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I found the following problem in Austrian Mathematical Olympiad: 2016 - Final Round - Q4:

"Demonstrate that any whole number concluding with the digits $133$ must include a prime divisor exceeding $7$."

So far, I have tried to do a proof by contradiction by assuming $n=3^x 7^y$

I would appreciate it if someone could walk me through the solution or point out the correct approach.

Ruchin Himasha
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6 Answers6

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The remaining other possibility is that the number, call it $n$, has the form $3^a7^b$.

If we know the three last digits of a natural number, we also know its residue class modulo $8$. As $133\equiv5\pmod8$, it follows easily that both $a$ and $b$ must be odd ($3^2\equiv7^2\equiv1\pmod8$ and $5\equiv3\cdot7\pmod 8$). So we must have $$n=21\cdot9^k\cdot 49^\ell$$ with $a=2k+1, b=2\ell+1$. But this means that the last digit of $n$ would be either $1$ or $9$.

Jyrki Lahtonen
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    Endorsing Chris Lewis's solution. Posting this to describe a labor saving mechanism taking advantage of the known third digit. I have no idea what the hint about the totient function means – Jyrki Lahtonen Dec 06 '24 at 11:11
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    This is a much better method than I used! I'll leave my answer up as an alternative but yours may well explain why the question was "show that no $3^a 7^b$ ends in $133$" rather than "show that no $3^a 7^b$ ends in $33$". I also don't really know what the "hint" about the totient function might have been referring to. – Chris Lewis Dec 06 '24 at 11:14
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    Thank you so much for your time. This is a great solution. – Ruchin Himasha Dec 06 '24 at 11:23
  • @Chris It can be done even more simply, e.g. see my answer. – Bill Dubuque Dec 06 '24 at 18:43
  • @Jyrki Could you please show in your answer the details of "it follows easily that ...". It is not clear exactly which method you intend to use, – Bill Dubuque Dec 06 '24 at 18:53
  • very clever indeed! +1 – Mike Dec 06 '24 at 20:12
  • @Mike Disproving congruences via modular checks is bread-and-butter basic number theory - rarely is anything "clever" needed, e.g. to disprove the congruence $,3^a 7^b \equiv 133\pmod{!10^3}$ it suffces to disprove it mod some (small) factor of $10^3.,$ Jyrki chose factors $,8,10,$ but it is simpler to use $,5,4,$ as in my answer. $\ \ $ – Bill Dubuque Dec 06 '24 at 21:51
  • @BillDubuque "If we know the three last digits of a natural number, we also know its residue class modulo 8."...How ? – Sophie Clad Dec 06 '24 at 22:38
  • @Sophie The last $3$ digits of $,n,$ give its remainder $,r,$ modulo $ 10^3,,$ i.e. $,n = r + 10^3 q = r + \color{#c00}8(125q),,$ so $,n\equiv r\pmod{!\color{#c00}8},$ follows from $,n\equiv r\pmod{!\color{#c00}8\cdot 125},,$ see congruences persist mod $\rm\color{#c00}{factors}$ of the modulus. $\ \ $ – Bill Dubuque Dec 06 '24 at 22:55
  • @BillDubuque No idea, I got a mystery downvote as well. May be somebody feels that the question is sub-par, and we should know better than to answer it? Shrug. – Jyrki Lahtonen Dec 09 '24 at 10:39
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Not much case analysis is actually necessary here.

Say $n$ ends in $133$. It certainly isn't divisible by $2$ or $5$; so we need to show it can't be of the form $3^a 7^b$.

You can do this by looking at powers of $3$ and $7$ modulo $1000$ and checking if any of their products end in $133$. But as has been pointed out, this leads to a lot of calculations.

The trick here is that if $n$ ends in $133$, it also ends in $33$.

Powers of $7$ modulo $100$ cycle through the values $7$, $49$,$43$,$1$. These of course also list their own inverses; for instance, if $43a \equiv b$ then $a\equiv 7b$.

You only have to check that none of $\{1,7,43,49\}\cdot 33\equiv \{33,31,19,17\}$ is a power of $3$ modulo $100$.

