Let $a$ be a positive integer and let $b$ be obtained from a by moving the initial digit of $a$ to the end.
Prove that it is impossible to have $b=5a$.
Prove that it is impossible to have integers $b=5a$ under a digit re-ordering
3
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number-theory
elementary-number-theory
recreational-mathematics
problem-solving
decimal-expansion
2 Answers
9
Let us assume there exists such naturals $a,b$ that $b=5a$. Note that $b$ cannot have more digits than $a$ from the definiton of $b$.
We have from that $b=5a$ that $b\equiv 0 \pmod {5}$, or the last digit of $b$ must be $5$ or $0$. However, the leading digit of $a$, which is the last digit of $b$ is non-zero. Thus the last digit of $b$ must be $5$.
So,the first digit of $a$ is $5$. However, because of carrying, this implies that $b=5a$ must have more digits than $a$. We are done.
6
If $a$ has $n$ digits, the first of which is $d$, then $b=10(a-10^{n-1}d)+d=10a-(10^n-1)d$. If $b=5a$, then
$$5a=(10^n-1)d$$
Clearly $5\not\mid(10^n-1)$, so we must have $5\mid d$, which implies $d=5$ or $d=0$. But the lead digit of a positive integer cannot be $0$, which leave $d=5$, in which case $a=10^n-1$, all of whose digits are $9$'s. We conclude that $b=5a$ is impossible.