I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let’s call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ […]

We know that $1331$ is divisible by $11$. As per the $11$ divisibility test, we can say $1331$ is divisible by $11$. However we cannot get any quotient. If we subtract each unit digit in the following way, we can see the quotient when $1331$ is divided by $11$. $1331 \implies 133 -1= 132$ $132 […]

How do you prove the following without induction: 1)$\prod\limits_{k=1}^n\left(\frac{2k-1}{2k}\right)^{\frac{1}{n}}>\frac{1}{2}$ 2)$\prod\limits_{k=1}^n \frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$ 3)$\prod\limits_{k=1}^n2k-1<n^n$ I think AM-GM-HM inequality is the way, but am unable to proceed. Any ideas. Thanks beforehand.

I’m solving a problem, and the solution makes the following statement: “The common difference of the arithmetic sequence 106, 116, 126, …, 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3.” Why is this statement true? Where does it come from? Is it generalizable […]

Let $a,b \in \mathbb{N}^{*}$. Prove that $\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational. If $(a,b)=(k,k\cdot6)$, then $\sqrt{13a^2+b^2}$ is rational, but I do not know if those are the only solutions.

This question already has an answer here: How do I compute $a^b\,\bmod c$ by hand? 7 answers

Given a number $n$, in how many ways can you choose two factors that are relatively prime to each other (that is, their greatest common divisor is 1)? Also, am I going in the correct direction by saying if $n$ is written as $p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$, where $p_i$ is a prime and $a_i\geq 1$, then the […]

If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$ , then is it true that $a=b=c=0$ ? I was thinking of infinite descent but can’t actually proceed , please help. Thanks in advance

I need to prove there are constants $c_1,c_2 > 1$ such that $$ c_1^n < \mathrm{lcm}(1,\dots,n) < c_2^n $$ for $n$ integer, $n \geq 2$. I tried to use that $\mathrm{lcm}(1,\dots,n)=\prod_{p \leq n} p ^{b_p}$ where $b_p$ is the greatest integer such that $p^{b_p} \leq n$. Maybe $c_2 = 2$? No ideas about the value […]

Sloane’s A053614 implies that $2, 5, 8, 12, 23$, and $33$ are the only natural numbers $n \geq 1$ which cannot be written as the sum of distinct triangular numbers (i.e., numbers of the form $\binom{k}{2}$, beginning $1,3,6,10,15,\ldots$). Question: How to prove that any natural number $n \geq 34$ can be written as the sum […]

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