For which positive integers $m$ does there exist an infinite arithmetic sequence of integers $a_1, a_2, . . .$ and an infinite geometric sequence of integers $g_1, g_2, . . .$ satisfying the following properties? [list] [*] $a_n - g_n$ is divisible by $m$ for all integers $n \ge 1$; [*] $a_2 - a_1$ is not divisible by $m$. [/list] [i]Holden Mui[/i]

Define a sequence of real numbers $ a_1$, $ a_2$, $ a_3$, $ \dots$ by $ a_1 = 1$ and $ a_{n + 1}^3 = 99a_n^3$ for all $ n \ge 1$. Then $ a_{100}$ equals $ \textbf{(A)}\ 33^{33} \qquad \textbf{(B)}\ 33^{99} \qquad \textbf{(C)}\ 99^{33} \qquad \textbf{(D)}\ 99^{99} \qquad \textbf{(E)}\ \text{none of these}$

Points $B$,$D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$, $E$, $I$ are midpoints of the sides of triangle , etc. If the dividing and shading process is done 100 times (the first three are shown) and $ AC=CG=6 $, then the total area of the shaded triangles is nearest [asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S);[/asy] $ \text{(A)}\ 6\qquad\text{(B)}\ 7\qquad\text{(C)}\ 8\qquad\text{(D)}\ 9\qquad\text{(E)}\ 10 $

It is given that $\log_{6}a+\log_{6}b+\log_{6}c=6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b-a$ is the square of an integer. Find $a+b+c.$

The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$ $\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21$

A line initially $ 1$ inch long grows according to the following law, where the first term is the initial length. \[ 1 \plus{} \frac {1}{4}\sqrt {2} \plus{} \frac {1}{4} \plus{} \frac {1}{16}\sqrt {2} \plus{} \frac {1}{16} \plus{} \frac {1}{64}\sqrt {2} \plus{} \frac {1}{64} \plus{} \cdots. \]If the growth process continues forever, the limit of the length of the line is: $ \textbf{(A)}\ \infty \qquad\textbf{(B)}\ \frac {4}{3} \qquad\textbf{(C)}\ \frac {8}{3} \qquad\textbf{(D)}\ \frac {1}{3}(4 \plus{} \sqrt {2}) \qquad\textbf{(E)}\ \frac {2}{3}(4 \plus{} \sqrt {2})$

Given a geometric sequence with the first term $ \neq 0$ and $ r \neq 0$ and an arithmetic sequence with the first term $ \equal{}0$. A third sequence $ 1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is: $ \textbf{(A)}\ 978 \qquad \textbf{(B)}\ 557 \qquad \textbf{(C)}\ 467 \qquad \textbf{(D)}\ 1068 \\ \textbf{(E)}\ \text{not possible to determine from the information given}$

If the distinct non-zero numbers $x ( y - z),~ y(z - x),~ z(x - y )$ form a geometric progression with common ratio $r$, then $r$ satisfies the equation $\textbf{(A) }r^2+r+1=0\qquad\textbf{(B) }r^2-r+1=0\qquad\textbf{(C) }r^4+r^2-1=0$ $\qquad\textbf{(D) }(r+1)^4+r=0\qquad \textbf{(E) }(r-1)^4+r=0$

Find all triples $ (x,y,z) $ of natural numbers that are in geometric progression and verify the inequalities $$ 4016016\le x<y<z\le 4020025. $$

We say that $(a,b,c)$ form a [i]fantastic triplet[/i] if $a,b,c$ are positive integers, $a,b,c$ form a geometric sequence, and $a,b+1,c$ form an arithmetic sequence. For example, $(2,4,8)$ and $(8,12,18)$ are fantastic triplets. Prove that there exist infinitely many fantastic triplets.

If $P$ is the product of $n$ quantities in Geometric Progression, $S$ their sum, and $S'$ the sum of their reciprocals, then $P$ in terms of $S$, $S'$, and $n$ is $\textbf{(A) }(SS')^{\frac{1}{2}n}\qquad\textbf{(B) }(S/S')^{\frac{1}{2}n}\qquad\textbf{(C) }(SS')^{n-2}\qquad\textbf{(D) }(S/S')^n\qquad \textbf{(E) }(S/S')^{\frac{1}{2}(n-1)}$

An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black? [asy] unitsize(36); fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white); draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1)); fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white); fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white); fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white); fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white); draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1)); label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S); label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S); [/asy] $\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}$

Find a triple $(l, m, n)$ of positive integers $(1<l<m<n)$, such that $\sum_{k=1}^{l}k, \sum_{k=l+1}^{m}k, \sum_{k=m+1}^{n}k$ form a geometric sequence in order.

Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had form a geometric progression. Alex eats 5 of his peanuts, Betty eats 9 of her peanuts, and Charlie eats 25 of his peanuts. Now the three numbers of peanuts that each person has form an arithmetic progression. Find the number of peanuts Alex had initially.

Triangle $AB_0C_0$ has side lengths $AB_0 = 12$, $B_0C_0 = 17$, and $C_0A = 25$. For each positive integer $n$, points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$, respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$. The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$, where $p$ and $q$ are relatively prime positive integers. Find $q$.

The limit of the sum of an infinite number of terms in a geometric progression is $ \frac {a}{1 \minus{} r}$ where $ a$ denotes the first term and $ \minus{} 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is: $ \textbf{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textbf{(E)}\ \text{none of these}$