Find the smallest natural number $n$ such that rightmost digit is $6$ and when we deleted that digit $6$ and add it to the left of the number we get $4n$.
Example of the operation:
$123456$ becomes $612345$
$16$ becomes $61$
$66$ becomes $66$
[Math] Find the smallest natural number $n$
decimal-expansionelementary-number-theorynumber-systems
Related Solutions
Using a quick search with my computer, here’s what I found so far :
Looking only at the last digit, some power of $n$ must be congruent to $1$ modulo $10$, so $n$ must be congruent to $1,3,7$ or $9$ modulo $10$.
Looking at the last two digits, some power of $n$ must be congruent to $11$ modulo $100$, so $n$ must be congruent to $11,31,71$ or $91$ modulo $100$.
Looking at the last three digits, ome power of $n$ must be congruent to $111$ modulo $1000$, so $n$ must be congruent to one of $31, 71, 111, 191, 231, 271, 311, 391, 431, 471, 511, 591, 631, 671, 711, 791, 831, 871, 911, 991$ modulo $1000$.
Looking at the last four digits, ome power of $n$ must be congruent to $1111$ modulo $10000$, so $n$ must be congruent to one of $71, 1031, 2071, 3031, 4071, 5031, 6071, 7031, 8071, 9031$ modulo $10000$.
Do you see the pattern in those sequences ?
The Chinese remainder theorem is about relatively prime modulos. Having all the modulos from $2$ to $13$ is redundant.
$x\equiv 1 \mod 2; x \equiv 3\mod 4; x\equiv 7\mod 8$ are redundant and can be replaced with just $x \equiv 7 \pmod 8$.
Assuming the question is legitimate, we need not consider any $x \equiv j \pmod {2^km}$ and have to consider only $x\equiv l\pmod m$
i.e., this question can be reduced to.
$x\equiv 7\pmod 8$
$x \equiv 8\pmod 9$
$x \equiv 4 \pmod 5$
$x \equiv 6\pmod 7$
$x \equiv 10 \pmod {11}$
$x \equiv 0 \pmod {13}$.
All the rest are redundant, as the solution to the above is unique.
Note the solution to all but $x \equiv 0 \pmod {13}$ is $x \equiv -1 \pmod n$ for $n = 8,9,5,7,11$ so
$x\equiv -1 \pmod{8*9*5*7*11=27720}$ and $x \equiv 0 \pmod {13}$.
So you need to solve. $x = 27720k -1 = 13m$
$27720 \equiv 4 \pmod {13}$
So $x \equiv 27720k - 1\equiv 0 \pmod {13}$
$\equiv 4k -1 \equiv 0 \pmod {13}$
So $4k \equiv 1 \pmod {13}$ so we just have to find the inverse of $4 \mod {13}$
And $4 \times 10 = 40 \equiv 1 \pmod{13}$ and so $k = 10$
and $x \equiv 277199 \pmod {27720*13}$
Best Answer
If $n$ ends with 6, then $4n$ ends with 4 (because $6\cdot 4 = 24$). So the tail looks like
Then we know $n$ ends with 46 – and so on:
Done.
An example of a larger number with the same property is 153846153846153846.