with length 5.
So this cannot be the shorter length of two parallel segments with slope $\tan 15^\circ$
Denote by $2i + 1$ the shorter length of two parallel segments with slope $\tan 15^\circ$ .
Given $i$ , the number of pairs of two parallel segments is $1 + 2 \left( 3 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 3}{2}$
Therefore, for $k = \tan 15^\circ$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^1 \left( 1 + 2 \left( 3 - 2 i \right) \right)
\binom{2i + 3}{2}
& = 51 .
\end{align*}
The number of rectangles for $k = \tan 45^\circ$ and $\tan 75^\circ$ are the same.
Therefore, the total number of rectangles in this case is $51 \cdot 3 = 153$
Putting all cases together, the total number of rectangles is $162 + 153 = \boxed{315}$


Answer: 163.0

Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$

Solution:
Let $f(x) = a(x-r)(x-s)$ . Then $ars=2010=2\cdot3\cdot5\cdot67$ . First consider the case where $r$ and $s$ (and thus $a$ ) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$ $r$ , and $s$ . However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$ , we have $r=s$ . The other $80$ cases are double counting, so there are $40$
We must now consider the various cases of signs. For the $40$ cases where $|r|\neq |s|$ , there are a total of four possibilities, For the case $|r|=|s|=1$ , there are only three possibilities, $(r,s) = (1,1); (1,-1); (-1,-1)$ as $(-1,1)$ is not distinguishable from the second of those three.
You may ask: How can one of ${r, s}$ be positive and the other negative? $a$ will be negative as a result. That way, it's still $+2010$ that gets multiplied.
Thus the grand total is $4\cdot40 + 3 = \boxed{163}$

Solution:
We use Burnside's Lemma . The set being acted upon is the set of integer triples $(a,r,s)$ such that $ars=2010$ . Because $r$ and $s$ are indistinguishable, the permutation group consists of the identity and the permutation that switches $r$ and $s$ . In cycle notation, the group consists of $(a)(r)(s)$ and $(a)(r \: s)$ . There are $4 \cdot 3^4$ fixed points of the first permutation (after distributing the primes among $a$ $r$ $s$ and then considering their signs. We have 4 ways since we can keep them all positive, first 2 negative, first and third negative, or last two negative) and $2$ fixed points of the second permutation ( $r=s=\pm 1$ ). By Burnside's Lemma, there are $\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}$ distinguishable triples $(a,r,s)$ .
Note: The permutation group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$

Solution:
The equation can be written in the form of $k(x-a)(x-b)$ , where $|k|$ is a divisor of $2010$ . Factor $2010=2\cdot 3\cdot 5\cdot 67$ . We divide into few cases to study.
Firstly, if $|k|=1$ , the equation can be $-(x-a)(x+b)$ or $(x-a)(x-b)$ or $(x+a)(x+b)$ , there are $2^4+2^4=32$ ways
If $|k|\in \{2,3,5,67\}$ . Take $|k|=2$ as an example, follow the procedure above, there are $2^3+2^3=16$ ways, and there are $\binom {4}{1}=4$ ways to choose $|k|$ , so there are $16\cdot 4=64$ ways
If $|k|$ is the product of two factors of $2010$ , there are $8$ ways for each. Thus there are $\binom {4}{2}\cdot 8=48$ ways
If $|k|$ is the product of three factors of $2010$ , there are $\binom {4}{3}\cdot 4=16$ ways
In the end, $|k|=2010$ , only $2010(x-1)(x-1); 2010(x+1)(x+1); -2010(x-1)(x+1)$ work, there are $3$ ways
Thus, $32+64+48+16+3=\boxed{163}$


Answer: 728.0

Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$

Solution: Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$ . Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$ , and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. However, some sets have $2$ numbers that are the same, namely the ones in the form $1,1,x$ and $3,3,x$ , which are each counted $3$ times, and each other set is counted $6$ times, so the desired answer is $\dfrac{729 \cdot 6-6}{6} = \boxed{728}$

