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This is a problem from BdMO $2012$ Dhaka region Question Paper:

The product of a number with itself is called its square. For example, $2$ multiplied by $2$ is $4$, so $4$ is the square of $2$. If you take a square number and multiply it with itself, what will be the largest possible remainder if the product is divided by $10$?

I came up with this: $$x^4 \mod {10}$$

I know that the modulus (%) operator calculates the remainder of a division. And that it can be used to see, suppose, whether $N$ is a multiple of $M$ or not. Nothing more than that. I am much familiar with mod because of my programming experience with mid-level languages like C and C++. It was not until later that I came to know that modulus is used in mathematics as well.

Now, how to use the 'modulus' operator? How can I use this to go further into solving this problem?

Vidyanshu Mishra
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Soha Farhin Pine
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    modulo $10$, there are only $10$ different values $x \pmod {10}$ can take. You can just calculate fourth power of them case by case – user160738 Dec 25 '16 at 11:30
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    Now that a couple of hours have passed, I don't think you have to wait for any other answers to come along; it's fine to accept one. If they're all equally good, pick one at random, so people can see that the question has an accepted answer. – David K Dec 25 '16 at 14:12
  • I'm still thinking... Can't decide... each of the answers has an specialty – Soha Farhin Pine Dec 25 '16 at 14:15
  • @DavidK I did, just now, following your last advice. – Soha Farhin Pine Dec 25 '16 at 14:17
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    Note that using "mod" as an operator is common in programming, but less usual in mathematics. The standard notation is using it as a "modifier" of a relation symbol $\equiv$ or $=$: so we write "$a \equiv b \mod 10$". – Federico Poloni Dec 25 '16 at 22:36
  • @FedericoPoloni Thanks for helping me on mod. – Soha Farhin Pine Dec 26 '16 at 05:36

4 Answers4

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$1$. The remainder will be the unit digit of the number you are dividing. For example, Remainder when $16$ is divided by $10$ is $6$.

Proof: If you have got a $n$ digit number then you can write it as $10^{n-1}a_0+10^{n-2}a_1+........+10a_{n-2}+a_{n-1}$ where $a_{n-1}$ is the unit digit. Notice that all the terms in the sum are divisible by $10$, the only suspect is $a_{n-1}$.

$2$. Notice that unit digit of a square number can be $0,1,4,5,6,9$ and corresponding unit digits of $4th$ powers can be $0,1,6,5,6,1,$.

So, largest remainder is $6$.

Vidyanshu Mishra
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By Fermat's Little Theorem, $x^4$ has remainder $1$ when divided by $5$; so, when divided by $10$, it could only have as remainder $1$ or $6$. The latter is achievable, e.g., by $2^4$ mod $10 \equiv 6$ as the max. QED.

Benjamin Dickman
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We know that the unit digits of the squares always belong in the set of these numbers : $A = \{ 1,4,9,6,5,0\}$.

The unit digits of the squares of the numbers $\in A$ gives us the possible unit digits of the fourth powers of all numbers. Thus, we create a a set $B = \{1,6,5,0\}$ where the elements in $B$ denotes the possible modulo of the fourth powers of a number $10$.

Out of these, we can see that the maximum possible remainder is $6$. On a side note, this is achieved for all even numbers except multiples of $10$. Hope it helps.

  • It's so hard for me to accept an answer. All are the best in their own way! – Soha Farhin Pine Dec 25 '16 at 11:50
  • Thanks for this answer so much! It really helps me. – Soha Farhin Pine Dec 25 '16 at 11:53
  • @SohaFarhinPine You can leave the decision for an hour or two, which will probably get more people visiting your question - maybe even more answers. No rush. – Joffan Dec 25 '16 at 11:53
  • @Joffan You are right. I should wait for a couple of hours before deciding to accept one particular answer. This will allow more traffic on this question. – Soha Farhin Pine Dec 25 '16 at 11:56
  • This problem partly requires trial and error. After that, the repeating pattern will be prominent. – Soha Farhin Pine Dec 25 '16 at 11:59
  • @SohaFarhinPine It is actually not trial and error. It is the analysis of a pattern among numbers. Trial and error means juggling between various solutions when you are unsure of how to solve a problem. There is a huge difference between the two. –  Dec 25 '16 at 12:03
  • @Rohan That is it. You need to do some practical work here to figure out the pattern first, right? This is trying. That is what I meant. Actually, trial and error wasn't the right term to use in this context. – Soha Farhin Pine Dec 25 '16 at 12:07
  • Yes, you are correct. Don't use that word trial and error. –  Dec 25 '16 at 12:12
  • Yep, learnt a lesson! Definitely wouldn't opt for that in such context! – Soha Farhin Pine Dec 25 '16 at 12:13
  • Could you answer this question of mine: http://math.stackexchange.com/questions/2070615/formula-for-calculating-the-total-number-of-solution-pairs-to-an-equation? – Soha Farhin Pine Dec 25 '16 at 12:16
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So you can simply try out all $10$ possible options:

  • $x\equiv0\pmod{10} \implies x^4\equiv0^4\equiv 0\equiv0\pmod{10}$
  • $x\equiv1\pmod{10} \implies x^4\equiv1^4\equiv 1\equiv1\pmod{10}$
  • $x\equiv2\pmod{10} \implies x^4\equiv2^4\equiv 16\equiv6\pmod{10}$
  • $x\equiv3\pmod{10} \implies x^4\equiv3^4\equiv 81\equiv1\pmod{10}$
  • $x\equiv4\pmod{10} \implies x^4\equiv4^4\equiv 256\equiv6\pmod{10}$
  • $x\equiv5\pmod{10} \implies x^4\equiv5^4\equiv 625\equiv5\pmod{10}$
  • $x\equiv6\pmod{10} \implies x^4\equiv6^4\equiv1296\equiv6\pmod{10}$
  • $x\equiv7\pmod{10} \implies x^4\equiv7^4\equiv2401\equiv1\pmod{10}$
  • $x\equiv8\pmod{10} \implies x^4\equiv8^4\equiv4096\equiv6\pmod{10}$
  • $x\equiv9\pmod{10} \implies x^4\equiv9^4\equiv6561\equiv1\pmod{10}$
barak manos
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