Chris Lewis
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    You have a stronger result here! $n\equiv 33\bmod 100$ implies it has a prime factor greater than $7$. Nice. – Joshua Tilley Dec 06 '24 at 11:23
  • Thank you so much. This solution is really clever. I would have never think about considering 33 instead of 133 would work. Even though Jyrki had the most elegent solution, yours is the cleverest. – Ruchin Himasha Dec 06 '24 at 11:24
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    Actually, you can even do it mod 20. The powers of 3 mod 20 cycle through $3, 9, 7, 1$ and the powers of 7 mod 20 cycle through $7, 9, 3, 1$. No product of those is congruent to 13 mod 20. – Michael Lugo Dec 06 '24 at 19:44
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    @MichaelLugo That could be the most illuminating answer. – Jyrki Lahtonen Dec 07 '24 at 05:30
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A number with no prime factors greater than $7$ must have only prime factors in $\{2,3,5,7\}$. Such a number takes the form $2^a3^b5^c7^d$. If $n\equiv 133\bmod 1000$ takes this form, it must have no $2$s or $5$s as $133$ is not a multiple of $2$ or $5$. We just need to show that $133$ is not congruent to $3^b7^d$ mod $1000$. This is now a brute force search. We only need to check for $0\le b,d< \phi(1000)=400$.

At this point I used Python. However, as Jyrki points out in the comments, doing it by hand you won't actually have to brute force search the entire $400\times 400$ sized search space. One should check first the order of $3,7$ modulo $1000$, and check for relations of the form $3^k7^l\equiv 1 \bmod 1000$, giving a smaller $200$ sized search space. Still a bit much in my opinion by hand.

Joshua Tilley
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Proof$_{\:\!1}$ $\!\bmod 20\!:\, 7\equiv 3^{-1}$ so $\,3^a7^b\!\equiv 3^{a-b}\!\in S\equiv \{1,3,9,7\}\,$ but $\,133\!\equiv\! 13\not\in S$ $\,\ \bf\small QED$

Proof$_{\:\!2}$ $ \bmod 4\!:\, \ \ \ \color{#c00}1\equiv 133\ \equiv\,\ 3^{\large a} \color{0af}7^{\large b}\ \equiv \color{#c00}{(-1)^{\color{0af}{\large a+b}}} \Rightarrow\,a\!+\!b\,\ \rm\color{#0a0}{even}$
$\qquad\quad\ \bmod 5\!:\, \color{#c00}{{-}1}\equiv 133^2\!\equiv 3^{\large 2a}7^{\large 2b^{\phantom{|^.}}}\!\!\!\equiv \color{#c00}{(-1)^{\large \color{0af}{a+b}}}\Rightarrow\, a\!+\!b\,\ \rm\color{#0af}{odd},\,$ contra above. $\, \ \small\bf QED$

Update $ $ In reply to comments (& the $5$ downvotes!) I expand on the (fundamental) method used. If $\:3^a7^b\equiv 133\pmod{\!10^3}\,$ has a solution $\,a,b\in\Bbb N\,$ then it remains a solution $\!\bmod n\,$ for every factor $\,n\,$ of $\,10^3.\,$ We start by testing the smallest factors: $\,2,4,5\ldots\ $ First $\!\bmod 2\,$ it's always true $\,1^a 1^b\equiv 1.\,$ Next, $\bmod 4,\,$ we get $\,\color{#c00}{(-1)^{a+b}\equiv 1},\,$ thus $\,a\!+\!b\,$ is $\rm\color{#0a0}{even}.\,$ Next $\!\bmod 5\,$ it's $\,3^a 2^b\!\equiv 3;\,$ squaring, by $\,3^2,2^2\equiv -1\,$ we get $\,\color{#c00}{(-1)^{a+b}\equiv -1},\: $ so $\ a\!+\!b\,$ is $\rm\color{#0af}{odd},\,$ contra, above. [The first proof is an optimization - we further require that the two generators $\:\!7,3\:\!$ collapse to one via $\,7^k\equiv 3.\,$ Quickly we find a small value $\,k=-1\,$ that works, where $\,7^{-1}\equiv 3\!\iff\! 1\equiv 21\!\iff\! 20\equiv 0$].

Summarizing: no solution exists $\!\bmod 10^3$ since assuming so leads to a $\rm\color{#0a0}{parity}$ $\rm\color{#0af}{contradiction}$ when reducing the purported solution $\!\bmod 4\,$ and $\!\bmod 5.\,$ Such modular reduction obstructions are one of the simplest and most widely used methods to prove Diophantine equations unsolvable.

