[tex]\displaystyle \textsf{Dividend - Divisor(Quotient) = Remainder}\\\displaystyle \textsf{Dividend = Divisor(Quotient) + Remainder}[/tex]
[tex]\\[/tex]
[tex]\displaystyle \textsf{Let } x\textsf{ be the number, }q_1, q_2, q_3\textsf{ are the quotients where } q_1, q_2, q_3\in \mathbb{W}\\[/tex]
[tex]\displaystyle x = 11(q_1) + 1\\\displaystyle x = 13(q_2) + 11\\\displaystyle x = 17(q_3) + 4\\[/tex]
[tex]\displaystyle x=11(q_1)+1\\\displaystyle 11(q_1) + 1 = 13(q_2) + 11\\\displaystyle 13(q_2) + 11 = 17(q_3) + 4\\[/tex]
[tex]\displaystyle x=11(q_1)+1\\\displaystyle q_1 = \frac{13(q_2) +10}{11}\\\displaystyle q_2 = \frac{17(q_3)-7}{13}\\[/tex]
[tex]\displaystyle \textsf{We search for }q_2,q_3\textsf{ such that }q_1,q_2,q_3 \in \mathbb{W}\\[/tex]
[tex]\displaystyle \textsf{We find } q_3\textsf{ through brute force or by trying the numbers in } \mathbb{W}\texsf{: }\\[/tex]
[tex]\displaystyle q_3 = \{ 0, 1, 2, 3, 4, \boxed{5}, 6, 7, 8, 9, 10, \dots \}\\\\\displaystyle \textsf{By careful substitution, 5 is a valid solution: }\\[/tex]
[tex]\displaystyle q_2 = \frac{17(q_3) -7}{13}\\\displaystyle q_2 = \frac{17(5) -7}{13}\\\displaystyle q_2 = \frac{78}{13}\\\\\displaystyle \boxed{ q_2 = 6}\\[/tex]
[tex]\displaystyle q_1 = \frac{13(q_2)+10}{11}\\\displaystyle q_1 = \frac{13(6)+10}{11}\\\displaystyle q_1 = \frac{88}{11}\\\\\displaystyle \boxed{q_1 = 8}\\[/tex]
[tex]\displaystyle x = 11(q_1)+1\\\displaystyle x = 11(8)+1\\\displaystyle x = 88+1\\\\\displaystyle x = \boxed{ 89 }\\[/tex]
[tex]\\[/tex]
[tex]\displaystyle \textsf{There we go, the answer is \boxed{ 89 }.}\\\\\displaystyle \textsf{You may wonder, is 89 the only answer? No, 89 is the smallest number }\\\displaystyle \textsf{that satisfies the problem and there are more numbers out there.}[/tex]
