Geometric proof (irrationality of 2)

Euclid's Algorithm

Congruence Proofs

Properties of Prime Numbers

Consecutive numbers divisible by squares

Sundaram's sieve

Stirling numbers of 2nd kind

Pythagorean Triples (Consecutive odd integers as one side)

Basic Facts on Primes for middle school

I saw this example and i think i have understood how Chinese Remainder Theorem Works. First lets take three numbers which are relative prime $2,3$ and $5$. Lets try to find a number which gives the remainder $1,2$ and $3$ when divided by those 3 relative primes we chose. Lets write the original number are $x = 15x_1+10x_2+6x_3$ now the nice thing about this $x$ is when we divide by $\frac{x}{2} = \frac{15x_1}{2}=1$ and the smallest value which satisfies this equation is when $x_1 = 1$, Similarly $\frac{x}{3}=\frac{10x_2}{3}\Rightarrow x_2=2$ and $\frac{x}{5}=\frac{6x_3}{5} \Rightarrow x_3 =3$. Thus our number $x = 15x_1+10x_2+6x_3 = 15(1)+10(2)+6(3)=53$. Is this correct ? well this does satisfy the original condition of the remainders, however this is not the smallest number. If we multiply the original numbers we get $2*3*5=30$, now this divisible by all $2,3$ and $5$. So if we subtract the multiples of this number from any number the remainders will not change and hence we can obtain a smaller number which is $53-30=23$

At least one of 3 consecutive odd numbers is a multiple of 3

The way to approach this problem is to show that there will always be a number among three consecutive odd numbers that is divisible by $3$. Let three consecutive odd numbers be $2n+1,2n+3,2n+5$. If $2n+1$ is a multiple of $3$ then we are done. Else the remainder when divided by $3$ is either $1$ or $2$. Suppose its $1$ then it means $2x$ is divisible by $3$ which means $2x+3$ is divisible by $3$. Or suppose the remainder is $2$, which means $2x-1$ is divisible by $3$ which means $2x+2$ and $2x+5$ are divisible by $3$. Hence proved.