Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

Solution: The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$ . The quotient of the two is -
$\frac{(2)(n!)}{(n)(n+1)}$
which simplifies to $\frac{(2)((n-1)!)}{n+1}$ . Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some number in $n!$ , cancelling out $n+1$ if it is even, which leaves us with the answer choices $n = 996$ and $n = 998$ . Notice that $n + 1$ must also be prime as otherwise there will be a factor of $n + 1$ in $2$ $n!$ somewhere. So either $997$ or $999$ must be prime - $999$ is obviously not prime as it is divisible by 9, so our answer should be $n$ where $n + 1 = 997$ , and so our answer is $n = 996$ or $\boxed{996}$

Answer: B 996
Problem: A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$

Solution: The number of tiles the bug visits is equal to $1$ plus the number of times it crosses a horizontal or vertical line.  As it must cross $16$ horizontal lines and $9$ vertical lines, it must be that the bug visits a total of $16+9+1 = \boxed{26}$ squares.

Answer: C 26
Problem: A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$

Solution: We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ feet sections.
[asy] unitsize(20); for(int i =0; i<= 7; ++i) { for(int j =0; j<= 4; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle); }  for(int k =0; k<= 4; ++k) { draw((8,k)--(8.5,k)--(8.5,k+1)--(8,k+1)--cycle); }  }  draw((0,5)--(8.5,0)--cycle); [/asy]
Though it may appear that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count the number of squares the line passes through using the diagram, we get $13$ squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of $2 \cdot 13 = \boxed{26}$

Answer: C 26
Problem: How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$

Solution: Prime factorizing $201^9$ , we get $3^9\cdot67^9$ . 
A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$ . This yields $5\cdot5 = 25$ perfect squares.
Perfect cubes must have multiples of $3$ for each of their prime factors' exponents, so we have either $0, 3, 6$ , or $9$ for both $3$ and $67$ , which yields $4\cdot4 = 16$ perfect cubes, for a total of $25+16 = 41$
Subtracting the overcounted powers of $6$ $3^0\cdot67^0$ $3^0\cdot67^6$ $3^6\cdot67^0$ , and $3^6\cdot67^6$ ), we get $41-4 = \boxed{37}$

Answer: C 37
Problem: How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$

Solution: Observe that $201 = 67 \cdot 3$ . Now divide into cases:
Case 1 : The factor is $3^n$ . Then we can have $n = 2$ $3$ $4$ $6$ $8$ , or $9$
Case 2 : The factor is $67^n$ . This is the same as Case 1.
Case 3 : The factor is some combination of $3$ s and $67$ s.
This would be easy if we could just have any combination, as that would simply give $6 \cdot 6$ . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for $n$
$n = 2$ is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
$n = 3$ is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of $3$
$n = 4$ is a "square".
$n = 6$ is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.
$n = 8$ is a "square"
$n = 9$ is a "cube".
Now let's consider subcases:
Subcase 1 : The squares are with each other.
Since we have $3$ square terms, and they would pair with $3$ other square terms, we get $3 \cdot 3 = 9$ possibilities.
Subcase 2 : The cubes are with each other.
Since we have $2$ cube terms, and they would pair with $2$ other cube terms, we get $2 \cdot 2 = 4$ possibilities.
Subcase 3 : A number pairs with $n=6$
Since any number can pair with $n=6$ (as it gives both a square and a cube), there would be $6$ possibilities. Remember however that there can be two different bases ( $3$ and $67$ ), and they would produce different results. Thus, there are in fact $6 \cdot 2 = 12$ possibilities.
Finally, summing the cases gives $6+6+9+4+12 = \boxed{37}$

Answer: C 37
Problem: How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$

Solution: We first prime factorize $201^9 = 3^9 \cdot 67^9$ . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of $3$ in the exponent. The largest square for $3$ can be $3^8$ , so their must be $\dfrac {8}{2} = 4$ ways. The largest cube is $3^9$ , so there must be $\dfrac{9}{3} = 3$ . Minus one $3^6$ due to overlapping and we get $4 + 3 -1 = 6$ ways for $3$ to be a cube/square. We can see that this same thing happens for $67^9$ due to the same exponent. Adding $0$ as a case, we have our answer; $6 \cdot 6 + 1 = \boxed{37}$

