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One of my friends recently gave a mock test of a math exam in which he was asked this horrific question. He asked the same to me and I was totally blank on looking at it. So, it will be a huge help if any of you can solve this. Here is the question:

A 79 digit long number is made by writing the natural numbers from 1 to 44 in order like 1234567891011121314.....424344. What is the remainder when this number is divided by 45?

Thanks in advance! Enjoy Solving!

Udit Jethva
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    Well, clearly your number is $4\pmod 5$, so you just need to compute it $\pmod 9$. That you can do by summing the digits. Or, more easily, by arguing that you just need to compute $\sum_{n=1}^{44} n \pmod 9$ – lulu Jul 30 '19 at 12:27
  • I am sorry @lulu but I didn't get what you're trying to say? – Udit Jethva Jul 30 '19 at 12:29
  • @UditJethva: look up https://en.wikipedia.org/wiki/Chinese_remainder_theorem – Vasili Jul 30 '19 at 12:30
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    Where are you confused? If you know the remainder modulo $5,9$ you can use the Chinese remainder theorem to compute it $\pmod {45}$. – lulu Jul 30 '19 at 12:30
  • https://math.stackexchange.com/questions/3152587/largest-multiple-of-7-lower-than-some-78-digit-number/3152669#3152669 might help a bit, if you know enough math. phi(45)=32 –  Jul 30 '19 at 12:31

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Hint: I'd use the Chinese remainder theorem, which says that the mapping $${\Bbb Z}_{45} \rightarrow {\Bbb Z}_5\times {\Bbb Z}_9: x\mapsto (x\mod 5, x\mod 9)$$ is a ring isomorphism.

Thus you can separately divide the large number by 5 and by 9. These remainders can then be combined to obtained the remainder modulo 45.

Modulo 5 the situation is simple, just look at the last digit of the number.

Modulo 9, observe that $10 \equiv 1 \mod 9$ and so $10^n\equiv 1 \mod 9$ for each $n\geq 1$. So modulo 9 you just look for the ''Quersumme'' (cross sum).

So you have to find a number $x$ between 0 and 44 such that $x$ is congruent 4 modulo 5 and congruent $Q$ modulo 9, where $Q$ is the cross sum of your number.

This number is congruent 4 mod 5 and so is one of the following: 4, 9, 14, 19, 24, 29, 34, 39, 44.

Wuestenfux
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    With all due respect to the OP, I doubt that bringing ring isomorphisms into the picture will help them to solve their problem. – TonyK Jul 30 '19 at 12:41
  • I would agree: answers shouldn't be more complicated than necessary. – H Huang Jul 30 '19 at 12:44
  • @TonyK: It solves the problem. Please check. – Wuestenfux Jul 30 '19 at 12:45
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    Reading the OP's question, I would guess that Udit can't tell a ring isomorphism from a hole in the ground. So it's not much of a hint, is it? – TonyK Jul 30 '19 at 12:52
  • @TonyK I've now described the solution procedure which should be doable by somebody who has no access to ring theory. The down voting is unjustified. – Wuestenfux Jul 30 '19 at 13:00
  • Downvoting is justified by "This answer is not useful". That is a stronger condition than "This answer is not useful for the OP". If the answer is okay than at least others might have profit. – drhab Jul 30 '19 at 13:02
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    Come on you serious mathematicians, think about my solution. Its the easiest one can think of. And this holds for every large number the OP can present. – Wuestenfux Jul 30 '19 at 13:03
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    Upvoted. What's the matter with the downvoters? Even if it is too difficult for the OP, it can be useful for someone else! – J. De Ro Jul 30 '19 at 13:43
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Let $x$ be your number, let $y$ be its remainder when divided by 45.

Since $x$ and $y$ differ by a multiple of $45$, and each multiple of $45$ is also a multiple of $5$ and a multiple of $9$, $x$ and $y$ have the same remainders when divided by $5$, and also when divided by $9$.

So what you need to do is find the remainders of $x$ when divided by $5$ and by $9$, and then find the number $y$ between $0$ and $44$ that has the same remainders.

Magma
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  • I can find out the remainder when it's divided by 5 but how am i going to find it when it's divided by 9? Please help. – Udit Jethva Jul 30 '19 at 12:38
  • @UditJethva: do you know divisibility by $9$ rule? Find the sum of digits and divide it by 9, the remainder will be the same when you divide the original number – Vasili Jul 30 '19 at 12:40
  • Yes I do know that but the problem is that how am I going to find the sum of the digits? – Udit Jethva Jul 30 '19 at 12:42
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    By adding them up. There's only 79 of them, it isn't that hard. You can rearrange them to make it go faster (10 twos sum to 20, for example, and the digits 1 to 9 sum to 45). – Magma Jul 30 '19 at 12:46
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    @UditJethva: Modulo $9$, the sum of the digits in the long number is also the sum of the digits in the sum $1+2+3+\dots+44$. There's a formula for that. – awkward Jul 30 '19 at 12:49
  • Sorry to say @awkward, but that is incorrect as for example adding numbers like 44 would mean that 40 + 4 is added instead of 4 + 4 – Udit Jethva Jul 30 '19 at 13:11
  • Yes, it's not the same sum, but it still has the same remainder when divided by 9, and that's all you need anyway. – Magma Jul 30 '19 at 13:14
  • @UditJethva If you don't know the formula you can use Gauss's trick as in my answer. This exploitation of reflection symmetry is widely applicable. – Bill Dubuque Jul 30 '19 at 13:40
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    @UditJethva I said the results are the same modulo 9. In your example, note that $4+4$ and $44$ both are $8$, modulo $9$. – awkward Jul 30 '19 at 14:34
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$\color{#c00}{n \equiv 4}\pmod{5}$ by its unit digit $= 4$, and $\,n\equiv 0\pmod{9}\,$ by casting out nines as below

$$\begin{align} 1 &+\ \ 2 + \cdots + 22\\ +\ 44 &+ 43 + \cdots +23\\ \hline 45 &+ 45 + \cdots +45\end{align}\qquad\qquad $$

Thus $\ n\bmod 45 = 9 (\color{#c00}n/9 \bmod 5) = 9(\color{#c00}4/4\bmod 5) = 9\,$ by the mod Distributive Law.

J. W. Tanner
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Bill Dubuque
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