Solution: Again, notice that the prime factors of the product are $3, 3, 7, 11, 17, 31, 41, 61$ . In this problem, we are asked to partition this set of distinct(ish) factors into three smaller indistinct sets. To do this, we can use Stirling numbers of the second kind, but Stirling numbers of the second kind would assume no empty parts, which isn't what we want. However, this is easy to fix. Denote Stirling numbers of the second kind with $S(n,k)$ . We may start at the situation when all three sets are nonempty. Then, the number of partitions is simply $S(8,3)$ . However, we are overcounting, since every time the two threes are in different sets, (unless they are both individually in their own sets), each one is double counted. To fix this, we can see that they were in different sets $S(8, 3) - S(7,3)$ times by complementary counting, but one of these times they were in their own individual sets, so the total overcount is $\dfrac{S(8,3) - S(7,3) - 1}{2}$ . Similarly, we can do the cases for if two of the sets are nonempty and one of the sets are nonempty, (but in those last two cases the threes cannot be individually in their own sets). We then find that the "answer" is given by: \[S(8,3) - \dfrac{S(8,3) - S(7,3) - 1}{2} + S(8,2) - \dfrac{S(8,2) - S(7,2)}{2}+ S(8,1) - \dfrac{S(8,1) - S(7,1)}{2}\] \[= \dfrac{S(8,3)+S(8,2) + S(8,1) + S(7,3) + S(7,2) + S(7,1) + 1}{2}\] Drawing out the Stirling number triangle and evaluating yields $730$ for the last expression. However, throughout the entire solution, we ignored the fact that the three numbers $a$ $b$ , and $c$ needed to be distinct. However, this is easy to fix, since there are only two sets in which they are not, namely $(1,1,x)$ and $(3,3,x)$ . Thus, the actual answer is $730 - 2 = \boxed{728}$


Answer: 196.0

Solution: The number of subsets of a set with $n$ elements is $2^n$ . The total number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ is equal to $2^8$ . The number of sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ can be found using complementary counting. There are $2^5$ subsets of $\{1, 2, 3, 4, 5\}$ and $2^5$ subsets of $\{4, 5, 6, 7, 8\}$ . It is easy to make the mistake of assuming there are $2^5+2^5$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , but the $2^2$ subsets of $\{4, 5\}$ are overcounted. There are $2^5+2^5-2^2$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , so there are $2^8-(2^5+2^5-2^2)$ subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ $2^8-(2^5+2^5-2^2)=\boxed{196}$

Solution: Upon inspection, a viable set must contain at least one element from both of the sets $\{1, 2, 3, 4, 5\}$ and $\{4, 5, 6, 7, 8\}$ . Since 4 and 5 are included in both of these sets, then they basically don't matter, i.e. if set A is a subset of both of those two then adding a 4 or a 5 won't change that fact. Thus, we can count the number of ways to choose at least one number from 1 to 3 and at least one number from 6 to 8, and then multiply that by the number of ways to add in 4 and 5. The number of subsets of a 3 element set is $2^3=8$ , but we want to exclude the empty set, giving us 7 ways to choose from $\{1, 2, 3\}$ or $\{6, 7, 8\}$ . We can take each of these $7 \times 7=49$ sets and add in a 4 and/or a 5, which can be done in 4 different ways (by adding both, none, one, or the other one). Thus, the answer is $49 \times 4=\boxed{196}$

Solution: This solution is very similar to Solution $2$ . The set of all subsets of $\{1,2,3,4,5,6,7,8\}$ that are disjoint with respect to $\{4,5\}$ and are not disjoint with respect to the complements of sets (and therefore not a subset of) $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ will be named $S$ , which has $7\cdot7=49$ members. The union of each member in $S$ and the $2^2=4$ subsets of $\{4,5\}$ will be the members of set $Z$ , which has $49\cdot4=\boxed{196}$ members. $\blacksquare$

