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I was looking at this problem: The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers? I was wondering if you could solve this using Chinese Remainder Theorem, I am new to using Chinese Remainder Theorem but I'm fairly certain you need the GDC for all mods equal to 1 and clearly GDC(4,6)=2, so I don't know what to do.

  • Call the answer $n$. From $n\equiv 1\pmod 6$, this breaks into both $n\equiv 1\pmod 2$ and $\pmod 3$. – peterwhy Jan 13 '24 at 03:13
  • Notice that if $x \equiv 1 \pmod {nm}$ then $nm|x-1$ so $n|x-1$ and $m|x-1$ so $x\equiv 1\pmod n$ and $x \equiv 1 \pmod m$. So, .... sure, $4$ and $6$ are not relatively prime but $4$ and $3$ are and as $x\equiv 1 \pmod 6$ you have $x\equiv 1 \pmod 3$. So you have $x\equiv 1 \pmod 4$ and $x \equiv 1 \pmod 5$ and $x\equiv 1\pmod 3$. Chinese Remainder away till you are blue in the face. – fleablood Jan 13 '24 at 03:32
  • @peterwhy, I don't really understand how? –  Jan 13 '24 at 03:58
  • Applying CCRT twice: $\ x\equiv a\pmod {!m\ &\ n\ &\ k}\iff x\equiv a\pmod{!\ell},,\ \ell = {\rm lcm}(m,n,k)$ – Bill Dubuque Jan 13 '24 at 09:03
  • Generally for noncoprime moduli a slight tweak of the CRT formula works, see general Easy CRT. In particular this method uses only fast gcds (not slow prime factorizations), which is crucial when the numbers are large. – Bill Dubuque Jan 13 '24 at 09:08

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If $x$ is $1$ more than a multiple of $4$, and of $5$, and of $6$, then it is $1$ more than a multiple of their least common multiple (LCM).

Bill Dubuque
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 13 '24 at 09:04