Let $(u,v)\in\mathbb{Z}^2\setminus(0,0)$ and $p$ a prime number congruent to 1 modulo 4. Show that if $p$ does not divide $u^2+v^2$, then $$|L_{\mathrm{prim}}(u,v,p)| \leq 2 \quad \text{and} \quad |L_{\mathrm{prim}}(u,v,p^2)| \leq 1.$$
Let $d$ be an odd integer. Show that $$|S_{\mathrm{prim}}(d^2)| \geq d^2\, 2^{-|\mathcal{P}(d)|}.$$ Deduce that, for all $\varepsilon>0$, we have $$d^{2-\varepsilon} = \underset{d\rightarrow+\infty}{o}\left(S_{\mathrm{prim}}(d^2)\right).$$
Let $(u,v)\neq(0,0)$. Show the existence of a constant $C$ (depending on $(u,v)$) such that for all $d>0$, $$|L_{\mathrm{prim}}(u,v,d)| \leq C\, 2^{|\mathcal{P}(d)|}.$$ Deduce that, for all $\varepsilon>0$, we have $$|L_{\mathrm{prim}}(u,v,d)| = \underset{d\rightarrow+\infty}{o}\left(d^\varepsilon\right).$$
Let $x$ be a positive real number greater than or equal to 1 and $q \in \mathbb{N}^{*}$. Justify that the quantity $$\operatorname{Card}\{n \in \mathbb{N} \cap [1,x] : n \equiv 0 \pmod{q}\} - \frac{x}{q}$$ is bounded in absolute value by a real number independent of $x$ and $q$.
For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ We define $\mathscr{S} = \left\{n \geqslant 3 : \left|\frac{\omega(n) - \ln_{2}(n)}{\sqrt{\ln_{2}(n)}}\right| \geqslant \left(\ln_{2}(n)\right)^{1/4}\right\}$. Show that $$\lim_{x \rightarrow +\infty} \frac{1}{x} \operatorname{Card}\{n \leqslant x : n \in \mathscr{S}\} = 0$$ One may begin by writing $\operatorname{Card}(\mathscr{S} \cap [1,x]) \underset{x \rightarrow +\infty}{=} \operatorname{Card}(\mathscr{S} \cap [\sqrt{x}, x]) + O(\sqrt{x})$ and note that in the sum on the right-hand side, the difference $\left|\ln_{2}(n) - \ln_{2}(x)\right|$ remains bounded.
Let $x$ be a positive real number greater than or equal to 1 and $q \in \mathbb{N}^*$. Justify that the quantity $$\operatorname{Card}\{n \in \mathbb{N} \cap [1,x] : n \equiv 0 \pmod{q}\} - \frac{x}{q}$$ is bounded in absolute value by a real number independent of $x$ and $q$.
For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ We set $\mathscr{S} = \left\{n \geqslant 3 : \left|\frac{\omega(n) - \ln_2(n)}{\sqrt{\ln_2(n)}}\right| \geqslant \left(\ln_2(n)\right)^{1/4}\right\}$. Show that $$\lim_{x \rightarrow +\infty} \frac{1}{x} \operatorname{Card}\{n \leqslant x : n \in \mathscr{S}\} = 0$$ One may begin by writing $\operatorname{Card}(\mathscr{S} \cap [1,x]) \underset{x \rightarrow +\infty}{=} \operatorname{Card}(\mathscr{S} \cap [\sqrt{x}, x]) + O(\sqrt{x})$ and note that in the sum on the right-hand side, the difference $\left|\ln_2(n) - \ln_2(x)\right|$ remains bounded.
144- If $S$ is a subset of natural numbers with 115 elements, when dividing the elements of $S$ by 27, at least how many elements certainly have the same remainder?
a) Show that among any 2-coloring (Red/Blue) of the vertices of a regular hexagon and its center, there must exist a monochromatic equilateral triangle. b) Extend the argument to show the result holds for a larger configuration of points.
a) $n$ lines are drawn through a point $A$ inside a circle, creating chords. Show that the number of regions created inside the circle is $(n+1)^2$, assuming no three lines meet at any point other than $A$. b) Using the result from part a), find the total number of regions when additional lines are drawn, and show the total is $(n+1)^2 + (2n+1)n$.