Solution:
Consider that we are trying to figure out how many subsets are possible of $\{1,2,3,4,5,6,7,8\}$ that are not in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ . Assume that the number of numbers we pick from the subset $\{1,2,3,4,5,6,7,8\}$ is $n$ . Thus, we can compute this problem with simple combinatorics:
If $n=1$ $\binom{8}{1}$ $-$ $\binom{5}{1}$ $\binom{5}{1}$ $-  2$ ) [subtract $2$ to eliminate the overcounting of the subset $\{4\}$ or $\{5\}$ ] = $8 - 8$ $0$
If $n=2$ $\binom{8}{2}$ $-$ $\binom{5}{2}$ $\binom{5}{2}$ $-  1$ ) [subtract $1$ to eliminate the overcounting of the subset $\{4,5\}$ ] = $28 - 19$ $9$
If $n=3$ $\binom{8}{3}$ $-$ $\binom{5}{3}$ $\binom{5}{3}$ ) = $56 - 20$ $36$
If $n=4$ $\binom{8}{4}$ $-$ $\binom{5}{4}$ $\binom{5}{4}$ ) = $70 - 10$ $60$
If $n=5$ $\binom{8}{5}$ $-$ $\binom{5}{5}$ $\binom{5}{5}$ ) = $56 - 2$ $54$
If $n>5$ , then the set $\{1,2,3,4,5,6,7,8\}$ is never in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ . Thus,
If $n=6$ $\binom{8}{6}$ $28$
If $n=7$ $\binom{8}{7}$ $8$
If $n=8$ $\binom{8}{8}$ $1$
Adding these together, our solution becomes $0$ $9$ $36$ $60$ $54$ $28$ $8$ $1$ $=\boxed{196}$


Answer: 235.0

Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution:
Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\]
We see that $f(1) = 2$ $f(2) = 3$ , and $f(n) = f(n-1) + f(n-2)$ . This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$ th Fibonacci number.
This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}$

Solution:
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
$\textbf{Case 1: (1,2)}$
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us $8 \choose {0}$ . We can also pick one number, giving us $7 \choose {1}$ because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us $6 \choose {2}$ . This pattern continues for each amount of numbers, yielding $5 \choose {3}$ for 3 numbers and $4 \choose {4}$ for four numbers. Adding these up, we have $8 \choose {0}$ $7 \choose {1}$ $6 \choose {2}$ $5 \choose {3}$ $4 \choose {4}$ $\textbf{34}$
$\textbf{Case 2: (2,3)}$
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have $7 \choose {0}$ $6 \choose {1}$ $5 \choose {2}$ $4 \choose {3}$ $\textbf{21}$
$\textbf{Case 3: (3,4)}$
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say $2 \choose {0}$ for choosing no numbers and $1 \choose {1}$ for choosing 1, leaving us with $2 \choose {0}$ $1 \choose {1}$ = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have $6 \choose {0}$ $5 \choose {1}$ $4 \choose {2}$ $3 \choose {3}$ = 13. With 2 choices on one side and 13 choices on the other side, we have $2\cdot13$ $\textbf{26}$ combinations in all.
$\textbf{Case 4: (4,5)}$
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have $3 \choose {0}$ $2 \choose {1}$ = 3 choices (1, 2, or neither) and for the numbers 6-10 we have $5 \choose {0}$ $4 \choose {1}$ $3 \choose {2}$ = 8 choices. With 3 choices on one side and 8 choices on the other side, we have $3\cdot8$ $\textbf{24}$ combinations in all.
$\textbf{Case 5: (5,6)}$
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices and for the numbers 5-10 we have the same $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices. $5\cdot5$ $\textbf{25}$ combinations in all.
$\textbf{Rest of the cases}$
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = $\boxed{235}$

Solution:
Denote by $N_1 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by $N_0 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers.
Denote by $a$ the smallest number in set $S$
First, we compute $N_1 \left( m \right)$
Consider $m \geq 3$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 1 \right)$
Case 2: A subset contains $a$ but does not contain $a + 1$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 2 \right)$
Case 3: A subset contains $a$ and $a + 1$
To have exactly one pair of consecutive integers, this subset cannot have $a + 2$ , and cannot have consecutive integers in $\left\{ a+3, a+4, \cdots , a + m - 1 \right\}$
Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \left( m - 3 \right)$
Therefore, for $m \geq 3$ \[N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .\]
For $m = 1$ , we have $N_1 \left( 1 \right) = 0$ .
For $m = 2$ , we have $N_1 \left( 2 \right) = 1$
Second, we compute $N_0 \left( m \right)$
Consider $m \geq 2$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has no consecutive integers is $N_0 \left( m - 1 \right)$
Case 2: A subset contains $a$
To avoid having consecutive integers, the subset cannot have $a + 1$
Thus, the number of subsets that has no consecutive integers is $N_0 \left( m - 2 \right)$
Therefore, for $m \geq 2$ \[N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right)  .\]
For $m = 0$ , we have $N_0 \left( 0 \right) = 1$ .
For $m = 1$ , we have $N_0 \left( 1 \right) = 2$
By solving the recursive equations above, we get $N_1 \left( 10 \right) = \boxed{235}$

