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What is the smallest whole number such that, when divided by each of $10,9,8,7,...,2$ gives a remainder of $9, 8,7,6,...,1$, respectively?

One can use the Chinese remainder theorem for each pair of $3$ or $4$ numbers, but for $9$ numbers it'll take quite a while. Is there another way or formula to do this problem? Thanks.

Vinyl_cape_jawa
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user526427
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    If $n\equiv k-1\bmod k$, then $n\equiv-1\bmod k$. So, you are looking for a number that is equivalent to $-1$ mod each of these numbers. So, $n=\operatorname{lcm}(2,3,4,5,6,7,8,9,10)-1=2519$. – Rushabh Mehta Mar 04 '20 at 16:20
  • This problem is not about knowing more theorems like the CRT in Number Theory but noticing and exploiting a visible pattern. It won't serve as a rigorous proof sometimes to exploit patterns but seeing these patterns gives you the solution. Like the above Comment, There's nothing new than just see that pattern in the given question and applying the right considerations. – RishiNandha Vanchi Mar 04 '20 at 16:23
  • Note many of the equations are redundant. $x\equiv 8\pmod 9\implies x\equiv 2\pmod 3$ and solving for $x\equiv a\pmod{89}$ makes solving for $x\equiv 8\pmod 9$ and $x\equiv 5\pmod 6$ redundant. Only have to solve for the $x\equiv p^k-1\pmod{p^k}$ for the primes. That is $x\equiv 8\pmod 9$, $x\equiv 7\pmod 8;x\equiv 6\pmod 7;x\equiv 4\pmod 5$. Use CRT to solve $x\equiv a\pmod{95}$ and $x\equiv b\pmod{87}$ and Use CRT again to solve $x\equiv N\pmod{9587}$. That's not long or hard. (I fit it in a single comment.....) – fleablood Mar 04 '20 at 16:47
  • Also.... Heh, heh..... For $m = 10.....2$ we have $x \equiv -1 \pmod m$ so $x\equiv -1 \pmod {958*7}$. – fleablood Mar 04 '20 at 16:49

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