property of a 6n digit number divisible by 7

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property of a 6n digit number divisible by 7

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Source: Mathematics Regional Olympiad of Mexico Center Zone 2011 P4

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#1 h Nov 11, 2021, 6:33 AM

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Show that if a $6n$ -digit number is divisible by $7$ , then the number that results from moving the ones digit to the beginning of the number is also a multiple of $7$ .

This post has been edited 2 times. Last edited by parmenides51, Nov 11, 2021, 3:02 PM

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#2 h Nov 11, 2021, 10:27 AM

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parmenides51 wrote:

Show that if a $6n-$ digit number is divisible by $7$ , then the number that results from moving the ones digit to the beginning of the number is also a multiple of $7$ .

can someone pls give an example of 6n digit number? is it like 123456 is a 6n digit number

This post has been edited 1 time. Last edited by TheMath_boy, Nov 11, 2021, 10:28 AM

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TheCollatzConjecture

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TheCollatzConjecture

#3 h Nov 11, 2021, 11:35 AM • 1 Y

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Let the unit digit of the $6n$ digit number be $a_0$ , and let the number itself be $7k$ for some integer $k$ . Now the new number is the same as $a_0 \cdot 10^{6n-1} +\frac{7k -a_0}{10}$ . We just need to prove that this new expression is $\equiv 0 (mod ~7)$ . Now $a_0 \cdot 10^{6n-1} \equiv a_0 \cdot 3^{6n-1} \equiv a_0 \cdot \frac{3^{6n}}{3} \equiv \frac{a_0}{3}$ (Since $729 \equiv 1$ ). $\frac{7k -a_0}{10} \equiv \frac{-a_0}{3}$ . So the whole expression is $\equiv \frac{a_0 -a_0}{3} \equiv 0$ , and we're done.
@above I think it just means that the number of digits in the number is a multiple of $6$ .

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