# Move the last digit of $10k+9$ to the front, and obtain $70k+63$

"N" is a number and its last digit is 9. Delete the last digit (9) and write (9) to first digit, new number is 7 times "N". so whats that "N" ?

Could anyone show me a way, how to solve such tasks?

EDIT:

These are my thoughts so far:

N := 10 x + 9 ( 1 )

where 10 x is the ( integer ) value of anything but the last 9. Now put the 9 in front:

N ' := 9 * 10 ^ ( d - 1 ) + x = 7 N = 70 x + 63 ( 2 )

but i do not know how to continue from here

EDIT2:

d is the number of digits of N.

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The paper "Multiplicative twins and primitive roots" by Bernstein (1968) identifies all numbers where moving the last digit to the first position results in a multiple of the number. (doi 10.1007/BF01135448, MR0225709) – Kundor Apr 4 '14 at 20:45

You have made a good start. Your equation 2 is $9\cdot 10^{d-1}+x=70x+63$ or $x=\frac 9{69}(10^{d-1}-7)$ so you need $10^{d-1} \equiv 7 \pmod {69}$ Now you can just start looking. Increase $d$, multiply by $10$ and reduce the result $\pmod {69}$ It starts like this $$\begin {array}{c | c}d-1&10^{d-1}\\ \hline 1&10\\2&10\cdot 10 =100 \equiv 31 \pmod {69}\\ 3&10 \cdot 34 = 310 \equiv 34 \pmod {69}\end{array}$$ Keep going until you hit $7$

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would you please take a look at my answer and say if i found the correct result? – Felix Lahmer Apr 2 '14 at 13:33

N' := 7*N

if the last digit of N = 9 then the last digit of N is like the last digit of 9*7, which is 3 (Remember 6 carry). The 3 is also the penultimate digit of N. 3*7+6 = 27 -> 7 is the next number (2carry)

... and so on, till the next Digit is 9 and carry is 0.

N Ü N'
9 6 3
3 2 7
7 5 1
1 1 2
2 1 5
5 3 6
6 4 5
5 3 9
9 6 6
6 4 8
8 6 0
0 0 6
6 4 2
2 1 8
8 5 7
7 5 4
4 3 3
3 2 4
4 3 0
0 0 3
3 2 1
1 0 9

N = 1304347826086956521739
N' = 9130434782608695652173

N2 = 13043478260869565217391304347826086956521 739
N2' = 91304347826086956521739130434782608695652 173


If i say that N is a dezimal number between 0 and 1 i get the same digits but faster

n/10 + 0,9 = 7n + ε

n + 9 = 70n + ε

n = (9-ε)/69 = (3-ε)/23 = 3/23 - ε

n = 0,(Period)1304347826086956521739 - ε

n = 0,1304347826086956521739

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Yes, this is a fine way to do it. – Ross Millikan Apr 2 '14 at 13:48

Let us call the original number $N$, and the modified number $M$. Thus $N$ ends with a 9 and $M$ begins with a 9. Since $N = M\div 7$, we can set up a short division problem that looks like this:

$$\require{enclose} \begin{array}{rl} & \ \,1\ldots\\ 7&\enclose{longdiv}{9\ldots} \end{array}$$

Here $N$ is the quotient and $M$ the dividend. We see that $N$ begins with the digit 1, so this is the second digit of $M$:

$$\begin{array}{rl} & \ \,1\ldots\\ 7&\enclose{longdiv}{91\ldots} \end{array}$$

Now that we know the second digit of $M$, we can divide one more step, computing the next digit of $N$:

$$\begin{array}{rl} & \ \,13\ldots\\ 7&\enclose{longdiv}{913\ldots} \\ \\ & \ \,130\ldots\\ 7& \enclose{longdiv}{9130\ldots} \\ \\ & \ \,1304\ldots\\ 7& \enclose{longdiv}{91304\ldots} \\ \\ \end{array}$$

We must continue in this fashion until we reach a point where there the remainder is 0 and the quotient ends with 9. This first occurs at:

$$\begin{array}{rl} & \ \,1304347826086956521739\\ 7& \enclose{longdiv}{9130434782608695652173} \\ \\ \end{array}$$

This is the smallest answer. By continuing to divide until the next stopping point, we can reach larger solutions, such as 91304347826086956521739130434782608695652173.

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thanks, that was my final way of going – Felix Lahmer Apr 2 '14 at 13:54

We require $$y=7x$$ or in other words $$9a_1a_2\dots a_n = 7(a_1a_2\dots a_n 9)$$ where the $a_i$ represent decimal digits.

Multiplying both sides by $10$, we have $$9a_1a_2\dots a_n 0 = 70x$$ or $$9\times 10^{n+1} + (x - 9) = 70x$$ or $$69x = 9\times 10^{n+1} - 9\,.$$ The answer is not unique. For example, $$x = 1,304,347,826,086,956,521,739$$ and $$x = 13,043,478,260,869,565,217,391,304,347,826,086,956,521,739$$ both have the desired property (these correspond to $n=21$ and $n=42$, respectively.

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yubb, you are right I took the smallest natural number N from this set in my solution. But thanks for pointing it out. – Felix Lahmer Apr 2 '14 at 13:56