Special Numbers

Logic Level 3

Suppose $M$ is a four-digit positive integer such that $4M$ is also a four-digit number and $4M$ has the same digits as $M$ , but in the reverse order. Find the sum of all possible $M$ .

As an illustration, if $M$ is of the form $\overline{abcd}$ , then $4M$ is of the form $\overline{dcba}$ , where $a$ , $b$ , $c$ and $d$ are digits.

The answer is 2178.

1 solution

Guillermo Templado
Jul 13, 2016

$1000 < M < 2500$ becuase if $M \ge 2500$ then $4M \ge 10000$ , and M isn't able to start for 1, because there doesn't exist any integer number $M$ such that $4M$ ends at 1. So, $2000 \leq M < 2500$ . Therefore, with the same previous reasoning $M = 2bc8$ because $M$ only can end at 8 or 9, but there doesn't exist any integer $M$ such that $4M$ ends at 9 and this implies that $b = 0 \text{ or } 1 \text{ or } 2$ because if $M \ge 2308$ then $4M \ge 9232$ ( and this is a contradiction). With hit and trial about $c$ the only possibility is $c = 7$ ... And hence, the only number is $2178$ ,,,,,, $4\times 2178 =8712$

Yes,that is the way I approached the problem.Nice solution.

Indraneel Mukhopadhyaya - 4 years, 11 months ago

Special Numbers

The answer is 2178.

1 solution

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