Excellent question! The answer is $127$ and $128$... but why? If you wanted to find a number divisible by $1,2,3,4$ you might first multiply these numbers and say $24$. However, you soon realize $4$ is already a multiple of $2$; you can use just $3\times4$ to get $12$. Therefore, you need only multiply the largest powers of the primes that factor all of the digits from $2$ to $200$ to get a number that is divisible by all of the integers from $1$ to $200$.
If you do this; you will find the number is $2^7\cdot3^4\cdot5^3\cdot7^2\cdot11^2\cdot13^2\cdot17\cdot19\cdot23\cdot29\cdot\ldots$(the rest of the primes up to $199$) = a very large number.
Next we need to find a restriction to eliminate two consecutive numbers. One of the two numbers must be even. The only way to remove an even number from the above calculation without modifying any of the other primes is to reduce the power of $2^7$ to $2^6$; this removes the number $128$ from the list. Since $127$ is also a prime number, it can also be removed from the list without affecting any of the other primes in the list...
I hope this helps.
Your final answer is fine, but your method looks highly suspect at one particular line which needs clarification. You write:
The number must be the lowest common multiple of $3x+3$ , $4x+6$ and
$5x+10$, which is $60x + 30$.
But it’s not clear what you mean by this, as the lowest common multiple of those three numbers is not $60x+30$ (try it for $x=2$). Maybe what you’re actually doing to arrive at $60x+30$ is correct, or maybe it isn’t — but at a minimum you need to explain what you’re doing because “lowest common multiple” doesn’t describe it adequately, and the reuse of the same $x$ for three different values is a big red flag.
My hunch is that you’re taking the separate LCMs of the coefficients $(3,4,5)$ and of $(3,6,10)$ and then gluing them back together. If so then this that’s definitely wrong. Consider the slight variation where 3,4,5 are replaced by 2,3,5. Then you’d have $2x+1, 3x+3, 5x+10$ combining to $30x+30$, which never works (it’s even, so it can never be the sum of two consecutive integers). Being a multiple of $2x+1$ is very different from being. equal to $2x+1$.
This broken method lucks out sometimes because of the fact that it basically works for odd coefficients (when $k$ is odd, being the sum of $k$ consecutive integers is the same as being divisible by $k$). But if that’s really what you’re doing then it is broken and should not be part of your solution.
Best Answer
The brute force method: as $m^3−3m^2+2m=m(m−1)(m−2)$, and $79,83$ are prime, you can just solve the following nine congruences: $m\equiv\alpha\pmod{79}$, $m\equiv\beta\pmod{83}$, where $\alpha,\beta\in\{0,1,2\}$. This is possible as per Chinese Remainder Theorem, and the smallest of the nine $m$'s you will get (greater than $2$) is the solution.
It is easy to solve all those congruences simultaneously: per Wikipedia, we first express $1$ as $1=79u+83v$, where $u,v$ can be found using Euclidean algorithm. In this case, as $4=83-79$ and $1=20\cdot 4-79$, we have $1=20\cdot 83-21\cdot 79$.
Now, $m\equiv\alpha\pmod{79}$ and $m\equiv\beta\pmod{83}$ resolves as $m\equiv 20\cdot 83\alpha-21\cdot 79\beta\pmod{79\cdot 83}$, i.e. $m\equiv 1660\alpha-1659\beta\pmod{6557}$. This gives us the following table:
$$\begin{array}{r|r|r|r}\alpha&\beta&m\pmod{6557}&\text{smallest }m\gt 2\\\hline0&0&0&6557\\0&1&4898&4898\\0&2&3239&3239\\1&0&1660&1660\\1&1&1&6558\\1&2&4899&4899\\2&0&3320&3320\\2&1&1661&1661\\2&2&2&6559\end{array}$$
so the smallest solution seems to be $m=1660$.