Answer: C 37
Problem: How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$

Solution: Notice that $201=3 \cdot 67$ . We factorize $201^9$ to get $3^9 \cdot 67^9$ . We then list perfect squares and cubes. $3^2$ $3^4$ $3^6$ $3^8$ $3^3$ $3^6$ $3^9$ $67^2$ $67^4$ $67^6$ $67^8$ $67^3$ $67^6$ $67^9$ . Notice that the powers of $6$ overlap. We must not forget $1$ though. Of course, all of these factors already work. This gives us $15-2=3$ . Next, we count the perfect squares. Since there are $4$ options we have $4 \cdot 4=16$ . We do the same for the perfect cubes except with 3 options this time, and we have $3 \cdot 3=9$ . However, we accidentally overcounted $3^6 \cdot 67^6$ . We add our answers and subtract $1$ to get $13+16+9-1 = \boxed{37}$

Answer: C 37
Problem: Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Solution: [asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));[/asy]
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$ . We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$
\[\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}\]
\[\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}\]
Then, we take triangle $BFC$ , and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{110}.$

Answer: D 110
Problem: Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Solution: Alternatively, we could have used similar triangles. We start similarly to Solution 1.
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Therefore, \[\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.\]
So, $\triangle BEF \sim BDA$ by AA Similarity, since $\angle EBF = \angle DBA$ and $\angle BEC = 90^{\circ} = \angle BDA$ . 
Thus, we know \[\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.\]
Finally, we deduce \[\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{110}.\]

Answer: D 110
Problem: Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Solution: Through the property of angles formed by intersecting chords, we find that \[m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}\]
Through the Outside Angles Theorem, we find that \[m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}\]
Adding the two equations gives us \[m\angle BFC + m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB\]
Since $\overarc{BC}$ is the diameter, $m\overarc{BC}=180^{\circ}$ , and because $\triangle ABC$ is isosceles and $m\angle ACB=40^{\circ}$ , we have $m\angle CAB=70^{\circ}$ . Thus \[m\angle BFC=180^{\circ}-70^{\circ}=\boxed{110}\]

Answer: D 110
Problem: Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Solution: Notice that if $\angle BEC = 90^{\circ}$ , then $\angle BCE$ and $\angle ACE$ must be $20^{\circ}$ . Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20^{\circ}$ . Thus $\angle CBF = 70 - 20 = 50^{\circ}$ , and so $\angle BFC = 180 - 20 - 50 = 110^{\circ}$ , which is $\boxed{110}$

Answer: D 110
Problem: Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Solution: $\triangle{ABC}$ is isosceles so $\angle{CAB}=70^{\circ}$ . Since $CB$ is a diameter, $\angle{CDB}=\angle{CEB}=90^{\circ}$ . Quadrilateral $ADFE$ is cyclic since $\angle{ADF}+\angle{AEF}=180^{\circ}$ . Therefore $\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110}$

Answer:  110
Problem: A child builds towers using identically shaped cubes of different colors. How many different towers with a height $8$ cubes can the child build with $2$ red cubes, $3$ blue cubes, and $4$ green cubes? (One cube will be left out.)
$\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320$

Solution: Arranging eight cubes is the same as arranging the nine cubes first, and then removing the last cube. In other words, there is a one-to-one correspondence between every arrangement of nine cubes, and every actual valid arrangement. Thus, we initially get $9!$ . However, we have overcounted, because the red cubes can be permuted to have the same overall arrangement, and the same applies with the blue and green cubes. Thus, we have to divide by the $2!$ ways to arrange the red cubes, the $3!$ ways to arrange the blue cubes, and the $4!$ ways to arrange the green cubes. Thus we have $\frac {9!} {2! \cdot 3! \cdot  4!} = \boxed{1,260}$ different possible towers.

Answer: D 1,260
Problem: A child builds towers using identically shaped cubes of different colors. How many different towers with a height $8$ cubes can the child build with $2$ red cubes, $3$ blue cubes, and $4$ green cubes? (One cube will be left out.)
$\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320$

Solution: We can divide the problem into three cases, each representing one cube to be excluded:
Case 1 : The red cube is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The number of possible arrangements is $\frac{8!}{4!\cdot3!}=280$ . Note that we do not need to multiply by the number of red cubes because there is no way to distinguish between the first red cube and the second.
Case 2 : The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The number of possible arrangements is $\frac{8!}{2!\cdot2!\cdot4!}=420$
Case 3 : The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The number of possible arrangements is $\frac{8!}{2!\cdot3!\cdot3!}=560$
Adding up the individual cases from above gives the answer as $280+420+560=\boxed{1,260}$

Answer: D 1,260
Problem: For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution: We can expand the fraction $0.\overline{23}_k$ as follows: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots$
Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\]
By summing the geometric series and simplifying, we have $\frac{2k+3}{k^2-1} = \frac{7}{51}$ . Solving this quadratic equation (or simply testing the answer choices) yields the answer $k = \boxed{16}$