Solution:
Let $a_n$ be the number of subsets of the set $\{1,2,3,\ldots,n\}$ such that there exists exactly 1 pair of consecutive elements.
Let $b_n$ be the number of subsets of the set $\{1, 2, 3\ldots, n\}$ such that there doesn't exist any pair of consecutive elements.
First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either:
1. $\{n-1, n\}\subseteq S,$ 2. $n\not\in S$ , or
3. $n-1 \not\in S$ and $n\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, \[a_n = a_{n-1} + a_{n-2} + b_{n-3}.\] Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either:
1. $n \not\in T,$ or
2. $n \in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, \[b_n = b_{n-1} + b_{n-2}.\] Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$ th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values.
\[\begin{array}{r|l} n & a_n \\ \hline 1 & 0 \\ 2 & 1\\ 3 & 2\\ 4 & 2 + 1 + F_2 = 5\\5 & 5 + 2 + F_3 = 10\\6 & 10 + 5 + F_4 = 20 \\ 7 & 20 + 10 + F_5 = 38\\ 8 & 38 + 20 + F_6 = 71 \\ 9& 71 + 38 + F_7 = 130\\ 10 & 130 + 71 + F_8 = 235 \end{array}\] Our answer is $a_{10} = \boxed{235}.$ ~AtharvNaphade

Solution:
Note: This is a very common stars and bars application.
Casework on number of terms, let the number of terms be $n$ . We can come up with a generalized formula for the number  of subsets with n terms.
Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$
Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$ . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$ $1 + d_2 + ... + d_{n-1} = k$
To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$ $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers.
Finally, we add in $D_0$ , the distance between 0 and the first term of the set, and $D_n$ , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by $D_0$ , k, and $D_N$ \[D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11\] \[D_0 + D_2 + D_3 + ... + D_n = 12 - n\] There are n positive terms , by Balls and Urns , there are ${(12-n)-1 \choose (n)-1} = {11-n \choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence $(n-1){11-n \choose n-1}$ for $n > 2$
Case 1: Two terms $n = 2$ , so $(2-1){11-2 \choose 2-1} = 1{9 \choose 1} = 9$
Case 2: Three terms $n = 3$ , so $(3-1){11-3 \choose 3-1} = 2{8 \choose 2} = 56$
Case 3: Four terms $n = 4$ , so $(4-1){11-4 \choose 4-1} = 3{7 \choose 3} = 105$
Case 4: Five terms $n = 5$ , so $(5-1){11-5 \choose 5-1} = 4{6 \choose 4} = 60$
Case 5: Six terms $n = 6$ , so $(6-1){11-6 \choose 6-1} = 5{5 \choose 5} = 5$
We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=\boxed{235}$

Solution:
Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.)
Now, notice that if we take $i$ and $i+1$ as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\{1, 2, ... , i-2 \}$ is consecutive with any element of the set $\{ i+3, i+4, ... , 10 \}$
Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has $i-2$ consecutive integer elements and the second set has $8-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$
Summing this over for all $1 \leq i \leq 9$ yields \[\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \cdot 34 + 1 \cdot 21 + 2 \cdot 13 + 3 \cdot 8) + 5^2 = 2 \cdot 105+25 = \boxed{235}.\]

Solution:
The problem is the same as laying out a line of  polynomoes to cover spots $0,1,...10$ : 1 triomino ( $RGG$ ), $n$ dominoes ( $RG$ ), and $8-2n$ monominoes ( $R$ ).  The $G$ spots cover the members of the subset.
The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
$R + RG + RGG =$ Polyominoes $\rightarrow$ Orderings
$0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5$
$2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60$
$4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105$
$6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56$
$8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9$
The sum is $\boxed{235}$

Solution:
Let $dp(i)$ be the number of subsets of a set $i$ consecutive integers such that the maximum value in the set is $i$ and there exists exactly one pair of consecutive integers. Define $dp2(i)$ similarly, but without any pair of consecutive integers. The base cases are $dp2(1)=dp2(2)=1$ $dp(1)=0$ , and $dp(2)=1$
The transitions are:
\[dp2(i)=\sum_{j=1}^{i-2}(dp(j))+1\]
\[dp(i)=dp2(i-1)+\sum_{j=1}^{i-2}dp(j)\]
Note that $dp2$ is the Fibonacci numbers.
\[\begin{array}{rcl} i & dp & dp2\\ \hline 1 & 0 & 1 \\ 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 3 & 3 \\ 5 & 5 & 5 \\ 6 & 10 & 8 \\ 7 & 18 & 13 \\ 8 & 33 & 21 \\ 9 & 59 & 34 \\ 10 & 105 & 55 \\ \end{array}\]
Summing over $dp$ yields $1+1+3+5+10+18+33+59+105=\boxed{235}$


