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Find the smallest integer $a > 2$ such that $2|a, 3|a + 1, 4|a + 2, 5|a + 3, 6|a + 4$

The question is from David Burton's book on Number Theory. The solution which I have found is like this:

Since $2|a$ is equivalent to $4|a + 2$ and $6|a + 4$ is equivalent to $3|a + 1$, so the given problem reduces in finding the smallest integer $a > 2$ such that $3|a + 1, 4|a + 2, 5|a + 3$. Thus, we have to find the least value of $a > 2$ such that $a \equiv 2 \mod 4$, $a \equiv 2 \mod 5$ and $a \equiv 2 \mod 6$. Since 4 and 6 are not co-prime, so Chinese Remainder Theorem is not applicable. However, we observe that $a = 62$ serves as a solution. This is the smallest solution because $a \equiv 2 \mod 6$ implies that if $2 < a < 62$, then possible values for $a$ are 8, 14, 20, 26, 32, 38, 44, 50 and 56. But $8 \not \equiv 2 \mod 5$, $14 \not \equiv 2 \mod 5$, $20 \not \equiv 2 \mod 5$, $26 \not \equiv 2 \mod 5$, $32 \not \equiv 2 \mod 4$, $38 \not \equiv 2 \mod 5$, $44 \not \equiv 2 \mod 5$, $50 \not \equiv 2 \mod 5$, and $56 \not \equiv 2 \mod 5$. Thus, the desired least value of $a$ is 62.

Khalid Naeem
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  • As to how @barakmanos found it, hint: think about the $LCM(2,3,4,5,6)$ – Shraddheya Shendre Jan 16 '17 at 18:08
  • It holds for $a=60n+2$, where $n$ is a positive integer. – barak manos Jan 16 '17 at 18:10
  • Would you please give some detail about how you knew that the solution is $a = 60n + 2$? – Khalid Naeem Jan 16 '17 at 18:28
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    Not sure why this was voted to be close. But $2|a$ is equivalent to $4|a+2$???? What if a =, say, 16? $6|a+4$ equivalent to $3|a+1$? What if $a$ is even, say $a = 5$ $3|5+1$ but $6|5+4$. Not equivalent at all. – fleablood Jan 17 '17 at 20:43
  • But they are conditional. If $4|a+2$ then $2|a+2$ and $2|a$ (but the reverse direction does not work). If $6|a+4$ then $3|a+4$ so $3|a+1$ (but the reverse direction does not work).... Actually $6|a+4 \implies$ both $3|a+1$ and $2|a$. – fleablood Jan 17 '17 at 20:46

1 Answers1

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$\ \begin{align}{\bf Hint}\qquad\quad &a\equiv 2\! \pmod{ 2,3,4,5,6}\\[.1em] \iff\ &2,3,4,5,6\mid a\!-\!2\\[.1em] \iff\ &60\mid a\!-\!2, \ \ {\rm by} \ \ 60 = {\rm lcm}(2,3,4,5,6)\\[.1em] \iff\ &a\equiv2\! \pmod{60}\end{align}$

Bill Dubuque
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