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Using Induction, prove that $107^n-97^n$ is divisible by $10$

We need to prove the basis first, so let $ n = 1 $

$107^1-97^1$

$107-97 = 10$

This statement is clearly true when $ n = 1 $

Now let's use $P(k)$

$107^k-97^k$

So far so good... next I have to use $P(k+1)$ and there is one part that is driving me nuts. The induction hypothesis implies that $107^k-97^k = 10 m$

$107^{k+1}-97^{k+1}$

$107^k * 107-97^k * 97$

I know that $107^k = 10m+97^k$

$(10m+97^k) * 107 -97^k * 97$

I am lost at this line. What do I do next? Please explain very clearly because I tried to look everywhere online and there aren't very good explanations for after this step.

usukidoll
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3 Answers3

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You are very close to the solution. Look at the last line, and multiply out the brackets

$(10m\times107) + (97^k\times 107) - (97^k\times97)$

The first term is divisible by $10$. How can you simplify the second and third terms to a multiple of $10$?

Alijah Ahmed
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  • arghh I don't know. that's where I'm lost. umm assumption on the second and third terms? sorry prefer step by step . – usukidoll Feb 08 '14 at 08:17
  • For the second and third terms, the the common factor is $97^k$, so we have $97^k\times(107-97)$, which is $97^k \times 10$ and hence divisible by $10$. . – Alijah Ahmed Feb 08 '14 at 08:19
  • so factor out the $97^k$ and that would be $97^k * 107-97$/// $97^k * 10$.... ummm what about the first term. Do we leave it alone? – usukidoll Feb 08 '14 at 08:22
  • No, we add the first term to the simplified second and third terms. So the total result is $(10m\times107)+(97^k\times 10)$, which is divisible by $10$ as it equals $10\times(107m+97^k)$ – Alijah Ahmed Feb 08 '14 at 08:25
  • ok so the simplified second and third terms are $97^k * 10$ and the result would be $(10m * 107) +(97^k * 10)$ ...since it's divisible by 10....wait why is the m in $107$ ? – usukidoll Feb 08 '14 at 08:28
  • I have simply factorised your result, so that $(10m\times107)+(97^k\times10)=10\times(107m+97^k)$ which is divisible by $10$, as $10m\times107$ is the same as $10\times107m$. – Alijah Ahmed Feb 08 '14 at 08:34
  • ????? so factoring out $10$, there's $10 (m * 107) +97^k$ ?! – usukidoll Feb 08 '14 at 08:36
  • Factoring out $10$, the answer is $10(($m$\times 107) + 97^k)$ - note the term $97^k$ is in the brackets. – Alijah Ahmed Feb 08 '14 at 08:41
  • yeah but... did you move the $m$ from $10$ to $107$? Or does $10m * 107$ and $107m * 10$ have the same meaning regardless of where the $m$ is – usukidoll Feb 08 '14 at 08:44
  • let us continue this discussion in chat – Alijah Ahmed Feb 08 '14 at 08:45
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You are almost done at the step where you have

$$107^k\cdot107 - 97^k\cdot97$$

Note that you can rewrite this as

$$107^k\cdot 10 + (107^k\cdot97-97^k\cdot97)$$

The first part is a obvious multiple of 10, the second part is a multiple of 10 by assumption.

user127.0.0.1
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Note that $$107^{k+1}-97^{k+1}\\=107^k\cdot 107-97^k\cdot 97\\ =107^k\cdot (10+97)-97^k\cdot 97\\ =107^k\cdot 10+107^k\cdot 97-97^k\cdot 97\\ =107^k\cdot 10+(107^k-97^k)\cdot 97$$ where the first term is divisible by $10$, and the second term is also divisible by $10$ by induction.

Paul
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  • slow down... where did you get the (10+97) .... why are they split up? Why is $107^k$ in the middle on the fourth line :/ – usukidoll Feb 08 '14 at 08:20
  • From the second line to third line, I write $107$ as $10+97$. – Paul Feb 08 '14 at 08:22
  • From the third line to fourth line, we have $a\cdot(b+c)=a\cdot b+a\cdot c$, where $a=107^k$, $b=10$, and $c=97$. – Paul Feb 08 '14 at 08:23
  • how do you know that we need the $10+97 =107$.... unless we take the fact that's it's divisible by 10 or from the basis step from earlier? – usukidoll Feb 08 '14 at 08:25