JEE Novice - (13)

I don't really know but the way I got the answer was first trying to reduce 286 to a number so that I'd be divisible by seven (finding $c$ ) and then doing the same thing for $\frac{286-c}{7}$ to find $a$ and $b$ ,Is this also valid?

Extra solution exists when $a \le 2$ and $b \le 1$ , because $49\times 1 = 7 \times 7$ , when we decrease $a$ by $1$ , we can add back by increasing $b$ by $7$ . Similarly, when $b = 1$ and $c \le 2$ , we can use $b=0$ and $c+7$ . For the above case of $49a+7b+c=105$ , $\overline{207}$ is another extra solution.

[Response to Challenge Master Note]

There's a unique solution $(a,b,c)$ to the equation $49a+7b+c=x$ where $x$ is a non-negative integer $\leq 342$ if we put the restriction that $a,b,c$ are non-negative integers less than $7$ .

The reason for this is very obvious once we realize that the LHS is actually the base $7$ expansion of $x$ when $a,b,c$ are non-negative integers less than $7$ . So, in general, with that restriction imposed on $a,b,c$ , the equation is:

$\overline{abc}_7=x$

For the problem here, we have $x=286$ . We can get the unique solution by the general method for getting the base $7$ value which is dividing $x$ by $7$ and collecting the remainders (continuing till we get a value less than $7$ ). Then, concatenate the remainders (including the last obtained value) going bottom to top.

The "extra solution" $180$ in the shown case is introduced of this:

$\begin{aligned}\overline{210}_7&=2\times 49+7\times 1+0\\&=1\times 49 + \underbrace{49}_{7\times 7}+1\times 7+0\\&=1\times 49+(7+1)\times 7+0=1\times 49+8\times 7+0\end{aligned}$

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Here's how we get the unique solution (I'm explaining the reason for uniqueness at the bottom of this comment) for the original problem:

$\begin{array}{c|c|c}7&286&6\\ \hline 7&40&5\\ \hline &5\end{array}$

The final column is the remainder column. We concatenate them from bottom to top, getting $\overline{556}_7=\overline{286}_{10}$ .

It can easily be verified that, for the original problem posted by Nihar, this solution $(556)$ is unique by noting the fact that we have the ten's and one's digit greater than $2$ which implies that we can't "increase those digit values" by "contribution from the higher position digits" which was possible in the case of $\overline{210}_7$ .

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Just for added fun, we can analyze it further and get that extra solutions (solutions other than the base $7$ value) exist iff the base $7$ representation of $x$ , which is, say $\overline{pqr}_7$ satisfies one or both of the conditions stated below:

$\begin{cases}r\leq 2~\land~q\geq 2\\ q\leq 2~\land~p\geq 2\end{cases}$

The explanation for this is quite intuitive (trivial) and is left as an exercise to the reader (this comment is already way too long. :P )