Henry Ernest Dudeney/Modern Puzzles/57 - A Misunderstanding/Solution

by : $57$

 * A Misunderstanding

Solution

 * $\dfrac {857142} 3 = 285714$

Proof
We are being asked to find a number:


 * $N = \sqbrk {a_k a_{k - 1} \ldots a_2 a_1}_{10}$

such that:


 * $\dfrac N 3 = \sqbrk {a_1 a_k a_{k - 1} \ldots a_2}_{10}$

where $a_1$ is required to be $2$.

We extract the general case below:

Let $q$ be the rational number which can be expressed as:


 * $q = 0 \cdotp \dot a_k a_{k - 1} \ldots a_2 \dot a_1$

such that:
 * $\dfrac q 3 = 0 \cdotp \dot a_1 a_k a_{k - 1} \ldots \dot a_2$

Then:

From the construction of the problem, the possible values for $a_1$ are $1$ and $2$, giving:

as possible answers.

From 7 is Cyclic Number:


 * $q = 6 \times 0 \cdotp \dot 14285 \dot 7 = 0 \cdotp \dot 85714 \dot 2$

which leads us to:
 * $N = 857142$

So the specific answer required was:
 * $\dfrac {857142} 3 = 285714$

while we also have:
 * $\dfrac {428571} 3 = 142857$