Answer: D 16
Problem: For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution: Let $a = 0.2323\dots_k$ . Therefore, $k^2a=23.2323\dots_k$
From this, we see that $k^2a-a=23_k$ , so $a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$
Now, similar to in Solution 1, we can either test if $2k+3$ is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is $\boxed{16}$

Answer: D 16
Problem: For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution: Just as in Solution 1, we arrive at the equation $\frac{2k+3}{k^2-1}=\frac{7}{51}$
Therefore now, we can rewrite this as $\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}$ . Notice that $2k+3=2(k+1)+1=2(k-1)+5$ . As $17$ is a prime number, we therefore must have that one of $k-1$ and $k+1$ is divisible by $17$ . Now, checking each of the answer choices, this will lead us to the answer $\boxed{16}$

Answer: D 16
Problem: For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution: Assuming you are familiar with the rules for basic repeating decimals, $0.232323... = \frac{23}{99}$ . Now we want our base, $k$ , to conform to $23\equiv7\pmod k$ and $99\equiv51\pmod k$ , the reason being that we wish to convert the number from base $10$ to base $k$ . Given the first equation, we know that $k$ must equal 9, 16, 23, or generally, $7n+2$ . The only number in this set that is one of the multiple choices is $16$ . When we test this on the second equation, $99\equiv51\pmod k$ , it comes to be true. Therefore, our answer is $\boxed{16}$

Answer: D 16
Problem: For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution: Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},\] from which we see our equation becomes \[\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).\]
Note that $17$ therefore divides $k^2 - 1,$ but as $17$ is prime this therefore implies \[k \equiv \pm 1 \pmod{17}.\] (Warning: This would not be necessarily true if $17$ were composite.) Note that $\boxed{16}$ is the only answer choice congruent satisfying this modular congruence, thus completing the problem. $\square$

Answer: D 16
Problem: What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$ , which can be simplified to $(x^2+5x+5)^2-1+2019$ . Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$ , the answer is $\boxed{2018}$

Answer: B 2018
Problem: What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: Let $a=x+\tfrac{5}{2}$ . Then the expression $(x+1)(x+2)(x+3)(x+4)$ becomes $\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)$
We can now use the difference of two squares to get $\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)$ , and expand this to get $a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}$
Refactor this by completing the square to get $\left(a^2-\tfrac{5}{4}\right)^2-1$ , which has a minimum value of $-1$ . The answer is thus $2019-1=\boxed{2018}$

Answer: B 2018
Problem: What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$
Letting $y=x^2+5x$ , we get the expression $(y+4)(y+6)+2019$ . Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function:
$\frac{d}{dx}(y^2+10y+24)=0$
$2y+10=0$
$2y=-5$
$y=-5,0$
To minimize the result, we use $y=-5$ . Hence, the minimum is $(-5+4)(-5+6)=-1$ , so $-1+2019 = \boxed{2018}$

Answer: B 2018
Problem: What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x < -3$ . Plugging in $x = -\frac32$ or $x = -\frac72$ yields $-\frac{15}{16}$ , which is very close to $-1$ . Thus the answer is $-1 + 2019 = \boxed{2018}$

Answer: B 2018
Problem: What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: Answer choices $C$ $D$ , and $E$ are impossible, since $(x+1)(x+2)(x+3)(x+4)$ can be negative (as seen when e.g. $x = -\frac{3}{2}$ ). Plug in $x = -\frac{3}{2}$ to see that it becomes $2019 - \frac{15}{16}$ , so round this to $\boxed{2018}$

Answer: B 2018
Problem: What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: We can ignore the $2019$ and consider it later, as it is a constant. By difference of squares, we can group this into $\left((x+2.5)^2-0.5^2\right)\left((x+2.5)^2-1.5^2\right)$ . We pull a factor of $4$ into each term to avoid dealing with decimals:
\[\dfrac{\left((2x+5)^2-1\right)\left((2x+5)^2-9\right)}{16}.\]
Now, we let $a=2x+5$ . Our expression becomes:
\[\dfrac{(a-1)(a-9)}{16}=\dfrac{a^2-10a+9}{16}.\]
Taking the derivative, we get $\dfrac{2a-10}{16}=\dfrac{a-5}8.$ This is equal to $0$ when $a=5$ , and plugging in $a=5$ , we get the expression is equal to $-1$ and therefore our answer is $2019-1=\boxed{2018}.$

Answer: B 2018