Answer: 601.0

Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
\begin{equation*}
a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
\end{equation*}

Solution:
$ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000$
Note $(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0$ . Thus, $a/b/c=100$ . There are $201$ cases for each but we need to subtract $2$ for $(100,100,100)$ . The answer is $\boxed{601}$

Solution:
We have
\begin{align*}
& a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \\
& = ab \left( a + b \right) + bc \left( b + c \right) + ca \left( c + a \right) \\
& = ab \left( 300 - c \right) + bc \left( 300 - a \right) + ca \left( 300 - b \right) \\
& = 300 \left( ab + bc + ca \right) - 3 abc \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 10^4 \left( a + b + c \right) + 10^6
\right) \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 2 \cdot  10^6
\right) \\
& = 6 \cdot 10^6 .
\end{align*}
The first and the fifth equalities follow from the condition that $a+b+c = 300$
Therefore,
\[
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .
\]
Case 1: Exactly one out of $a - 100$ $b - 100$ $c - 100$ is equal to 0.
Step 1: We choose which term is equal to 0. The number ways is 3.
Step 2: For the other two terms that are not 0, we count the number of feasible solutions.
W.L.O.G, we assume we choose $a - 100 = 0$ in Step 1. In this step, we determine $b$ and $c$
Recall $a + b + c = 300$ . Thus, $b + c = 200$ .
Because $b$ and $c$ are nonnegative integers and $b - 100 \neq 0$ and $c - 100 \neq 0$ , the number of solutions is 200.
Following from the rule of product, the number of solutions in this case is $3 \cdot 200 = 600$
Case 2: At least two out of $a - 100$ $b - 100$ $c - 100$ are equal to 0.
Because $a + b + c = 300$ , we must have $a = b = c = 100$
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of solutions is $600 + 1 = \boxed{601}$

Solution:
We will use Vieta's formulas to solve this problem. We assume $a + b + c = 300$ $ab + bc + ca = m$ , and $abc = n$ . Thus $a$ $b$ $c$ are the three roots of a cubic polynomial $f(x)$
We note that $300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n$ , which simplifies to $100m - 2000000 = n$
Our polynomial $f(x)$ is therefore equal to $x^3 - 300x^2 + mx - (100m - 2000000)$ . Note that $f(100) = 0$ , and by polynomial division we obtain $f(x) = (x - 100)(x^2 - 200x - (m-20000))$
We now notice that the solutions to the quadratic equation above are $x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}$ , and that by changing the value of $m$ we can let the roots of the equation be any pair of two integers which sum to $200$ . Thus any triple in the form $(100, 100 - x, 100 + x)$ where $x$ is an integer between $0$ and $100$ satisfies the conditions.
Now to count the possible solutions, we note that when $x \ne 100$ , the three roots are distinct; thus there are $3! = 6$ ways to order the three roots. As we can choose $x$ from $0$ to $99$ , there are $100 \cdot 3! = 600$ triples in this case. When $x = 100$ , all three roots are equal to $100$ , and there is only one triple in this case.
In total, there are thus $\boxed{601}$ distinct triples.

Solution:
Let's define $a=100+x$ $b=100+y$ $c=100+z$ . Then we have $x+y+z=0$ and $6000000 = \sum a^2(b+c)$
$= \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) = \sum (20000 - 10000 x + x(40000-x^2))$
$= \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3$ , so we get $x^3 + y^3 + z^3 = 0$ . Then from $x+y+z = 0$ , we can find $0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz$ , which means that one of $a$ $b$ $c$ must be 0. There are 201 solutions for each of $a=0$ $b=0$ and $c=0$ , and subtract the overcounting of 2 for solution $(200, 200, 200)$ , the final result is $201 \times 3 - 2 = \boxed{601}$


Answer: 7.0

Find the number of two-digit positive integers whose digits total $7$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: The numbers are $16, 25, 34, 43, 52, 61, 70$ which gives us a total of $\boxed{7}$


Answer: 331.0

Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.

