A number when divided by 11 leaves a remainder of 1, when divided by 13 leaves a remainder of 11, when divided by 17 leaves a remainder of 4. What is the number?, Please answer it correctly, show your step by step solution. If you do not know the answer, do not answer. As simple as that.

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A number when divided by 11 leaves a remainder of 1, when divided by 13 leaves a remainder of 11, when divided by 17 leaves a remainder of 4. What is the number?


Please answer it correctly, show your step by step solution. If you do not know the answer, do not answer. As simple as that.

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\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}\\

\displaystyle x = 11(q_1) + 1\\\displaystyle x = 13(q_2) + 11\\\displaystyle x = 17(q_3) + 4\\

\displaystyle x=11(q_1)+1\\\displaystyle 11(q_1) + 1 = 13(q_2) + 11\\\displaystyle 13(q_2) + 11 = 17(q_3) + 4\\

\displaystyle x=11(q_1)+1\\\displaystyle q_1 = \frac{13(q_2) +10}{11}\\\displaystyle q_2 = \frac{17(q_3)-7}{13}\\

\displaystyle \textsf{We search for }q_2,q_3\textsf{ such that }q_1,q_2,q_3 \in \mathbb{W}\\

\displaystyle \textsf{We find } q_3\textsf{ through brute force or by trying the numbers in } \mathbb{W}\texsf{: }\\

\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: }\\

\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}\\

\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}\\

\displaystyle x = 11(q_1)+1\\\displaystyle x = 11(8)+1\\\displaystyle x = 88+1\\\\\displaystyle x = \boxed{ 89 }\\

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\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.}

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