Coerced Divisors

Relevant wiki: Lowest Common Multiple - Word Problems

$\begin{aligned}k+1\mid A+k&\implies A+k\equiv 0 \pmod{k+1} \\ &\implies A\equiv -k\quad\pmod{k+1} \\&\implies A\equiv 1\qquad\pmod{k+1}\quad \text{by adding (k+1)}\end{aligned} \\ \therefore A\equiv 1\pmod2\equiv 1\pmod3\equiv \ldots \equiv 1\pmod8 \\ \equiv 1\pmod{lcm(2,3,...,8)} \\ \equiv 1\pmod{840} \\=\boxed{841}\quad \text{only 3 digit number that works}$

Relevant wiki: Lowest Common Multiple - Word Problems

$k+1|A+k$ $k+1|A-1+(k+1)$ $k+1|A-1$ Hence $A-1$ is divisible by $2,3,\ldots, 8$ .

Their LCM is $8*7*5*3=840$ and any other multiple of $840$ is not a $3$ -digit number.

Thus $A=\boxed{841}$ .

Due to $A +1$ is divisible by $2 \Rightarrow A$ is an odd number.

Due to $A + 4$ is divisible by $5 \Rightarrow A$ finishes at 1, because if $A +4$ ended in $0$ , $A$ wouldn't be an odd number(contradiction with the previous argument). So, so far, the 3- digits number $A$ ends in $1$ .

Now, due to $A + 6$ ends in $7$ and is divisible by $7$ , the 2 first digits of $A$ have to be a multiple of $7$ (Realize that there isn't other possibility) . So, the only possibilities for $A$ are $141, 211, 281, 351, 421, 491, 561, 631, 701, 771, 841, 911$ or $981$ .

Using hit and trial: $A$ can't be $141$ because $143$ is not divisible by $3$ .

$A$ can't be $211$ because $214$ is not divisible by $4$ .

$A$ can't be $281$ because $283$ is not divisible by $3$ ...

Keeping on like this , we arrive to $841$ which fullfills all the requisites...

Note that the unit digit of the number can only be 1, because adding 4 to others doesn't suffice the condition for divisibility by 5. Therefore the number when added 6 to has 7 and in its unit place and the multiples of 7 have 7 in their unit place according to the pattern 7, 77, 147, 217, 287 and so on. Therefore the possible candidates are 141, 211, 281 and so on till 981. Note that among these only 841 satisfies all the conditions. Hence the answer.

From the question it is clear that, we are looking for a number A such that. $\frac{A+1}{2}$ $\frac{A+2}{3}$ $\frac{A+3}{4}$ $\frac{A+4}{5}$ $\frac{A+5}{6}$ $\frac{A+6}{7}$ $\frac{A+7}{8}$ are perfectly divisble.

Now, lets say ( A + 6 ) is a multiple of 7. So there exist some "m", such that ( A + 6) = 7 m. There also exist some (n+1) , such that (A+6) = 7(n +1) , where n is an Integer.

So, A + 6 = 7n + 7

$\Rightarrow$ A - 1 = 7n.

$\Rightarrow$ (A - 1) must be a mulitple of 7.

Applying same logic for all other values , we will find that , (A-1) must be a multiple of 2,3,4,5,6,7 & 8.

Now LCM of 2,3,4,5,6,7,8 is 840. So (A - 1)= 840.

So A can be 841 or 840 x 2 + 1 or 840 x 3+1 ... and so on. But since the number is a 3 digit number.

A can be 840+1 Hence A = 841