This is how I have attempted this problem: $37abc$,$37bca$ and $37cba$ have to be multiples of $37$. Since they all begin with $37$(which is a multiple of $37$), we need to find a number $abc$ such that $bca$ and $cba$ are also multiples of $37$. Using the following proof, we can conclude that for all multiples of $37$ less than $1000$, $bca$ and $cba$ are also multiples of $37$: Let us say $abc$ is a multiple of $37$.

$$abc= 100a+10b+c\;,\tag1$$


Subtracting $2$ from $1$,

$$abc-bca=100a-a+10b-100b+c-10c=99a-90b-9c=9(11a-10b-c)\;,$$ $$9(11a-10b-c)\equiv 9(-26a-10b-c)\equiv -9(26a+10b+c) \equiv -9(100a+10b+c) \equiv -9(0) \equiv 0\mod 37\;.$$

Hence, $abc-bca$ is divisible by $37$. Since, $abc$ is divisible by $37$, $bca$ should also be divisible by $37$. We can prove similarly that $cab$ is also always divisible by $37$ whenever $abc$ is divisible by $37$. Hence for all $abc$'s that are a multiple of $37$, $bca$ and $cab$ are a multiple of $37$. My doubt is what would be the final answer? Is it $27$?

  • $\begingroup$ $37111,37222,37333,37444,37555,37666,37888, 37999$ $\endgroup$ – N.S.JOHN Jun 18 '16 at 7:06

Hint $\,{\rm mod}\ 37\!:\ \color{#c00}{10^3\equiv 1}\,$ so multiplication by $10\,$ does a cyclic shift $\, abc \mapsto bca,\,$ i.e.

$$\begin{eqnarray} abc_{\,10} &=&\, 10^2 a +\, 10\ b + c\\[.3em] \Rightarrow\ 10\,abc_{\,10} &\equiv&\, \underbrace{\color{#c00}{10^3}}_{\large\color{#c00}1} a + 10^2 b + 10 c \,\equiv\, bca_{\,10}\end{eqnarray}$$

Therefore the problem reduces to counting the number of multiples of $\,37\,$ in the interval $\,[37000,37999].\,$ But your count is one low.


$27$ is almost but not quite right. The first of the numbers you found is $37000$, and the last is $37999$, so there are $\frac{37999-37000}{37}+1=9\cdot\frac{111}{37}+1=28$ of them.

A shorter proof of the divisibility would have been

$$ bca=10\cdot abc-999\cdot a\equiv 10\cdot abc\mod37\;. $$


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