The number of triplets $(a, b, c)$ of integers such that $a < b < c$ and $a, b, c$ are sides of a triangle with perimeter 21 is (A) 7 (B) 8 (C) 11 (D) 12
The number of triplets $(a, b, c)$ of integers such that $a < b < c$ and $a, b, c$ are sides of a triangle with perimeter 21 is (A) 7 (B) 8 (C) 11 (D) 12
The number of triplets $(a, b, c)$ of integers such that $a < b < c$ and $a, b, c$ are sides of a triangle with perimeter 21 is (A) 7 (B) 8 (C) 11 (D) 12
A number is called a palindrome if it reads the same backward or forward. For example, 112211 is a palindrome. How many 6-digit palindromes are divisible by 495? (A) 10 (B) 11 (C) 30 (D) 45
An examination has 20 questions. For each question the marks that can be obtained are either $-1$ or $0$ or $4$. Let $S$ be the set of possible total marks that a student can score in the examination. Then, the number of elements in $S$ is (A) 93 (B) 94 (C) 95 (D) 96 .
There are 128 numbers $1,2 , \ldots , 128$ which are arranged in a circular pattern in clockwise order. We start deleting numbers from this set in a clockwise fashion as follows. First delete the number 2, then skip the next available number (which is 3 ) and delete 4 . Continue in this manner, that is, after deleting a number, skip the next available number clockwise and delete the number available after that, till only one number remains. What is the last number left ? (A) 1 (B) 63 (C) 127 (D) None of the above.
Prove that every positive rational number can be expressed uniquely as a finite sum of the form $$a _ { 1 } + \frac { a _ { 2 } } { 2 ! } + \frac { a _ { 3 } } { 3 ! } + \cdots + \frac { a _ { n } } { n ! } ,$$ where $a _ { n }$ are integers such that $0 \leq a _ { n } \leq n - 1$ for all $n > 1$.
A $3 \times 3$ magic square is a $3 \times 3$ rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square
Suppose $f : \mathbb { Z } \rightarrow \mathbb { Z }$ is a non-decreasing function. Consider the following two cases: $$\begin{aligned}
& \text { Case 1. } f ( 0 ) = 2 , f ( 10 ) = 8 \\
& \text { Case 2. } f ( 0 ) = - 2 , f ( 10 ) = 12
\end{aligned}$$ In which of the above cases it is necessarily true that there exists an $n$ with $f ( n ) = n$? (A) In both cases. (B) In neither case. (C) In Case 1. but not necessarily in Case 2. (D) In Case 2. but not necessarily in Case 1.
Let $A = \{ n \in [ 100,700 ] \cap \mathbb { N } : n$ is neither a multiple of 3 nor a multiple of $4 \}$. Then the number of elements in $A$ is (1) 290 (2) 280 (3) 300 (4) 310
Let $S = \{p_1, p_2, \ldots, p_{10}\}$ be the set of first ten prime numbers. Let $A = S \cup P$, where $P$ is the set of all possible products of distinct elements of $S$. Then the number of all ordered pairs $(x, y)$, $x \in S$, $y \in A$, such that $x$ divides $y$, is $\underline{\hspace{2cm}}$.
Let $n$ be a two-digit natural number such that the remainder of $n ^ { 3 }$ divided by 66 is $n$. We are to find the number of such $n$ 's and to find the prime numbers among them. From the conditions we have $$n ^ { 3 } = \mathbf { A B } p + n \quad ( 0 < n \leqq \mathbf { C D } ) ,$$ where $p$ is the integer quotient of $n ^ { 3 }$ divided by 66 . This can be transformed into $$n ( n - 1 ) ( n + 1 ) = \mathrm { AB } p$$ Since either $n - 1$ or $n$ has to be a multiple of $\mathbf { E }$ and either $n - 1 , n$ or $n + 1$ has to be a multiple of $\mathbf { F }$, and furthermore $\mathbf { E }$ and $\mathbf { F }$ are mutually prime, we know that $n ( n - 1 ) ( n + 1 )$ is a multiple of $\mathbf { G }$. (Write the answers in the order $1 < \square < \mathbf { E } < \mathbf { F } < \mathbf { G }$.) Hence one of $n - 1 , n$ and $n + 1$ must be a multiple of $\mathbf{HI}$. So, since $n \leqq \mathrm { CD }$, the number of $n$ 's where $n - 1$ is a multiple of $\mathbf{HI}$ is $\mathbf{J}$, where $n$ is a multiple of $\mathbf{HI}$ is $\mathbf { K }$, and where $n + 1$ is a multiple of $\mathbf{HI}$ is $\mathbf { L }$. Thus, the number of $n$ 's is $\mathbf { M N }$ and the prime numbers among them are $\mathbf { O P } , \mathbf { Q R }$, $\mathbf{ST}$, in ascending order.