Solution: Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$ . Continuing this pattern until $21$ coins in the first pile, we have the sum \begin{align*} 31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\ &= 176+155 \\ &= \boxed{331}

Solution:
We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\geq b\geq c.$ By stars and bars, we have $\dbinom{65}{2}=2080$ to assign positive integer solutions to $a + b + c = 66.$ Now we need to subtract off the cases where it doesn't satisfy the condition.
We start with $a = b.$ We can write that as $2b + c = 66.$ We can find there are 32 integer solutions to this equation. There are $32$ solutions for $b=c$ and $a = c$ by symmetry. We also need to add back $2$ because we subtracted $(a,b,c)=(22,22,22)$ $3$ times.
We then have to divide by $6$ because there are $3!=6$ ways to order $a, b,$ and $c.$ Therefore, we have $\dfrac{\dbinom{65}{2}-96+2}{6} = \dfrac{1986}{6} = \boxed{331}.$

Solution:
Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$ . Then, let $b = a + k_1$ , and $c = b + k_2$ , such that each $k_i \geq 1$ . The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66$ . This is simply the number of positive solutions to the equation $3x+2y+z = 66$ . Now, we take cases on $a$
If $a = 1$ , then $2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31$ . Each value of $k_1$ corresponds to a unique value of $k_2$ , so there are $31$ solutions in this case. Similarly, if $a = 2$ , then $2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29$ , for a total of $29$ solutions in this case. If $a = 3$ , then $2k_1 + k_2 = 57 \implies 1 \leq k_1 \leq 28$ , for a total of $28$ solutions. In general, the number of solutions is just all the numbers that aren't a multiple of $3$ , that are less than or equal to $31$
We then add our cases to get \begin{align*} 1 + 2 + 4 + 5 + \cdots + 31 &= 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) \\ &= \frac{31(32)}{2} - 3(55) \\ &= 31 \cdot 16 - 165 \\ &= 496 - 165 \\ &= \boxed{331} as our answer.


Answer: 45.0

Find the number of ways to place a digit in each cell of a 2x3 grid so that the sum of the two numbers formed by reading left to right is $999$ , and the sum of the three numbers formed by reading top to bottom is $99$ . The grid below is an example of such an arrangement because $8+991=999$ and $9+9+81=99$
\[\begin{array}{|c|c|c|} \hline 0 & 0 & 8 \\ \hline 9 & 9 & 1 \\ \hline \end{array}\]

Solution:
Consider this table:
$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f\\ \hline \end{array}$
We note that $c+f = 9$ , because $c+f \leq 18$ , meaning it never achieves a unit's digit sum of $9$ otherwise. Since no values are carried onto the next digit, this implies $b+e=9$ and $a+d=9$ . We can then simplify our table into this:
$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline 9-a & 9-b & 9-c \\ \hline \end{array}$
We want $10(a+b+c) + (9-a+9-b+9-c) = 99$ , or $9(a+b+c+3) = 99$ , or $a+b+c=8$ . Since zeroes are allowed, we just need to apply stars and bars on $a, b, c$ , to get $\tbinom{8+3-1}{3-1} = \boxed{045}$ . ~akliu


Answer: 47.0

Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]

Solution:
Since we are dealing with acute angles, $\tan(\arctan{a}) = a$
Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$ , by tangent addition. Thus, $\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}$
Applying this to the first two terms, we get $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}$
Now, $\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}$
We now have $\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}$ . Thus, $\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1$ ; and simplifying, $23n + 24 = 24n - 23 \Longrightarrow n = \boxed{47}$

Solution: Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\arctan\frac{1}{n}$ , is the argument of $n+i$ . The sum of these angles is then just the argument of the product \[(3+i)(4+i)(5+i)(n+i)\] and expansion give us $(48n-46)+(48+46n)i$ . Since the argument of this complex number is $\frac{\pi}{4}$ , its real and imaginary parts must be equal; then, we can we set them equal to get \[48n - 46 = 48 + 46n.\] Therefore, $n=\boxed{47}$


Answer: 312.0

Find the positive integer $n\,$ for which \[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\] (For real $x\,$ $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$

