For $x\neq 1$, find $X$ such that: $$x+1\mid 2$$ $$x+2\mid 3$$ $$x+3\mid 4$$ $$x+4\mid 5$$ Can someone help me with this ?
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$X=-179$ is one solution. – bof Dec 09 '20 at 13:52
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1Can you conclude something about some divisors of $X-1$? – Hagen von Eitzen Dec 09 '20 at 13:56
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What about $X = 1 + 120k$ for $k=1,2,3...$? – Infinity_hunter Dec 09 '20 at 14:10
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2@HagenvonEitzen Thank you! the divisors of $X-1$ are $3,4,5$ so $X=60*k+1$ – medamin Dec 09 '20 at 14:22
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Given that $$x+1\mid 2$$ $$x+2\mid 3$$ $$x+3\mid 4$$ $$x+4\mid 5$$
We rewrite it in modular form.
$$x+1\equiv 0 (\mod 2)$$ $$x+2\equiv 0 (\mod 3)$$ $$x+3\equiv 0 (\mod 4)$$ $$x+4\equiv 0 (\mod 5)$$
Subtracting so only $x$ is on the left, you get, by rules of modular arithmetic,
$$x\equiv 1 (\mod 2)$$ $$x\equiv 1 (\mod 3)$$ $$x\equiv 1 (\mod 4)$$ $$x\equiv 1 (\mod 5)$$ Therefore, $x-1$ is divisible by $2, 3, 4, 5$. The least common multiple of $2, 3, 4, 5$ is $60$, and so one solution is $x-1=60$, and so $x=61$ is a solution.
A more general solution is all numbers one more than a multiple of 60, usually written as $x=60k+1$ for $k\in \mathbb {Z} $ (But $k \neq 0$ because $x \neq 1$).
KingLogic
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