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With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures.

For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or $n+2$ digits of the form

$$\underbrace{4\cdot a}_{1\text{ or }2}~ \underbrace{aaa\ldots a}_{n-2} ~\underbrace{7\cdot a}_{2^{*}}.$$

$*$ if $a=1$, then $7 \cdot a$ is the sequence of digits $07$.

I am wondering whether the result above can be proven using some number-theory tool.

Thanks in advance!

  • 0
    Your formula for the initial and final digits is wrong. – 2017-01-04
  • 0
    You added the $\cdot$ after the fact, didn't you ? – 2017-01-04
  • 0
    I was right to be wrong as with your notation no separator means digit concatenation. – 2017-01-04
  • 0
    No problem, don't worry. – 2017-01-04

3 Answers 3

4

Nice observation!

This follows because $aaa\cdots a = a \cdot 111 \cdots 1$ and

$37 \cdot 11 = 407$

$37 \cdot 111 = 4107$

$37 \cdot 1111 = 41107$

$\cdots$

Indeed, let $u_n = 111 \cdots 1$ ($n$ ones). Then $u_n = \dfrac{10^n-1}{9}$ and

$$ \begin{align} 37 \cdot 111\cdots 1 \quad (n \text{ ones}) &= 37u_n\\&= 36u_n + u_n\\ &= 4\cdot9\cdot u_n+10u_{n-2}+11\\ &=4(10^n-1)+10u_{n-2}+4+7\\ &=4\cdot10^n+10u_{n-2}+7\\ &=4111\cdots107 \quad (n-2 \text{ ones}) \end{align} $$

4

This is not only true for $37$ but all $2$-digit numbers of form $[ab]$, where $a + b =10$.

Since $[ab]=10a+b$

$ \begin{align} &\ \ \ \ \ [ab]\times11\\ &=(10a+b)\times11\\ &=100a+10(a+b)+b\\ &=100(a+1)+b\\ &=[(a+1)0b]\\ \end{align} $

In a similar way,

$ \begin{align} &\ \ \ \ \ [ab]\times11...n\ times...11\\ &=10^{n}a+10^{n-1}(a+b)+10^{n-2}(a+b)...+10(a+b)+b\\ &=10^{n}(a+1)+10^{n-1}+...+100+b\\ &=[(a+1)11...n-2\ times...110b]\\ \end{align} $

Now for $[ab]\times [kk..n\ times..kk]$

$ \begin{align} &\ \ \ \ \ [ab]\times11...n\ times...11\times k\\ &=[(a+1)11...n-2\ times...110b]\times k\\ &=[(k\times(a+1))kk..n-2\ times..kk00]+(k\times b)\\ \end{align} $

This with some modification can be applied to all numbers of form $[abc...]$, where $a+b+c+...=10$.

  • 1
    Thanks for the generalization! :) – 2017-01-04
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Observe the patterns in the written calculation of these products

$$a=1\to\begin{matrix}3&7\\&3&7\\&&3&7\\&&&3&7\end{matrix}$$

$$a=2\to\begin{matrix}7&4\\&7&4\\&&7&4\\&&&7&4\end{matrix}$$

$$a=3\to\begin{matrix}1&1&1\\&1&1&1\\&&1&1&1\\&&&1&1&1\end{matrix}$$

$$a=4\to\begin{matrix}1&4&8\\&1&4&8\\&&1&4&8\\&&&1&4&8\end{matrix}$$

$$a=5\to\begin{matrix}1&8&5\\&1&8&5\\&&1&8&5\\&&&1&8&5\end{matrix}$$

$$\cdots$$

$$a=9\to\begin{matrix}3&3&3\\&3&3&3\\&&3&3&3\\&&&3&3&3\end{matrix}$$

A digit of the product, in a complete column, is the sum of the digits of $37\cdot a$ plus a possible carry from the previous column, i.e. the sum of the digits of the sum of the digits of $37\cdot a$.

The initial and final digit pairs correspond to partial sums.

We have

$$ 37\cdot1=37\to10\to1,\\ 37\cdot2=74\to11\to2,\\ 37\cdot3=111\to3\to3,\\ 37\cdot4=148\to13\to4,\\ 37\cdot5=185\to14\to5,\\ \cdots\\ 37\cdot9=333\to9\to9.$$

The phenomenon is explained by the fact that $37=4\cdot9+1\to10\to1$, and this occurs for all factors of the form $k\cdot9+1$. For example,

$$73\cdot66666666666666=4866666666666618.$$