Solution:
Note that if $2^x \le a<2^{x+1}$ for some $x\in\mathbb{Z}$ , then $\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x$
Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $\lfloor\log_2{a}\rfloor=x$ . So the sum of $\lfloor\log_2{a}\rfloor$ for all such $a$ is $x\cdot2^x$
Let $k$ be the integer such that $2^k \le n<2^{k+1}$ . So for each integer $j<k$ , there are $2^j$ integers $a\le n$ such that $\lfloor\log_2{a}\rfloor=j$ , and there are $n-2^k+1$ such integers such that $\lfloor\log_2{a}\rfloor=k$
Therefore, $\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994$
Through computation: $\sum_{j=0}^{7}(j\cdot2^j)=1538<1994$ and $\sum_{j=0}^{8}(j\cdot2^j)=3586>1994$ . Thus, $k=8$
So, $\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}$


Answer: 13.0

Solution:
We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
To simplify the equation, substitute $a = x^2 - 10x - 29$ (the denominator of the first fraction). We can rewrite the equation as $\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0$ . Multiplying out the denominators now, we get:
\[(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0\]
Simplifying, $-64a + 40 \times 16 = 0$ , so $a = 10$ . Re-substituting, $10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3)$ . The positive root is $\boxed{013}$


Answer: 748.0

Find the smallest integer $k$ for which the conditions
(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers
(2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$
(3) $a_9=k$
are satisfied by more than one sequence.

Solution:
From $(2)$ $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$
Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$
Since $\gcd(13,21)=1$ , the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$ . Then, $a_2=y_0-13$
By $(1)$ $a_2 \ge a_1 \Rightarrow y_0-13 \ge x_0+21 \Rightarrow y_0 \ge x_0+34 \ge 35$ . So the smallest value of $k$ is attained when $(x_0,y_0)=(1,35)$ which yields $(a_1,a_2)=(1,35)$ or $(22,22)$
Thus, $k=13(1)+21(35)=\boxed{748}$ is the smallest possible value of $k$


Answer: 44.0

Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms.

Solution: Using Simon's Favorite Factoring Trick , we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$ . Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 \ge 1996$ , the smallest square after $1996$ is $2025 = 45^2$ , so our answer is $45 - 1 = \boxed{044}$

Solution:
Notice that the coefficients in the problem statement have no effect on how many unique terms there will be in the expansion. Therefore this problem is synonymous with finding the amount of terms in the expansion of $(xy+x+y+1)^n$ (we do this to simplify the problem).
If we expand the exponent the expression becomes $\underbrace{(xy+x+y+1)\cdot(xy+x+y+1)\cdots(xy+x+y+1)}_{n\;times}$
This is equivalent to starting off with a $x^0y^0$ terms and choosing between $4$ options $n$ different times:
$\bullet$ Adding nothing to either exponent (choosing $1$ ).
$\bullet$ Adding $1$ to the $x$ exponent (choosing $x$ ).
$\bullet$ Adding $1$ to the $y$ exponent (choosing $y$ ).
$\bullet$ Adding $1$ to the $x$ exponent and adding $1$ to the $y$ exponent (choosing $xy$ ).
Doing this $n$ times, you can see that you end up with a term in the form $k\cdot x^i\cdot y^j$ where $k$ is some coefficient (which we don't care about) and $0\le i,j\le n$ and $i,j \in \mathbb{N}$
Repeating this for all possible combinations of choices yields $n+1$ options for each of $i$ and $j$ which means there are a total of $(n+1)^2$ possible terms in the form $x^i\cdot y^j$ . Therefore $(xy+x+y+1)^n$ has $(n+1)^2$ terms.
$(n+1)^2 \ge 1996$ which yields $n=44$ $\boxed{044}$


Answer: 1999.0

Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.

Solution:
We claim that $n = 1999$ is the smallest such number. For $n \le 1998$ , we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$ . We call a row or column filled if all $1000$ of its squares are colored. Then any of the remaining $999$ colored squares must share a column or row, respectively, with one of the colored squares in a filled row or column. These two squares, and any other square in the filled row or column, form a colored right triangle, giving us a contradiction. Hence, no filled row or column may exist.
Let $m$ be the number of columns with $1$ colored square. Then there are $1999-m$ colored squares in the remaining columns, and in each of these $< 1999-m$ columns that have a colored square must have at least two colored squares in them. These two colored squares will form a triangle with any other colored square in either of the rows containing the colored squares. Hence, each of the $1999-m$ colored squares must be placed in different rows, but as there are only $1000$ rows, the inequality $1999 - m \le 1000 \Longrightarrow m \ge 999$ holds. If $m = 1000$ , then each column only has $1$ colored square, leaving no place for the remaining $999$ , contradiction. If $m = 999$ , then each of the $1000$ rows has $1$ black square, leaving no place for the other $999$ , contradiction. Hence $n = \boxed{1999}$ is the minimal value.