Bill Dubuque
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  • @Downvoter If anything is not clear you are welcome to ask questions. It's all trivial mod arithmetic. – Bill Dubuque Dec 06 '24 at 19:07
  • I did not downvote this--I do tell why when I do downvote an answer as I appreciate the courtesy myself!--but with that said, this answer here is quite sparse and I am finding this hard-to-follow, especially in light of the above answers. – Mike Dec 06 '24 at 20:09
  • @Mike Where are you stuck? This is simpler than all the other answers. $\ \ $ – Bill Dubuque Dec 06 '24 at 20:19
  • Well, to be honest I didn't go through much effort of mentally compiling this . The reason being is that with the earlier answers above I hardly even had to. It's more sparsely written here but I'm not sure it is simpler. In general, it seems that later answers tend to get judged more harshly on here I think. I do think more words on this post would help. – Mike Dec 06 '24 at 20:31
  • In particular, what was meant by the $()^2$ above the $\implies$? – Mike Dec 06 '24 at 20:32
  • @Miike That square operation means: square the prior congruence (using $,3^2\equiv -1\equiv 7^2).,$ It is essential to grok modular arithmetic if you wish to master number theory. – Bill Dubuque Dec 06 '24 at 20:38
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    This is super elegant, well done. I'm in two minds about the style - I initially found it a little off-putting but actually following it through is quite satisfying (the fact you have to do a little work reading it feels a bit like solving it yourself!). I'm not sure it would be the most suitable style if this had been the only answer but among all the others I think it's great. – Chris Lewis Dec 06 '24 at 21:42
  • IMHO, this answer is a bit obtuse, or at least too sparse. I think you should explicitly state that in mod 4 a+b must be even for the congruence to hold, but in mod 5 we need a+b to be odd, hence there's a contradiction. – PM 2Ring Dec 06 '24 at 22:57
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    I see it now, @BillDubuque . Very elegant indeed but I do find it much clearer now after the edits +1 – Mike Dec 06 '24 at 22:57
  • My point is that I don't feel that your notation is self-explanatory. However, I was viewing your answer using my phone's forced-dark mode, and some of your text was invisible, making it appear even terser. ;) (The text 'a+b' before 'odd' and 'even' was invisible). – PM 2Ring Dec 06 '24 at 23:04
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    @PM2Ring $ $ Kudos - you correctly guessed the $TeX$nicolor parity contradiction hidden by dark mode! $\ \ \ $ I appended an update to the $2$-line proof. Now it should be more widely accessible. – Bill Dubuque Dec 07 '24 at 02:05
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    What a clever solution. Thank you so much for your time. – Ruchin Himasha Dec 07 '24 at 13:32
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Just to add some details to the nice answers already posted, the subgroup of $(\mathbb{Z}/1000)^{\times}$ generated by $3$ and $7$ has order 200, so it is of index $2$ ( since $|(\mathbb{Z}/1000)^{\times}|=\phi(1000)=400$). One could see this by noticing first that $3^{100}=7^{20}=3^{10}7^{18}= 1 \mod 1000$. So there exists a real Dirichlet character $\mod 1000$ $\mu$( that is, with values $\pm 1$) such that $\mu(3)=\mu(7) = 1$, but $\mu(133) = -1$. In Mathematica's notation, it will be the Dirichlet character with index $251$, as one can check, see also WA.

$\bf{Added:}$

Let's write explicitly a Dirichlet character $\mod 1000$ that is $1$ on $3$ and $7$, but $-1$ on $133$. From theory we know that a Dirichlet character $\mod 1000$ is the product of a Dirichlet character $\mod 8$ and a Dirichlet character $\mod 125$. This also works for real Dirichlet characters ( i. e. those with values $\pm 1$). Now, a Dirichlet character $\mod 8$ is a multiplicative function from $(\mathbb{Z}/8)^{\times} \to \{\pm 1\}$. Here is one that takes values $-1$ on $3$, $7$, but $1$ on $133=5 \mod 8$, it comes from

$$(\mathbb{Z}/8)^{\times}\to (\mathbb{Z}/4)^{\times}\simeq \{\pm 1\}$$

Now for the character $\mod 125$, it is

$$\mathbb(\mathbb{Z}/125)^{\times}\to (\mathbb{Z/5})^{\times} \to \{ \pm 1\}$$

where the last one is the Legendre character ( takes $3$, $7$, and $133 = 3\mod 5$ to $-1$ )

Now take the product of the above two Dirichlet characters. It takes $3$ and $7$ to $1$, but $133$ to $-1$.

orangeskid
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    Note that the question is tagged elementary-number-theory so it is highly unlikely that the OP knows anything about Dirichlet characters (and likely the same for group theory). As such, you might consider rephrasing it in more accessible terms. – Bill Dubuque Dec 08 '24 at 02:06
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    Of course, but the point of my comment was to nudge you to elaborate so that many more readers might find it useful. – Bill Dubuque Dec 10 '24 at 16:20
  • @Bill Dubuque: did add more details, will think some more about it – orangeskid Dec 10 '24 at 17:49
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If we look at the group generated by $3$ modulo $100$.

$3,9,27,81,43,29,$ etc.

There are $20$ members to this group. Every element has an even number in the $10$s digit. $7$ is a member of the group.

All numbers that can be expressed as $3^i7^j$ have an even number in the $10$'s digit.

user317176
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