Answer: 8.0

Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients.

Solution:
You can factor the polynomial into two quadratic factors or a linear and a cubic factor.
For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that
\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\]
Therefore, again setting coefficients equal, $a + c = 0\Longrightarrow a=-c$ $b + d + ac = 0\Longrightarrow b+d=a^2$ $ad + bc = - n$ , and so $bd = 63$
Since $b+d=a^2$ , the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$ . From this we find that the possible values for $n$ are $\pm 8 \cdot 62$ and $\pm 4 \cdot 2$
For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$
Therefore, the answer is $4 \cdot 2 = \boxed{008}$

Solution:
Let $x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$ . From this, we get that $bd=63\implies d=\frac{63}{b}$ and $a+c=0\implies c=-a$ . Plugging this back into the equation, we get $x^4-nx+63=(x^2+ax+b)\left(x^2-ax+\frac{63}{b}\right)$ . Expanding gives us $x^4-nx+63=x^4+\left(-a^2+b+\frac{63}{b}\right)x^2+\left(\frac{63a}{b}-ab\right)x+63$ . Therefore $-a^2+b+\frac{63}{b}=0$ . Simplifying gets us $b(a^2-b)=63$ . Since $a$ and $b$ must be integers, we can use guess and check for values of $b$ because $b$ must be a factor of $63$ . Note that $b$ cannot be negative because $a$ would be imaginary. After guessing and checking, we find that the possible values of $(a,b)$ are $(\pm 8, 1), (\pm 4, 7), (\pm 4, 9),$ and $(\pm 8, 63)$ . We have that $n=ab-\frac{63a}{b}$ . Plugging in our values for $a$ and $b$ , we get that the smallest positive integer value $n$ can be is $\boxed{008}$ .
-Heavytoothpaste


Answer: 159.0

Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$

Solution:
$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$
The period of the tangent function is $180^\circ$ , and the tangent function is one-to-one over each period of its domain.
Thus, $19x \equiv 141 \pmod{180}$
Since $19^2 \equiv 361 \equiv 1 \pmod{180}$ , multiplying both sides by $19$ yields $x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}$
Therefore, the smallest positive solution is $x = \boxed{159}$

Solution:
$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}$ which is the same as $\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141{^\circ}}$
So $19x = 141 +180n$ , for some integer $n$ .
Multiplying by $19$ gives $x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}$ .
The smallest positive solution of this is $x = \boxed{159}$

Solution: It seems reasonable to assume that $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}$ for some angle $\theta$ . This means \[\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}\] for some constant $\alpha$ . We can set $\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}$ .Note that if we have $\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\sin{45^{\circ}} = \cos{45^{\circ}} = \tfrac{\sqrt{2}}{2}$ , if $\alpha = \tfrac{\sqrt{2}}{2}$ we have \[\alpha (\cos{96^{\circ}} + \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \sin{45^{\circ}} + \sin{96^{\circ}} \cos{45^{\circ}} = \sin({45^{\circ} + 96^{\circ}}) = \sin{141^{\circ}}\] from the sine sum formula. For the denominator, from the cosine sum formula, we have \[\alpha (\cos{96^{\circ}} - \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \cos{45^{\circ}} + \sin{96^{\circ}} \sin{45^{\circ}} = \cos({96^{\circ}  + 45^{\circ}}) = \cos{141^{\circ}}.\] This means $\theta = 141^{\circ},$ so $19x = 141 + 180k$ for some positive integer $k$ (since the period of tangent is $180^{\circ}$ ), or $19 x \equiv 141 \pmod{180}$ . Note that the inverse of $19$ modulo $180$ is itself as $19^2 \equiv 361 \equiv 1 \pmod {180}$ , so multiplying this congruence by $19$ on both sides gives $x \equiv 2679 \equiv 159 \pmod{180}.$ For the smallest possible $x$ , we take $x = \boxed{159}.$